Quadratic Equations

A polynomial equation is a quadratic equation, if it is of the form ax² + bx + c = 0 such that a ≠ 0

(i)

It is a quadratic equation.

(ii)

It is a quadratic equation.

(iii)

⇒ x⁴ -5x² + 1 = 0

It is not a quadratic equation as the highest power of x is ‘4’.

(iv)

⇒ x² – 3 = x³

It is not a quadratic equation.

(v)

It is not a quadratic equation as √x is present instead of ‘x’.

(vi)

It is not a quadratic equation as an additional √x term is present.

(vii)

⇒ 2x² + 2x + 6 = 0

It is a quadratic equation.

(viii)

⇒ x² + 1 – x = 0

It is a quadratic equation.

(ix)

It is a quadratic equation.

(x)

⇒ x⁴ + 1 + 2x² = 3x³ + 3x + 4x²

It is not a quadratic equation.

(xi)

⇒ 6x² + 7x + 2 = 6x² – 18x + 12

⇒ 25x = 10

It is not a quadratic equation.

(xii)

⇒ x² + 1 = x³

It is not a quadratic equation.

(xiii)

⇒ 16x² – 3 = 10x² + 19x – 15

⇒ 6x² – 19x + 12 = 0

It is a quadratic equation.

(xiv)

⇒ x³ + 8 + 6x² + 12x = x³ – 4

⇒ 6x² + 12x + 12 = 0

It is a quadratic equation.

(xv)

⇒ x² + x + 8 = x² – 4

⇒ x = - 12

It is not a quadratic equation.

We will have to check for each value and see whether it satisfies the equation.

(i)

For x = 2,

2² – 3 × 2 + 2 = 0

⇒ 0 = 0

Thus, x = 2 is a solution.

For, x = 1

1² – 3 × 1 + 2 = 0

⇒ 0 = 0

Thus, x = 1 is a solution.

(ii)

For x = 0,

⇒ 0 + 0 + 1 = 0

⇒ 1 = 0 which is not true thus x = 0 is not a solution

For x = 1,

⇒ 1 + 1 + 1 = 0

⇒ 3 = 0 which is not true thus x = 1 is not a solution

(iii)

For x= √3

⇒ 3 – 3√3 × √3 + 6 = 0

⇒ 3 – 9 + 6 = 0

⇒ 0 = 0

Thus, x = √3 is a solution

For x = -2√3

⇒ (-2√3)² – 3√3 × -2√3 + 6 = 0

⇒ 4 × 3 + 18 + 6 = 0

⇒ 36 = 0 which is not true, thus x = -2√3 is not a solution

(iv)

For x = 5/6

⇒ 61 = 65 which is not true, thus x = 5/6 is not a solution

For x = 4/3

⇒ 25/12 = 13/6

⇒ 25 = 26 which is not true, thus x = 4/3 is not a solution

(v)

For x = 2,

⇒ 2 × 4 – 2 + 9 = 4 + 4 × 2 + 3

⇒ 15 = 15, thus x = 2 is a solution.

For x = 3

⇒ 2 × 9 – 3 + 9 = 9 + 4 × 3 + 3

⇒ 24 = 24, thus x = 3 is also a solution

(vi)

For x = -√2,

⇒ 2 - √2 × -√2 – 4 = 0

⇒ 2 + 2 – 4 = 0

⇒ 0 = 0

Thus, x = -√2 is a solution

For x = -2√2

⇒ 4 × 2 - √2 × -2√2 – 4 = 0

⇒ 8 + 8 – 4 = 0

⇒ 12 = 0 which is not true, thus x = -2√2 is not a solution

(vii)

For, x = a/b

⇒ a⁴/b² – 3a² + 2b² = 0 which is not true, thus x = a/b is not a solution

For x = b/a

⇒ b² – 3b² + 2b² = 0

⇒ 0 = 0 , thus x = b/a is a solution

For the given value to be a solution, it should satisfy the quadratic equation

(i)

⇒ 7 × 4/9 + k × 2/3 – 3 = 0

⇒ 2k/3 = 3 – 28/9 = - 1/9

⇒ k = -1/6

(ii)

⇒ a² – a(a + b) + k = 0

⇒ a² –a² – ab + k = 0

⇒ k = ab

(iii)

⇒ k × 2 + √2 × √2 – 4 = 0

⇒ 2k = 2

⇒ k = 1

(iv)

⇒ a² – 3a × a + k = 0

⇒ k = 2a²

Given: If are the roots of the equation

To find: the values of a and b.

Solution:

Quadratic equation in roots form:

(x – a)(x – b) = 0, where a and b are the roots

Given, are the roots of the equation

Quadratic equation is,

(x – 2/3)(x + 3) = 0

⇒ x² -2x/3 + 3x – 2 = 0

⇒ 3x² +7x – 6 = 0

On comparing with

We get, a = 3 and b = -6

For the given value to be a root, it should satisfy given equation

⇒ 0 + 0 = √10

Thus x = 3 does not satisfy the given equation and it is not a root of the equation.

Given: The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm.

To find: the lengths of these sides.

Solution:

We know


(Hypotenuse)² = (perpendicular)² + (base)² ..... (1)

Given, hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm

Let the base be ‘b’.

⇒ Perpendicular = b – 5


Put the known values in (1),

⇒ 25² = b² + (b – 5)²


Apply the formula (x – y)² = x² + y² - 2xy in (b – 5)².


Here x=b and y=5

⇒ 625 = b² + b² + 25 – 10b


⇒ 625 = 2b²+ 25 – 10b


⇒ 2b²+ 25 – 10b = 625


⇒ 2b²+ 25 – 10b - 625 = 0


⇒ 2b²– 10b - 600 = 0

Take out 2 common of the above equation.

⇒ b² – 5b – 300 = 0

⇒ b² – 20b + 15b – 300 = 0

⇒ b(b – 20) + 15(b – 20) = 0


⇒ (b - 20 ) ( b + 15 ) = 0


⇒ (b - 20 ) =0 and ( b + 15 ) = 0

⇒ b = 20

Perpendicular = 20 – 5 = 15

Hence,Sides are 15 and 20

Let the smaller leg be ‘a’ and longer leg be ‘b’.

Hypotenuse² = length² + breadth²

Given, hypotenuse of a right triangle is cm

⇒ 9 × 10 = a² + b²

⇒ a² + b² = 90 -------- (1)

Now, the smaller leg is tripled and the longer leg doubled, new hypotenuse is cm.

⇒ (3a)² + (2b)² = 81 × 5

⇒ 9a² + 4b² = 405 -------- (2)

Multiplying (1) by 4 and subtracting from eq 2

⇒ 5a² = 45

⇒ a² = 9

⇒ a = 3

Thus, 9 + b² = 90

⇒ b² = 81

⇒ b = 9

Let the distance of pole from gate A be ‘a’.

Difference of the distance of the pole from two diametrically opposite fixed gates A and B on the boundary is 7 metres.

Distance of pole from gate B = a – 7 m

Diameter of the park = 13 m

Hypotenuse² = length² + breadth²

⇒ 13² = a² + (a – 7)²

⇒ 169 = 2a² + 49 – 14a

⇒ a² – 7a – 60 = 0

⇒ a² – 12a + 5a – 60 = 0

⇒ a(a – 12) + 5(a – 12) = 0

⇒ (a + 5)(a – 12) = 0

⇒ a = 12 m

Thus distance of pole is 12 m from gate A and 5 m from gate B

Let the shorter side be ‘a’.

Given, diagonal of a rectangular field is 60 metres more than the shorter side

Diagonal = a + 60

Also, longer side is 30 metres more than the shorter side

Longer side = a + 30

Hypotenuse² = length² + breadth²

⇒ (a + 60)² = (a + 30)² + a²

⇒ a² + 120a + 3600 = a² + 60a + 900 + a²

⇒ a² – 60a – 2700 = 0

⇒ a² – 90a + 30a – 2700 = 0

⇒ a(a – 90) + 30(a – 90) = 0

⇒ (a + 30)(a – 90) = 0

⇒ a = 90m

Length of sides = 90m, 120 m

Perimeter of a rectangle = 2(l + b)

Area of the rectangle = l × b

Given, perimeter of a rectangular field is 82 m and its area is 400 m²

Let the breadth be ‘a’ m and length be ‘b’ m

⇒ 2(a + b) = 82

⇒ b = 41 – a

Also, a × b = 400

⇒ a × (41 – a) = 400

⇒ a² – 41a + 400 = 0

⇒ a² – 25a – 16a + 400 = 0

⇒ a(a – 25) – 16(a – 25) = 0

⇒ (a – 16)(a – 25) = 0

⇒ a = 16, 25

Assuming breadth to smaller, thus breadth = 16m

Let the breadth of the hall be ‘a’

Length = a + 5

Given, area of the floor of the hall is 84 m²

⇒ a(a + 5) = 84

⇒ a² + 5a – 84 = 0

⇒ a² + 12a – 7a – 84 = 0

⇒ a(a + 12) – 7(a + 12) = 0

⇒ (a – 7)(a + 12) = 0

⇒ a = 7 m

Length of the hall = 7 + 5 = 12 m

Area of a square = side × side

Given, squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm².

