Quadratic Equations

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2
Thus a quadratic equation is of the form,
ax² + bx + c = 0 , where a ≠ 0
(i) (x + 1)² = 2 (x - 3)
We know (a + b)²=a² + b² + 2 ab

⇒ x² + 2x + 1 = 2x - 6

⇒ x² + 2x + 1 - 2x + 6 = 0
⇒ x² + 7 = 0

Equation is in the form of ax²+bx+c = 0, with a ≠ 0

So, we can say that the given equation is a quadratic equation.

(ii) x² – 2x = (-2)(3 – x )

⇒ x² – 2x = -6 + 2x

⇒ x² – 2x +6 - 2x=0
⇒ x² - 4x + 6 = 0

Equation is of the form ax²+bx+c = 0 with a ≠ 0

Hence, the given equation is a quadratic equation.

(iii) (x – 2 )(x + 1) = (x – 1)(x + 3)

⇒ x² – x – 2 = x² + 2x – 3

⇒ 3x – 1 = 0

Equation is not in the form ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(iv) (x – 3 )(2x + 1) = x(x + 5)

⇒ x² – 10x – 3 = 0
Degree of the equation is 2
Equation is in the form ax² + bx + c = 0, with a ≠ 0
Hence, the given equation is a quadratic equation.

(v) (2x – 1)(x – 3 ) = (x + 5)(x – 1 )

⇒ x² – 11x + 8 = 0

Equation is in the form ax²+bx+c = 0, with a ≠ 0

Hence, the given equation is a quadratic equation.

(vi) x² + 3x + 1 = (x – 2 )²

⇒ x² + 3x + 1 = x² + 4 – 4x

⇒ 7x – 3 = 0

It is not of the form ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(vii) (x + 2)³ = 2x (x² - 1)
Apply the formula (a+b)³ = a³ + b³ + 3ab(a + b)

⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x

⇒ x³ – 14x - 6x² – 8 = 0

It is not of the form ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(viii) x³ – 4x² – x + 1 = (x - 2)³

⇒ x³ – 4x² – x +1 = x³ – 8 – 6x² +12x

⇒ 2x² – 13x + 9 = 0

It is of the form ax²+bx+c = 0, with a ≠ 0

Hence, the given equation is a quadratic equation.

(i)The area of a rectangular plot is 528 m².
Let the breadth of the plot be x m.
The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Thus, the length of the plot is (2x + 1) m.
Area of a rectangle = Length × Breadth

∴ 528 = x (2x + 1)

⇒ 2x² + x – 528 = 0 (required quadratic form)

(ii)The product of two consecutive positive integers is 306. We need to find the integers.
Let the consecutive integers be x and x + 1.

It is given that their product is 306.

∴ x (x+1) = 306

⇒ x² + x – 306 = 0 (required quadratic form)

(iii)
Let Rohan’s present age be x.

Given, Rohan's Mother is 26 years older than him
Hence, his mother’s age = x+26

3 years hence,

Rohan’s age = x + 3

Mother’s age = x + 26 + 3
= x + 29

It is given that the product of their ages after 3 years is 360.

∴ (x+3) (x+29) = 360

⇒ x² +32x – 273 = 0 (required quadratic form)

(iv)
Let the speed of train be x km/h.
As speed = distance / time

⇒ Time taken for travel 480 km =

In the second condition,

speed of train = (x – 8) km/h

Given that the train will take 3 hours more to cover the same distance.

Therefore, Time taken for traveling 480 km =

Speed × Time = Distance

= 480

⇒ 480 + 3x - - 24 = 480

⇒ 3x - = 