⇒ x² + (x + 4)² = 656

⇒ x² + x² + 8x + 16 = 656

⇒ x² + 4x – 320 = 0

⇒ x² + 20x – 16x – 320 = 0

⇒ x(x + 20) – 16(x + 20) = 0

⇒ (x – 16)(x + 20) = 0

⇒ x = 16 cm

The other side = 16 + 4 = 20 cm

Let the base of the triangle be ‘a’.

Given, altitude exceeds the base by 7 m.

⇒ Altitude = a + 7 m

Area of a right angled triangle = 1/2 × base × height

⇒ 1/2 × a(a + 7) = 165

⇒ a² + 7a – 330 = 0

⇒ a² + 22a – 15a – 330 = 0

⇒ a(a + 22) – 15(a + 22) = 0

⇒ (a – 15)(a + 22) = 0

⇒ a = 15 m

Altitude = 15 + 7 = 22 m

Let the breadth be ‘a’ m

Given, rectangular mango grove whose length is twice its breadth.

Length = 2a

Area = 800 m²

⇒ 2a × a = 800

⇒ a² = 400

⇒ a = 20 m

Thus, length = 40m

Let the length and breadth of park be ‘a’ and ‘b’ m

Given, rectangular park of perimeter 80 m and area 400 m².

⇒ 2(a + b) = 80

⇒ a + b = 40

⇒ a = 40 – b

⇒ a × b = 400

⇒ (40 – b)b = 400

⇒ b² – 40b + 400 = 0

⇒ (b – 20)² = 0

⇒ b = 20 m

Thus, it is possible to design a rectangular park with length = 20m and breadth = 20 m

Area of a square = s²

Perimeter of a square = 4s

Let the sides of the square be a and b respectively.

Given, sum of the areas of two squares is 640 m² and the difference of their perimeters is 64m.

⇒ a² + b² = 640 and

4a – 4b = 64

⇒ a – b = 16

⇒ a = 16 + b

⇒ (16 + b)² + b² = 640

⇒ 2b² + 32b + 256 = 640

⇒ b² + 16b – 192 = 0

⇒ b² + 24b – 16b – 192 = 0

⇒ (b – 8)(b + 24) = 0

⇒ b = 8 m

Thus, a = 24 m

Area of a square = s²

Perimeter of a square = 4s

Let the sides of the square be a and b respectively.

Given, sum of the areas of two squares is 400 cm² and the difference of their perimeters is 16 cm.

⇒ a² + b² = 400 and

4a – 4b = 16

⇒ a – b = 4

⇒ a = 4 + b

⇒ (4 + b)² + b² = 400

⇒ 2b² + 8b + 16 = 400

⇒ b² + 4b – 192 = 0

⇒ b² + 16b – 12b – 192 = 0

⇒ (b + 16)(b – 12 ) = 0

⇒ b = 12 m

Thus, a = 16 m

Given: The area of a rectangular plot is 528 m². The length of the plot (in metres) is one metre more then twice its breadth.
To find: the length and the breadth of the plot.
Solution:
Let the breadth be ‘a’ m

Given, length of the plot (in metres) is one metre more then twice its breadth.

⇒ length = (2a + 1) m

Now, area of a rectangular plot is 528 m²

⇒ a × (2a + 1) = 528

⇒ 2a² + a – 528 = 0
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 2a² + 33a – 32a – 528 = 0

⇒ a(2a + 33) – 16(2a + 33) = 0

⇒ (2a + 33)(a – 16) = 0
⇒ (2a + 33) = 0 and (a – 16) = 0
⇒ 2a = -33 and a = 16
⇒ a = -33/2 and a = 16

⇒ a = 16 m
length = (2a + 1) = 2 × 16 + 1 = 33 m

Hence breadth = 16 m and length = 33 m

Let the number of days in which B finishes the work be ‘b’.

∴ Number of days in which A finishes the work = b – 10

In 1 day,

B finishes 1/b of the work

A finishes 1/(b – 10) of the work

Now, both A and B together can finish the work in 12 days

⇒ 12(b – 10 + b) = b² – 10b

⇒ 24b – 120 = b² – 10b

⇒ b² - 34b + 120 = 0

⇒ b² -30b – 4b + 120 = 0

⇒ b(b – 30) – 4(b – 30) = 0

⇒ b = 4, 30 b can’t be 4 as A takes 10 days less than B

Thus number of days in which B alone finishes the work is 30 days.

Let the slower pipe fill the reservoir in ‘a’ hours

Faster pipe fills it in ‘a – 10’ hours.

Given, the two pipes will fill the reservoir together in 12 hours.

In 1 hour, part of reservoir filled = 1/12

⇒ 12(a + a – 10) = a² – 10a

⇒ 24a – 120 = a² – 10a

⇒ a² - 34a + 120 = 0

⇒ a² – 30a – 4a + 120 = 0

⇒ a(a – 30) – 4(a – 30) = 0

⇒ (a – 4)(a – 30) = 0

Value of a can’t be 4 as (a – 10) will be negative

Thus a = 30

Let the smaller diameter tap fill the reservoir in ‘a’ hours

Larger diameter tap fills it in ‘a – 10’ hours.

Given, two water taps together can fill a tank in = 75/8 hours.

In 1 hour, part of tank filled = 8/75

⇒ 75(a + a – 10) = 8a² – 80a

⇒ 150a – 750 = 8a² – 80a

⇒ 8a² – 230a + 750 = 0

⇒ 4a² – 115a + 375 = 0

⇒ 4a² – 100a – 15a + 375 = 0

⇒ 4a(a – 25) – 15(a – 25) = 0

⇒ (4a – 15)(a – 25) = 0

Value of a can’t be 15/4 as (a – 10) will be negative

Thus a = 25

Time taken by faster tap = 25 – 10 = 15 hours

Let the faster pipe fill the tank in ‘a’ min

Slower pipe fills it in ‘a + 5’ min.

Given, the pipes running together can fill a tank in = 100/9 minutes.

In 1 min, part of tank filled = 9/100

⇒ 100(a + a + 5) = 9(a² + 5a)

⇒ 200a + 500 = 9a² + 45a

⇒ 9a² – 155a - 500 = 0

⇒ 9a² – 180a + 25a - 500 = 0

⇒ 9a(a – 20) + 25(a – 20) = 0

⇒ (9a + 25)(a – 20) = 0

⇒ a = 20 mins

Slower pipe will fill it in 25 min

Let the larger diameter pipe fill it in ‘a’ hours

The smaller diameter pipe fills it in ‘a + 10’ hours

In 1 hour, larger diameter pipe fills 1/a part of the pool.

In 1 hour, smaller diameter pipe fills 1/(a + 10) part of the pool.

Given, the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled.

⇒ 4 × 1/a + 9 × 1/(a+10) = 1/2

⇒ 2(4a + 40 + 9a) = a² + 10a

⇒ 26a + 80 = a² + 10a

⇒ a² – 16a – 80 = 0

⇒ a² – 20a + 4a – 80 = 0

⇒ a(a – 20) + 4(a – 20) = 0

⇒ (a + 4)(a – 20) = 0

⇒ a = 20 hours

Time in which smaller diameter pipe fills the pool = 20 + 10 = 30 hours

Let the length of cloth be ‘a’ m.

Given, piece of cloth costs Rs. 35 and if the piece were 4 m longer and each metre costs Rs. 1 less, the cost remains unchanged.

Cost of 1m of cloth = 35/a

⇒ (a + 4)(35 – a) = 35a

⇒ 35a + 140 – a² – 4a = 35a

⇒ a² + 4a – 140 = 0

⇒ a² + 14a – 10a – 140 = 0

⇒ a(a + 14) – 10(a + 14) = 0

⇒ (a – 10)(a + 14) = 0

⇒ a = 10 m

Let the number of students who planned the picnic be ‘a’.

Budget for the food was Rs. 480

Cost of food for each member = 480/a

Given, eight of these failed to go and thus the cost of food for each member increased by Rs. 10

⇒ (a – 8)(480 + 10a) = 480a

⇒ 480a + 10a² – 3840 – 80a = 480a

⇒ a² – 8a – 384 = 0

⇒ a² – 24a + 16a – 384 = 0

⇒ a(a – 24) + 16(a – 24) = 0

⇒ (a + 16)(a – 24) = 0

⇒ a = 24

Number of students who attended the picnic = 24 – 8 = 16

Let the cost price be Rs a.

Given, the dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article.

It's given that he gains as much as the cost price of the article, thus, Gain% = a%

Gain% =

⇒

⇒ a² = (24-a) 100

⇒ a² + 100a – 2400 = 0

⇒ a² + 120a – 20a – 2400 = 0

⇒ (a + 120)(a – 20) = 0

⇒ a = 20 or -120

Since money cannot be negative so, negalecting -120, we get,

⇒ a = 20

Thus, the cost price of the article is Rs 20

Let the number of swans in the pond be ‘a’.

Given, out of a group of swans, 7/2 times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water.

⇒ 7√a = 2a – 4

Squaring both sides

⇒ 49a = 4a² + 16 – 16a

⇒ 4a² – 65a + 16 = 0

⇒ 4a² – 64a – a + 16 = 0

⇒ 4a(a – 16) – (a – 16) = 0

⇒ (4a – 1)(a – 16) = 0

⇒ a can’t be 1/4, thus a = 16

Let the original price of the toy be ‘a’.

Given, when the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360.

The number of toys he can buy at the original price for Rs. 360 = 360/a

According to the question,

⇒ 360a = (a – 2)(360 + 2a)

⇒ 360a = 360a + 2a² – 720 – 4a

⇒ a² – 2a – 360 = 0

⇒ a² – 20a + 18a – 360 = 0

⇒ a(a – 20) + 18(a – 20) = 0

⇒ (a + 18)(a – 20) = 0
⇒ a + 18 = 0 or a - 20 = 0
⇒ a = -18 or a = 20
As, price can't be negative, a = -18 is not possible
Therefore,a = Rs. 20

Let the original number of people be ‘a’.

Given, Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less.

Amount which each receives= 9000/a

⇒ 9000a = (9000 – 160a)(a + 20)

⇒ 9000a = 9000a + 180000 – 160a² – 3200a

⇒ a² + 20a – 1125 = 0

⇒ a² + 45a – 25a – 1125 = 0

⇒ a(a + 45) – 25(a + 45) = 0

⇒ (a – 25)(a + 45) = 0

⇒ a = 25

Let the number of students who planned the picnic be ‘a’.

Budget for the food was Rs. 500

Cost of food for each member = 500/a

Given, 5 of these failed to go and thus the cost of food for each member increased by Rs. 5

⇒ (a – 5)(100 + a) = 100a

⇒ 100a + a² – 500 – 5a = 100a

⇒ a² – 5a – 500 = 0

⇒ a² – 25a + 20a – 500 = 0

⇒ a(a – 25) + 20(a – 25) = 0

⇒ (a + 20)(a – 25) = 0

⇒ a = 25

Number of students who attended the picnic = 25 – 5 = 20

Let the distance of pole from gate A be ‘a’.

⇒ Difference of the distance of the pole from two diametrically opposite fixed gates A and B on the boundary is 7 metres.

⇒ Distance of pole from gate B = a – 7 m

⇒ Diameter of the park = 13 m

⇒ 13² = a² + (a – 7)²

⇒ 169 = 2a² + 49 – 14a

⇒ a² – 7a – 60 = 0

⇒ a² – 12a + 5a – 60 = 0

⇒ a(a – 12) + 5(a – 12) = 0

⇒ (a + 5)(a – 12) = 0

⇒ a = -5 or a = 12 m

Thus, distance of pole is 12 m from gate A and (12 - 7) = 5 meters from gate B

Given: In a class test, the sum of the marks obtained by P in Mathematics and science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180.


To find: his marks in the two subjects.


Solution:

Let the marks obtained in Mathematics by P be ‘a’.

Given, sum of the marks obtained by P in Mathematics and science is 28.

⇒ Marks obtained in science = 28 – a

Also, if he got 3 marks more in Mathematics and 4 marks less in Science, product of his marks, would have been 180.

⇒ (a + 3)(28 – a – 4) = 180


⇒ (a + 3)(24 – a ) = 180

⇒ -a² + 21a + 72 = 180


⇒ -a² + 21a + 72 - 180 = 0


⇒ -a² + 21a - 108 = 0

⇒ a² – 21a + 108 = 0

⇒ a² – 12a – 9a + 108 = 0

⇒ a(a – 12) – 9(a – 12) = 0

⇒ (a – 9)(a – 12) = 0

⇒ a = 9 or 12


If marks obtained in mathematics is 9, the marks obtained in science is 28-a = 28 - 9 = 19

⇒Marks in Mathematics = 9, Marks in Science = 19

Or

Marks in Mathematics = 12, Marks in Science = 16

Given : In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210.
To find: her marks in two subjects.
Solution:
Let the marks obtained in Mathematics by Shefali be ‘a’.

Given, sum of the marks obtained by Shefali in Mathematics and English is 30.

Marks obtained in english = 30 – a

Also, she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210.

⇒ (a + 2)(30 – a – 3) = 210
⇒ (a + 2)(27 – a ) = 210

⇒ -a² + 25a + 54 = 210
⇒ -a² + 25a + 54 - 210 = 0
⇒ -a² + 25a - 156 = 0

⇒ a² – 25a + 156 = 0

⇒ a² – 13a – 12a + 156 = 0

⇒ a(a – 13) – 12(a – 13) = 0

⇒ (a – 12)(a – 13) = 0

⇒ a = 12 or 13

Hence

Marks in Mathematics = 12, Marks in English = 18

Or

Marks in Mathematics = 13, Marks in English = 17

Let the number of article produced on the day be ‘a’.

Given, it was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day.

Cost of production of each article = 2a + 3

Given, cost of production was Rs. 90

⇒ a(2a + 3) = 90

⇒ 2a² + 3a – 90 = 0

⇒ 2a – 12a + 15a – 90 = 0

⇒ 2a(a – 6) + 15(a – 6) = 0

⇒ (2a + 15)(a – 6) = 0

⇒ a = 6

Cost of each article = 2 × 6 + 3 = Rs. 15

Quadratic equation has equal roots then d = b² – 4ac = 0

Here a = 1, b = k and c = 4

So b² – 4ac = 0

⇒ k² – 4 × 1 × 4 = 0

⇒ k² – 16 = 0

⇒ k = 4

If roots of given equation are real and distinct then D = b² – 4ac > 0

Here a = 1, b = 4 and c = k

So, 4² – 4 (1) (k) >0

16 – 4k > 0

16 > 4k

K< 4

Consider the equation 4x² – 12x – 9 = 0,


To check the roots of the equation we will check the value of d = b² – 4ac

Here a = 4, b = 12 and c = – 9

So d = (12)² – 4 x 4 x – 9

= 144 + 144

= 288>0 which is real

So the roots of the given equation are Real and distinct.

If roots of given equation are distinct then

d = b² – 4ac > 0

Here a = 1, b = a, c = 1

So, a² – 4(1) (1) >0

a² – 4 > 0

a² >4

|a| > 2

Given: the equation 9x² + 6kx + 4 = o has equal roots.

To find: the roots are both equal to ?

Solution:

If roots of given equation are equal then D = b² – 4ac = 0

⇒ (6k)² – 4(9)(4) = 0

⇒ 36k² – 144 = 0

⇒ 36k² = 144

⇒ K² = 4

⇒ K = 2

Case 1 :- when k = 2

In equation 9x² + 6 k x + 4 = 0

9x² + 6(2) x + 4 = 0

9x² + 12 x + 4 = 0

(3x)² + 2 × 2 × 3x + (2)² = 0

(3x + 2)² = 0

3 x + 2 = 0

3x = – 2

Case 2 :- when k = – 2

In equation 9x² + 6kx + 4 = 0

9x² + 6(– 2) x + 4 = 0

9x² – 12 x + 4 = 0

(3x) ² – 2 X 2 X 3x + (2)² = 0

(3x – 2)² = 0

3x – 2 = 0

3x = 2

So the roots of the given quadratic equation are

1 + √2 Is a root of quadratic equation with rational coefficients that is the sum of the roots is rational and the product of the roots is also rational.

Since the rational roots occurs in conjugate pairs so the other root of the equation is 1 – √2

If ax² + bx + c = 0 then x =

If D = b² − 4ac ≥ 0 then the values of x are real

If D = b² − 4ac < 0 then the values of x are complex

|z| is always a positive real number regardless of x being a real number or complex number.

Given eqn. is |x|² + 3|x| + 2 = 0 and a = 1, b = 3 and c = 2

|x|² + 3|x| + 2 = 0

⇒ |x| =

⇒ |x| = 2 or −1

But |x| cannot be negative

No real root for the equation.

Let the roots of given equation be m and n

According to the question M = n

Sum of roots = m + n = – b / a

Product of roots = m × n =

m² =

=

=

= c

Therefore c =

First of all, the equation is x² + |x| − 6 = 0

CASE 1: x>0 then |x| = x

⇒ x² + x− 6 = 0

⇒ x² + 3x – 2x – 6 = 0

⇒ (x + 3)(x−2) = 0

⇒ x = 2, – 3

CASE 2: x<0 |x| = – x

⇒ x²−x−6 = 0

⇒ x² − 3x + 2x−6 = 0

⇒ (x−3)(x + 2) = 0

⇒ x = −2, 3

So the sum of the roots is 2 + (– 3) + (– 2) + 3

= 0

If the roots of given equation are distinct then

d = b² – 4ac = 0

⇒ d = b² – 4ac = 0

⇒ 2² – 4(a) (a) = 0

⇒ 4 – 4a² = 0

⇒ 4a² = 4

⇒ a² = 1

⇒ a = 1

If the given equation x² + kx + 64 has real roots then D ≥ 0

D = b² – 4ac ≥ 0

Here a = 1, b = k, c = 64

d = K² – 4 (1) (64) ≥0

d = K² – 256 ≥ 0

K ≥ 16…… (1)

If the given equation x² – 8x + k = 0 has real roots then then

d≥0

D = b² – 4ac ≥0

8² – 4(1) (k) ≥0

64– 4k ≥ 0

64 ≥ 4k

K ≤ 16…… (2)

From (1) and (2) we can conclude that k = 16

Consider x² + ax – 1 = 0,

For the quadratic equation to have real roots D ≥ 0

Here a = 1, b = a and c = 1

In the given equation D = a² – 4 ≥ o

⇒ a² ≥ 4

So for all the real values of ‘a’ which are greater than or equal to 2 and – 2 the equation will have the real roots.

A quadratic equation has two real roots if discriminant = 0

For the given equation, we have:

d = b² – 4 a c

d = (2)² – 4 (1) (a² + 1)

d = 4 – 4(a² + 1)

d = 4(1 – a² – 1)

d = – 4a²

Now, D = 0 when a = 0. So, the equation will have real and equal roots if a = 0. And for all other values of a, the equation will have no real roots.

No, there is no real value of ‘a’ for which the given equation has real roots.

In given equation let x =

So, x = √(6 + x)

Now squaring both side

x² = 6 + x

X² – x – 6 = 0

X² – 3x + 2x – 6 = 0

x(x – 3) + 2(x – 3) = 0

(x – 3) (x + 2) = 0

x = 3 or – 2

x cannot be equal to – 2 as root can never be negative.

x = 3

For being the perfect square, the roots are equal

So, d = b² – 4ac = 0

Here a = 1, b = 4 and c = λ

⇒ d = 16 – 4 λ = 0

⇒ λ = 4

2 is the root of given equation x² + bx + 12 = 0

So 2² + 2b + 12 = 0

16 + 2b = 0

b = – 8…………..1

Now, d = b² – 4ac = 0 of second equation is

d = b² – 4 (1) (q) = 0 here a = 1, b = – 8 (from 1) and c = q

(– 8)² – 4q = 0

64 – 4q = 0

q = 16

Hence value of q is 16.

Given the roots of both the equations are real

For first equation ax² + 2bx + c = 0

Its discriminant; d ≥ 0

D = b² – 4ac

D = (2b)² – 4 × a × c ≥ 0

4b² ≥

b² ≥ ac …1

For second equation

bx² - 2x + b = 0

d = b² – 4ac ≥ 0

= (2)² – 4 b xb ≥ 0

= 4ac – 4 b² ≥

ac ≥ b² …2

From 1 and 2 we get only one case where b² = ac

If the roots are equal then d = b² – 4ac = 0

Here a = (a² + b²), b = 2 (ac + bd), c = (c² + d²)

D = b² – 4ac = 0

⇒ b² = 4ac

⇒ {– 2(ac + bd)}² = 4{(a² + b²) (c² + d²)}

⇒ 4(a²c² + b²d² + 2acbd) = 4(a²c² + a²d² + b²c² + b²d²)

⇒ 2acbd = a²d² + b²c²

⇒ a²d² + b²c² – 2abcd = 0

⇒ (ad – bc)² = 0

⇒ ad – bc = 0

⇒ ad = bc

To have the real roots D = b² – 4ac ≥ 0

Here a = 2, b = k and c = 8

D = k² – 4 x 2 x 8 ≥ 0

⇒ K²≥ 64

⇒ K ≥ 8

So for all the values of k greater than or equal to 8 and – 8, the given quadratic equation will have real roots.

The roots of the equation are equal so d = b² – 4ac = 0

Here a = (a² + b²), b = – 2b (a + c), c = (b² + c²)

d = (– 2b (a + c))² = 4 (a² + b²) (b² + c²)

⇒ b² (a² + 2ac + c²) = a²b² + a²c² + b⁴ + b²c²

⇒ (ac)² – 2(ac) (b²) + (b²)² = 0

⇒ (ac – b²)² = 0

⇒ (ac – b²) = 0

⇒ a c = b²

If the equation does not possess real roots then

d = b² – 4ac 0

Here a = 1, b = – b, c = 1

d = b² – 4 < 0

b² < 4

b < 2

– 2 < b < 2

The sum of the two zeros of the quadratic equation is given by –

Here it’s given = 2

The product of the quadratic equation is

Here = 2

the quadratic equation is of the form ax² + b x + c = 0

or x² + (sum of the roots) x + product of the roots = 0

= x² – 2 x + 2

f(x) = k(x² – 2 x + 2), where k is any real number

Since x = 1 is root of equations, it will satisfy both the equations.

Putting x = 1 in ax² + ax + 3 = 0

1a + a + 3 = 0

2a + 3 = 0

A = – 3/2

Putting x = 1 in x² + x + b = 0

1 + 1 + b = 0

b = – 2

ab = 3

To be the solution of the equation the value x = – 3 should satisfy the given equation x² + 6x + 9 = 0

Putting value of x on L.H.S

(– 3)² + 6 x (– 3) + 9

⇒ 9 – 18 + 9 = 0 = R.H.S

Hence x = – 3 is the solution of given equation.

To be solution of the equation x = – 2 should satisfy the given equation

L.H.S 3x (– 2)² + 13 x – 2 + 14

⇒ 12 – 26 + 14 = 0 = R.H.S

Hence x = – 2 is the solution of the equation.

Since p and q are roots of the equations then

Sum of the roots is p + q = = – (p) = – p

Here a = 1, b = p and c = q

Products of the root = p × q = = q

∴ p × q = q

p = 1

Putting value of ‘p’ in p + q = – p

1 + q = – 1

q = – 2

For quadratic equation to have real roots,

d≥ 0

b² – 4a ≥ 0

b² ≥ 4a

For a = 1, 4a = 4, b = 2, 3, 4 (3 equations)

With values of (a,b) as (1,2), (1,3), (1,4)

a = 2, 4a = 8, b = 3, 4 (2 equations)

With values of a,b as (2,3), (2,4)

a = 3, 4a = 12, b = 4 (1 equation)

With value of (a,b) as (3,4)

a = 4, 4a = 16, b = 4 (1 equation)

With values of (a,b) as (4,16)

Thus, total 7 equations are possible.

d = b² – 4 ac

Here a = 3

B = 10 and c =

D = (10)² – 4 (3)

D = 100 – 36

D = 64

Since x = is the solution of the equation it should satisfy the equation

Putting value of x in the given equation

3(– )² + 2 k () – 3 = 0

– k – 3 = 0

= k

The roots of the equation are real (given)

Let α and β be the two roots according to the given condition

α = α²

β = β²

Sum of the roots = α + β = α² + β²

Product of the roots = α β = α² β²

There are only two number who does not change on squaring them that is 0 and 1

So the number of equations could be 2 by being the roots as

(0,1) and (1,0)

Since the equation

(a² + b²) x² + 2 (ac + bd) x + c² + d² = 0 has no real root

D < 0

b² – 4ac < 0

b² < 4ac

Here a = (a² + b²), b = 2 (ab + bd), c = c² + d²

4(ac + bd)² – 4 (a² + b²)(c² + d²) < 0

4a²c² + 4b²d² + 8abcd – 4(a²c² + b²c² + a²d² + b²d²) < 0

– 4(a²d² + b²c² – 2abcd) < 0

– 4(ad + bc)²< 0

∴ d is always negative

And ad bc

equation is x² – x = λ (2x – 1)

x² – x – λ (2x – 1) = 0

x² – (2λ + 1)x + λ = 0

Here a = 1, b = – (2λ + 1), c = λ

Sum of the roots = – b/a

⇒ –(– (2λ + 1)) = 0

⇒ λ = – 1/2

Since x = 1 is root of equation

Then it satisfy the equation

Putting x = 1 in first equation

a + a + 2 = 0

2a + 2 = 0

a = – 1

Putting x = 1 in equation second

1 + 1 + b = 0

2 + b = 0

b = – 2

ab = – 1 x – 2

ab = 2

The equation has equal root which means d = 0

d = b² – 4ac = 0

Here a = a, b = 2b, c = c

(2b)² – 4 ac = 0

b² – ac = 0

c =

Equation x² + k (4x + k – 1) + 2 = 0 has equal roots

d = 0

d = b² – 4ac = 0

Here a = 1, b = 4k, c = k² – k + 2

⇒ 16 k² – 4(k² – k + 2) = 0

⇒ 12 k² + 4k – 8 = 0

⇒ 3k² + k – 2 = 0

⇒ (3k – 2) (k + 1) = 0

⇒ K = 2/3, – 1

In the given equation kx² + 6x + 4k = 0

Sum of the roots = product of the roots (given)

– b/a = c/a

Here a = k, b = 6 and c = 4k

=

K =

Equation ax² + bx + c = 0 has and as two roots

sin α + cos α =

sin α × cos α = c/a …eq(1)

…..eq (2)

But sin² α + cos² α = 1

∴ a² (1 + 2 sinα.cos α) = b²

Putting sin α × cos α = c/a, we get,

⇒ b² = a² + 2ac.

The given equation x² + ax + 12 = 0 has a root = 2

So it will satisfy the equation

4 + 2a + 12 = 0

2a + 16 = 0

a = – 8

Putting value of a in second equation, it becomes

x² + ax + q = 0

x² – 8x + q = 0

Roots are equal so d = 0

⇒ b² – 4ac = 0

⇒ 64 – 4q = 0

⇒ q = 64/4

⇒ q = 16

In the given equation x² – (k + 6)x + 2 (2k – 1) = 0

a = 1, b = – (k + 6), c = 2 (2k – 1)

Sum of the roots = 1/2 (product of roots) (given)

K + 6 = 2k – 1

K = 7

Given a and b are roots of the equation x² + a x + b = 0

Here a = 1, b = a, c = b

Sum of the roots = a + b = – a/1

b = – 2a

Product of the roots

ab = b

a = 1

∴ b = – 2 x 1 = – 2

Now a + b = 1 + (– 2) = – 1

Let Root of an equation = α = 2

Sum of the roots = α + β = 0, where α and β are two roots of the equation

β = – 2

α β = 2 x – 2 = – 4

the general equation is of the form

x² + (sum of the roots)x + product of the roots = 0

x² – 4 = 0 is the required equation

In the given equation ax² + bx + c = 0

Let α and β be the two roots

Given α = 3β ……..1

α β = (product of the roots)

3 β² = c / a ……….. by using 1

β² = …………….2

α + β =

4β =

Squaring both the sides

16 β² =

By using 2

16 x =

= b²/ac

In the given equation 2x² + kx + 4 = 0

Let α and β be the two roots

α = 2 (given)

here a = 2, b = k and c = 4

sum of the roots

⇒ α + β =

⇒ 2 + β =

β = – k

α β = 2

β = 2/2 = 1(putting value of α = 2)

Let the given equation has roots α and β

α = 1

Here a = 1, b = a and c = 3

Sum of the roots

α + β = – b/a = – a

product of the roots

α β = c/a = 3

1 β = 3

β = 3

Let α and β be the two roots of the given equation

4x² – 2x + (λ – 4) = 0

According to the given condition

α = 1/β

Here a = 4, b = – 2 and c = (λ – 4)

α + β = 2/4 = 1/2

1/β + β = 1/2

α β =

β =

K = 8

If y = 1 is root of both the equation it will staify both the equations

Putting y = 1 in first equation

ay² + ay + 3 = 0

2a + 3 = 0

a =

Putting value of y in second equation

2 + b = 0

B = – 2

Now ab = x – 2

ab = 3

Given: 16x² + 4kx + 9 = 0
To find:The values of k for which the quadratic equation 16x² + 4kx + 9 = 0 has real and equal roots.
Solution:
To have real and equal roots d = 0
Where d=b² – 4ac

⇒b² – 4ac = 0
Compare with the general equation of quadratic equation ax² + bx + c = 0, a≠0

here a = 16, b = 4k and c = 9

⇒b² – 4ac =(4k)² – 4 x 16 x 9 = 0

⇒16 k² – 576 = 0

⇒k² = 576 /16

⇒k = 24/ 4

k = ±6

Let the consecutive numbers be ‘a’ and ‘a + 1’ respectively.

Given, product of two consecutive positive integers is 306

a × (a + 1) = 306

⇒ a² + a – 306 = 0

Number of marbles John has is x.

Given, John and Jivanti together have 45 marbles.

Number of marbles which Jivanti has = 45 – x

Now, both of them lost 5 marbles each, and the product of the number of marbles they now have is 128.

⇒ (x – 5)(40 – x) = 128

⇒ 40x – 200 + 5x – x² = 128

⇒40x – 200 + 5x – x² - 128 = 0

⇒45x – 328 – x² = 0

⇒ x² – 45x + 328 = 0

Number of toys produced that day is ‘x’.

Cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day.

∴ Cost of production of each toy = 55 – x

Given, total cost of production = Rs. 750

⇒ x × (55 – x) = 750

⇒ -x² + 55x – 750 = 0

⇒ x² – 55x + 750 = 0

By Pythagoras theorem :

Hypotenuse² = perpendicular² + base²

Given, height of a right TRIANGLE IS 7 CM LESS THAN ITS BASE and the hypotenuse is 13 cm.

Let the base be ‘x’

⇒ 13² = (x – 7)² + x²

⇒ 169 = x² – 14x + 49 + x²

⇒ x² – 7x – 60 = 0

Given: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train.

To find: the quadratic equation to find the average speed of express train.

Solution:


Let the average speed of passenger train be ‘x’ km/hr and the time taken by passenger train be ‘t’ hr.

So, For the express train


average speed = x + 11, time taken = t - 1

Since,


Distance = speed × time

Given, total distance traveled = 132 km


For passenger train:

⇒ x × t = 132

⇒ t = 132/x ...... (1)

Also for express train

(x + 11) × (t - 1) = 132

⇒ - x² -11x +1452 = 0

⇒ x² + 11x – 1452 = 0

Required quadratic equation

Let the speed of the train be ‘a’ km/hr and the actual time taken be ‘t’

Given, train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 4 hour less for the same journey.

Distance = speed × time

⇒ 360 = a × t

⇒ t = 360/a

Also, 360 = (a + 5)(t – 4)

⇒ 360a = (a + 5)(360 – 4a)

⇒ 360a = 360a + 1800 – 4a² – 20a

⇒ a² + 5a – 450 = 0

is already factorized

⇒ x = 4, -2

is already factorized

⇒ x = -3/2, 7/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² –x -√3x + √3 = 0

⇒ x(x – 1) - √3(x – 1) = 0

⇒ (x - √3)(x – 1) = 0

⇒ x = √3, 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 9x² – 6x + 3x – 2 = 0

⇒ 3x(3x – 2) + (3x – 2) = 0

⇒ (3x + 1)(3x – 2) = 0

⇒ x = -1/3, 2/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3√5x² + 30x – 5x – 10√5 = 0

⇒ 3√5x(x + 2√5) – 5(x + 2√5) = 0

⇒ (x + 2√5)(3√5x – 5) = 0

⇒ x = -2√5, √5/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 6x² + 9x + 2x + 3 = 0

⇒ 3x(2x + 3) + (2x + 3) = 0

⇒ (3x + 1)(2x + 3) = 0

⇒ x = -1/3, - 3/2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 5x² – 5x + 2x – 2 = 0

⇒ 5x(x – 1) + 2(x – 1) = 0

⇒ (5x + 2)(x – 1) = 0

⇒ x = -2/5, 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 48x² – 16x + 3x – 1 = 0

⇒ 16x(3x – 1) + (3x – 1) = 0

⇒ (16x + 1)(3x – 1) = 0

⇒ x = -1/16, 1/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 5)(x + 2) = 0

⇒ x = -5/3, - 2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 25x² + 25x + 4 = 0

⇒ 25x² + 20x + 5x + 4 = 0

⇒ 5x(5x + 4) + (5x + 4) = 0

⇒ (5x + 1)(5x + 4) = 0

⇒ x = -1/5, -4/5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 16x² – 27x – 10 = 0

⇒ 16x² – 32x + 5x – 10 = 0

⇒ 16x(x – 2) + 5(x – 2) = 0

⇒ (16x + 5)(x – 2) = 0

⇒ x = -5/16, 2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ √3x² – 3√2x + √2x – 2√3 = 0

⇒ √3x(x - √6) + √2(x - √6) = 0

⇒ (√3x + √2)(x - √6) = 0

⇒ x = √6, -√(2/3)

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 4√3x² + 8x – 3x – 2√3 = 0

⇒ 4x(√3x + 2) – √3(√3x + 2) = 0

⇒ (4x - √3)(√3x + 2) = 0

⇒ x = √3/4, - 2/√3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ √2x² – 4x + x – 2√2 = 0

⇒ √2x(x – 2√2) + (x – 2√2) = 0

⇒ (√2x + 1)(x – 2√2) = 0

⇒ x = -1/√2, 2√2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ a²x² – 2abx – abx + 2b² = 0

⇒ ax(ax – 2b) – b(ax – 2b) = 0

⇒ (ax – b)(ax – 2b) = 0

⇒ x = b/a, 2b/a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² - √2x – x + √2 = 0

⇒ x(x - √2) - (x - √2) = 0

⇒ (x – 1)(x - √2) = 0

⇒ x = 1, √2

Given:

To find: The value of above equation.

Solution:

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

Consider,



Here,



apply the above formula and solve,

⇒ 9x² – 3(a² + b²)x + 3(a² – b²)x – (a² + b²)(a² – b²) = 0

⇒ 3x(3x – (a² + b²)) + (a² – b²)(3x – (a² + b²)) = 0

⇒ (3x + a² – b²)(3x – (a² + b²) = 0

⇒ x

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 4x² + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

⇒ 2x(2x + a + b) – (a – b)(2x – (a + b)) = 0

⇒ (2x – (a – b))(2x + a + b) = 0

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ ax² + 4a²x – 3bx – 12ab = 0

⇒ ax(x + 4a) -3b(x + 4a) = 0

⇒ (ax – 3b)(x + 4a) = 0

⇒ x = 3b/a, -4a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 2x² + 2ax – ax – a² = 0

⇒ 2x(x + a) – a(x + a) = 0

⇒ (2x – a)(x + a) = 0

⇒ x = a/2, -a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² – 3√2x - √2x + 6 = 0

⇒ x(x – 3√2) - √2(x – 3√2) = 0

⇒ (x - √2)(x – 3√2) = 0

⇒ x = √2, 3√2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 2x² – 9 + 3x = 3x² – 14 –x

⇒ x² – 4x – 5 = 0

⇒ x² – 5x + x – 5 = 0

⇒ (x – 5)(x + 1) = 0

⇒ x = -1, 5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 12x² – 57x + 60 = 25x² + 300 – 175x

⇒ 13x² - 118x + 240 = 0

⇒ 13x² – 78x – 40x + 240 = 0

⇒ 13x(x – 6) – 40(x – 6) = 0

⇒ (13x – 40)(x – 6) = 0

⇒ x = 6, 40/13

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 4(x² + 3x + (x – 1)(x – 2)) = 17(x² – 2x)

⇒ 4(x² + 3x + x² – 3x + 2) = 17(x² – 2x)

⇒ 8x² + 8 = 17x² – 34x

⇒ 9x² – 34x – 8 = 0

⇒ 9x² – 36x + 2x – 8 = 0

⇒ 9x(x – 4) + 2(x – 4) = 0

⇒ (9x + 2)(x – 4) = 0

⇒ x = -2/9, 4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

Taking L.C.M

⇒ 7(2 x² + 18) = 48(x² – 9)

⇒ -84x = 48x² – 432

⇒ 4x² + 7x – 36 = 0

⇒ 4x² + 16x – 9x – 36 = 0

⇒ 4x(x + 4) -9(x + 4) = 0

⇒ (4x – 9)(x + 4) = 0

⇒ x = 9/4, -4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x(x – 1 + 2x – 4) = 6(x² -3x + 2)

⇒ 3x² - 5x = 6x² – 18x + 12

⇒ 3x² - 13x + 12 = 0

⇒ 3x² - 9x - 4x + 12 = 0

⇒ 3x(x - 3) -4(x – 3) = 0

⇒ (3x – 4)(x – 3) = 0

⇒ x = 4/3, 3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 6((x + 1)² - (x – 1)²) = 5(x² – 1)

⇒ 6 × 4x = 5x² – 5

⇒ 5x² – 24x – 5 = 0

⇒ 5x² – 25x + x – 5 = 0

⇒ 5x(x – 5) + 1(x-5) = 0

⇒ (5x + 1)(x – 5) = 0

⇒ x = 5, -1/5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 2(x² - 2x + 1 + 4x² + 4x + 1) = 5(2x² – x – 1)

⇒ 10x² + 4x + 4 = 10x² – 5x – 5

⇒ 9x = -9

⇒ x = -1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3x² – 15x + x – 5 = 0

⇒ 3x(x – 5) + 1(x – 5) = 0

⇒ (3x + 1)(x – 5) = 0

⇒ x = 5, -1/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒

⇒ m²x² + n² = mn – 2mnx

⇒ m²x² + 2mnx – mn + n² = 0

⇒ (mx + n)² = mn

⇒ mx + n = �√mn

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

Let (x – a)/(x – b) = y and a/b = c ..... (1)

So (x – b)/(x – a) = 1/ y and b/a = 1/c .... (2)

⇒ c(y² + 1) = y(c²+1)

⇒ cy² + c = c²y + y

⇒ cy² – c²y –y + c = 0

⇒ cy(y – c) – 1(y – c) = 0

⇒ (cy – 1)(y – c) = 0

⇒ y = 1/c, c

From (1)

⇒ (x – a)/(x – b) = c and a/b = c



⇒ x² - bx - ax + ab = a/b

⇒ b(x² - bx - ax + ab) = a

⇒ bx² - b²x - abx + ab² = a

⇒ bx ( x - b ) - ab ( x - b ) = a

⇒ (bx - ab) ( x - b ) = a

⇒ (bx - ab) = a and ( x - b ) = a

So,

bx - ab = a

⇒ bx = a + ab

⇒ x =( a + ab ) / b

And

x - b = a

⇒ x = a + b

Hence x =( a + ab ) / b , (a+b)

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 6(x² + 12 – 7x + x² + 4 – 5x + x² – 3x + 2) = (x² – 3x + 2)(x² + 12 – 7x)
⇒ 6(3x² + 18 – 15x ) = x⁴ + 12x² – 7x³ – 3x³ – 36x + 21x² + 2x² + 24 – 14x

⇒ 18x² – 90x + 108 = x⁴ + 12x² – 7x³ – 3x³ – 36x + 21x² + 2x² + 24 – 14x

⇒ x⁴ – 10x³ + 17x² + 40x + 84 = 0

⇒ (x + 2)²(x – 7)² = 0
Therefore, possible value of 'x' are -2, -2, and -7, -7

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² – 11x + 30 = 25/(24)²

⇒ x² – 11x + 121/4 – 121/4 + 30 = 25/(24)²

⇒ x² – 11x + 121/4 = 25/(24)² + 1/4

⇒ x² – 11x + 121/4 = 676/(4 × 24²)

⇒ (x – 11/2)² = (13/24)²

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 7x² + 3 = 178x/5

⇒ 35x² – 178x + 15 = 0

⇒ 35x² – 175x – 3x + 15 = 0

⇒ 35x(x – 5) -3(x – 5) = 0

⇒ (35x – 3)(x – 5) = 0

⇒ x = 5, 3/35

Given:

To find: The value of x.

Solution:

In factorization, we write the coefficient of middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the product of these two factors will be equal to the product of the coefficient of x² and the constant term.

...... (1)


Take the LCM of the denominators.

LCM is ( x - a ) ( x - b ) ( x - c )

⇒ (ax – ab + bx – ab)(x – c) = 2c (x² – bx - ax +ab)

⇒ (ax – ab + bx – ab)(x – c) = 2cx² – 2cbx - 2cax +2cab

⇒ ((a + b)x – 2ab)(x – c) = 2cx² – 2c(a + b)x + 2abc

⇒ (a + b)x² – (a + b)cx – 2abx + 2abc = 2cx² – 2(a + b)cx +2abc

⇒ (a + b – 2c)x² + ((a + b)c – 2ab)x = 0

⇒x[(a + b – 2c)x + ((a + b)c – 2ab)] = 0

⇒ x = o

and (a + b – 2c)x + ((a + b)c – 2ab)=0

⇒(a + b – 2c)x = - [(a + b)c – 2ab]

⇒(a + b – 2c)x = - (ac + bc – 2ab)

⇒

⇒


Hence x = 0,

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² – (2a + b)x + 2ab = 0

⇒ x² – 2ax – bx + 2ab = 0

⇒ x(x – 2a) – b(x – 2a) = 0

⇒ (x – b)(x – 2a) = 0

⇒ x = b, 2a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (a + b)²x² - (a² + b² + 2ab – a² – b² + 2ab)x – (a – b)² = 0

⇒ (a + b)²x² – (a + b)²x + (a – b)²x – (a – b)² = 0

⇒ (a + b)²x(x - 1) + (a – b)²(x – 1) = 0

⇒ ((a + b)²x + (a – b)²)(x – 1) = 0

⇒ x = 1,

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ ax² + a – a²x – x = 0

⇒ ax² – x(a² + 1) + a = 0

⇒ ax(x – a) – 1(x – a) = 0

⇒ (ax – 1)(x – a) = 0

⇒ x = 1/a, a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² – a² – x – a = 0

⇒ (x + a)(x – a) – 1(x + a) = 0

⇒ (x + a)(x – a – 1) = 0

⇒ x = -a, a + 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ ax² +a²x + x + a =0

⇒ ax(x + a) + 1(x + a) = 0

⇒ (ax + 1)(x + a) = 0

⇒ x = -a, -1/a

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ abx² – acx + b²x – bc = 0

⇒ ax(bx – c) + b(bx – c) = 0

⇒ (ax + b)(bx – c) = 0

⇒ x = -b/a, c/b

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ b²x(a²x + 1) – (a²x + 1) = 0

⇒ (b²x – 1)(a²x + 1) = 0

⇒ x = -1/a², 1/b²

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(x– 1)(x – 4) + 3(x – 2)(x – 3) = 10(x – 2)(x – 4)

⇒ 3x² + 12 – 15x + 3x² + 18 – 15x = 10x² – 60x + 80

⇒ 4x² – 30x + 50 = 0

⇒ 2x² – 15x + 25 = 0

⇒ 2x² – 10x – 5x + 25 = 0

⇒ 2x(x – 5) – 5(x – 5) = 0

⇒ (2x – 5)(x – 5) = 0

Thus, x = 5/2, 5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (√3x)² – 2√6x + (√2)² = 0

⇒ (√3x - √2)² = 0

⇒ x =

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 7(x + 5 – x + 1) = 6(x – 1)(x + 5)

⇒ 42 = 6x² + 24x – 30

⇒ x² + 4x – 12 = 0

⇒ x² + 6x – 2x – 12 = 0

⇒ x(x + 6) – 2(x + 6) = 0

⇒ (x – 2)(x + 6) = 0

⇒ x = 2, -6

Given:


To find: Solve the given quadratic equations by factorization.

Solution:

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x-2 - x = 3 x ( x - 2 )

⇒ x – 2 – x = 3(x² – 2x)

⇒ x – 2 – x = 3x² – 6x

⇒ 3x² – 6x + 2 = 0

Now convert the terms in above equation involving 3,

⇒ 3x² – (3+3) x + (3-1) = 0

Now add and subtract √3 in the coefficient of x,

⇒ 3x² – (3+√3+3-√3) x + (3-1) = 0

Now 3 = (√3 )² and 1 = 1²

⇒ (√3x)² – [(3+√3) + (3-√3) ]x + (3-1) = 0


⇒ (√3x)² – (3+√3) x - (3-√3) x + (√3² - 1²)= 0

Apply the formula a² - b² = ( a+b) ( a-b) in (√3² - 1²)

⇒ (√3x)² – (3+√3) x - (3-√3) x +[ (√3-1) (√3+1)]= 0


⇒ (√3x)² – √3(√3+1) x - √3(√3-1) x +[ (√3-1) (√3+1)]= 0

⇒ (√3x)² – √3(√3+1) x - √3(√3-1) x +[ (√3-1) (√3+1)]= 0

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x² – 3x – 1 = 3x – 9

⇒ x² – 6x + 8 = 0

⇒ x² – 4x – 2x + 8 = 0

⇒ x(x – 4) – 2(x – 4) = 0

⇒ (x – 2)(x – 4) = 0

⇒ x = 2, 4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 30(x – 7 – x – 4) = 11(x + 4)(x – 7)

⇒ -330 = 11x² – 308 – 33x

⇒ 11x² – 33x + 22 = 0

⇒ x² – 3x + 2 = 0

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – (x – 2) = 0

⇒ (x – 1)(x – 2) = 0

⇒ x = 1, 2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ x((x – 2) + 2(x – 3)) = 8(x – 3)(x – 2)

⇒ 3x² – 8x = 8x² – 40x + 48

⇒ 5x² - 32x + 48 = 0

⇒ 5x² – 20x – 12x + 48 = 0

⇒ 5x(x – 4) – 12(x – 4) = 0

⇒ (5x – 12)(x – 4) = 0

⇒ x = 12/5, 4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 2ab(2x – 2a – b – 2x) = (2a + b)2x(2a + b + 2x)

⇒ 2ab(-2a – b)= 2(2a+ b)(2ax + bx + 2x²)

⇒ -ab = 2ax + bx + 2x²

⇒ 2x² + 2ax + bx + ab = 0

⇒ 2x(x + a) + b(x + a) = 0

⇒ (2x + b)(x + a) = 0

⇒ x = -a, -b/2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (4 – 3x)(2x + 3) = 5x

⇒ 8x + 12 – 6x² – 9x = 5x

⇒ 6x² + 6x – 12 = 0

⇒ x² + x – 2 = 0

⇒ x² + 2x – x – 2 = 0

⇒ x(x + 2) – (x + 2) = 0

⇒ (x – 1)(x + 2) = 0

⇒ x = 1, - 2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(x² – 11x + 28) + 3(x² – 11x + 30) = 10(x² – 12x + 35)

⇒ 4x² – 54x + 176 = 0

⇒ 2x² – 27x + 88 = 0

⇒ 2x² – 16x – 11x + 88 = 0

⇒ 2x(x – 8) – 11(x – 8) = 0

⇒ (2x – 11)(x – 8) = 0

⇒ x = 11/2, 8

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (16 – x)(x + 1) = 15x

⇒ -x² – x + 16x + 16 = 15x

⇒ x² = 16

⇒ x = �4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(x² – 7x + 10) + 3(x² – 7x + 12) = 10(x² – 8x + 15)

⇒ 4x² – 38x + 84 = 0

⇒ 2x² – 19x + 42 = 0

⇒ 2x² – 12x – 7x + 42 = 0

⇒ 2x(x – 6) – 7(x – 6) = 0

⇒ (x – 6)(2x - 7) = 0

⇒ x = 7/2, 6

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 4(25 + x² + 10x) – 4(25 + x² – 10x) = 15(25 – x²)

⇒ 15x² + 80x – 375 = 0

⇒ 3x² + 16x – 75 = 0

⇒ 3x² + 25x – 9x – 75 = 0

⇒ x(3x + 25) – 3(3x + 25) = 0

⇒ (x – 3)(3x + 25) = 0

⇒ x = 3, - 25/3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (5 - x)(3x – 1) = 4x + 4

⇒ -3x² + 16x – 5 = 4x + 4

⇒ 3x² - 12x + 9 = 0

⇒ 3x² – 3x – 9x + 9 = 0

⇒ 3x(x – 1) -9(x – 1) =0

⇒ (3x – 9)(x – 1) = 0

⇒ x = 1, 3

Given:

to find: Solution of the above quadratic equation.

Solution:

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(9x² + 1 – 6x) – 2(4x² + 9 + 12x) = 5(6x² – 3 + 7x)

⇒ 27x² + 3 – 18x - 8x² - 18 – 24x = 30x² - 15 + 35x

⇒ 19x² - 42x – 15 = 30x² - 15 + 35x

⇒19x² - 30x² - 42x – 35x - 15 + 15 = 0

⇒ -11x² - 77x = 0

⇒ 11x² + 77x = 0

⇒ 11x(x + 7) = 0

⇒ x = 0, - 7

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(7x + 1)² – 4(5x – 3)² = 11(5x – 3)(7x + 1)

⇒ 3(49x² + 1 + 14x) – 4(25x² + 9 – 30x) = 11(35x² – 3 – 16x)

⇒ 338x² – 338x = 0

⇒ x(x – 1) = 0

⇒ x = 0, 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ (3x – 3 + 4x + 4)(4x – 1) = 29(x² – 1)

⇒ (7x + 1)(4x – 1) = 29x² – 29

⇒ 28x² – 3x – 1 = 29x² – 29

⇒ x² + 3x – 28 = 0

⇒ x² + 7x – 4x – 28 = 0

⇒ x(x + 7) – 4(x + 7) = 0

⇒ (x – 4)(x + 7) = 0

⇒ x = 4, - 7

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized

⇒ 5x(4x – 8 + 3x + 3) = 46(x + 1)(x – 2)

⇒ 35x² – 25x = 46x² – 92 – 46x

⇒ 11x² - 19x - 92 = 0

⇒ 11x² - 44x + 23x – 92 = 0

⇒ 11x(x – 4) + 23(x – 4) = 0

⇒ (11x + 23)(x – 4) = 0

⇒ x = 4, - 23/11

Given:

To find: the roots of the following quadratic (if they exist) by the method of completing the square.

Solution:

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

Step 1: Make the coefficient of x² unity.

In the equation ,

The coefficient of x²is 1.

Step 2: Shift the constant term on RHS,



Step 3: Add square of half of coefficient of x on both the sides.



Step 4: Apply the formula, (a - b)² = a² - 2ab + b² on LHS and solve RHS,

Here a=x and











As RHS is positive, the roots exist.

Now,take square root on both sides,

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

Divide the equation by 2 to get,

Use the formula (a + b)² = a² + 2ab + b²



⇒ (x – 7/4)² = 25/16

⇒ x – 7/4 = ±5/4

⇒ x = 3, 1/2

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ (x + 11/6)² = 1/36

⇒ x + 11/6 = �1/6

⇒ x = -2, -5/3

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ x² + x/2 – 2 = 0

⇒ x² + 2 × 1/4 × x + (1/4)² - (1/4)² – 2 = 0

⇒ (x + 1/4)² = 33/16

⇒ x + 1/4 = � √33/4

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ x² + x/2 + 2 = 0

⇒ x² + 2 × 1/4 × x + (1/4)² - (1/4)² + 2 = 0

⇒ (x + 1/4)² = -31/16

Roots are not real.

Given: The quadratic equation

To find: the roots of the following quadratic (if they exist) by the method of completing the square.

Solution:

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

Step 1: Make the coefficient of x² unity.

In the equation ,

The coefficient of x²is 4.

So to make the coffecient of x²equals to 1.

divide the whole equation by 4.

The quadratic equation now becomes:






Step 2: Shift the constant term on RHS,

Step 5: As the RHS is zero, the roots exist.

Since the quadratic equations have 2 roots, in this case both roots will be same.

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ x – 3/2√2 = �5/2√2

⇒ x = 2√2, -1/√2

Given:

To find:the roots of the following quadratic (if they exist) by the method of completing the square.

Solution:

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

Step 1: Make the coefficient of x² unity.

In the equation ,

The coefficient of x²is .

So to make the coffecient of x²equals to 1.

divide the whole equation by .

The quadratic equation now becomes:





Step 2: Shift the constant term on RHS,



Step 3: Add square of half of coefficient of x on both the sides.



Step 4: Apply the formula, (a + b)² = a² + 2ab + b² on LHS and solve RHS,

Here a=x and










As RHS is positive, the roots exist.

Step 5: take square root on both sides,

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ x² – (√2 + 1)x + ((√2 + 1)/2)² - ((√2 + 1)/2)² + √2 = 0

⇒ (x - (√2 + 1)/2)² = (2 + 1 + 2√2)/4 - √2

⇒ (x – (√2 + 1)/2)² = (2 + 1 – 2√2)/4 = ((√2 – 1)/2)²

⇒ x - (√2 + 1)/2 = �(√2 – 1)/2

⇒ x = √2, 1

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

⇒ x² – 2 × 2ax + 4a² = b²

⇒ (x – 2a)² = b²

⇒ x – 2a = ±b

⇒ x = 2a + b, 2a – b

(i)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

⇒ D = 25 – 4 × 2 × 3 = 1

(ii)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

Given,

⇒ D = 4 – 4 × 4 × 1 = - 12

(iii)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

Given,

⇒ 2x² – 3x + 1 = 0

⇒ D = 9 – 4 × 2 × 1 = 1

(iv)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

⇒ D = 4 – 4 × 1 × k = 4 – 4k

(v)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

⇒ D = 8 – 4 × √3 × -2√3 = 32

(vi)

For a quadratic equation, ax² + bx + c = 0,

Discriminant, D = b² – 4ac

Given,

⇒ D = 1 – 4 × 1 = -3

(i)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

16x² – 24x – 1 = 0

⇒ D = 24 × 24 + 4 × 16 × 1 = 640

Roots are real.

(ii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 1 – 4 × 2 = - 7

Roots are not real

(iii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 100 + 4 × 8√3 × √3 = 196

Roots are real

⇒ x = (-10 � 14)/2√3

⇒ x = -4√3, 2/√3

(iv)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 4 – 4 × 2 × 3 = - 20

Roots are not real

(v)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 4 × 6 – 4 × 3 × 2 = 0

Roots are equal

(vi)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 64a²b² – 4 × 3a² × 4b² = 16a²b²

(vii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

D = 20 + 4 × 5 × 3 = 80

⇒ x = -√5, √5/3

(viii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 4 – 4 × 1 × 1 = 0

Roots are equal

x = (2 �√(4 – 4))/2 = 1

(ix)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 75 – 4 × 2 × 6 = 27

⇒ x = -2√3, -√3/2

(x)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 49 – 4 × 5√2 × √2 = 9

⇒ x = -5/√2 , -√2

(xi)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

D = (2√2)² – 4 × 2 × 1

⇒ D = 8 – 8 = 0

Roots are equal

x = (2√2 �0)/4

⇒ x = 1/√2, 1/√2

(xii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 25 – 4 × 3 × 2 = 1

x = (5 � √1)/6

⇒ x = (5 � 1)/6

⇒ x = 1, 2/3

(i)






LCM of the denominator is (x-2)(x-4)

Now further solve it,

⇒ 3(x– 1)(x – 4) + 3(x – 2)(x – 3) = 10(x – 2)(x – 4)

⇒ 3x² + 12 – 15x + 3x² + 18 – 18x = 10x² – 60x + 80

⇒ 4x² – 30x + 50 = 0

⇒ 2x² – 15x + 25 = 0

⇒ 2x² – 10x – 5x + 25 = 0

⇒ 2x(x – 5) – 5(x – 5) = 0

⇒ (2x – 5)(x – 5) = 0


⇒2x – 5 = 0



⇒ x – 5 = 0

x=5

Thus,

(ii)

⇒ x – 2 – x = 3x² – 6x

⇒ 3x² – 6x + 2 = 0

Using

(iii)

LCM of denominators is x.

⇒

⇒ x² + 1 = 3x

⇒ x² – 3x + 1 = 0

Using

(iv)

Use cross multiplication to get,


(16 – x)(x + 1) = 15x

⇒ 16x – x² + 16 – x = 15x

⇒ x² = 16

⇒ x = ±4

(i)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 9 – 4 × 5 × 2 = -31

Roots are not real.

(ii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 36 – 4 × 2 × 3 = 12

Roots are real and distinct.

(iii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 4/9 – 4 × 3/5 × 1 = -88/45

Roots are not real.

(iv)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 48 – 4 × 3 × 4 = 0

Roots are real and equal

(v)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 24 – 4 × 3 × 2 = 0

Roots are real and equal.

(vi)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ x² – (2a + 2b)x + 4ab = 4ab

⇒ x² – (2a + 2b)x = 0

D = (2a + 2b)² – 0 = (2a + 2b)²

Roots are real and distinct

(vii) ,

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 576a²b²c²d² – 4 × 16 × 9 × a²b²c²d² = 0

Roots are real and equal

(viii)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = 4(a + b)² – 4 × 2 × (a² + b²)

⇒ D = -4(a² + b²) + 2ab = -(a – b)² – 3(a² + b²)

Roots are not real

(ix)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real

If D > 0, roots are real and unequal

If D = 0, roots are real and equal

⇒ D = (a + b + c)² – 4a(b + c)

⇒ D = a² + b² + c² – 2ab – 2ac + 2bc

⇒ D = (a – b – c)²

Thus, roots are real and unequal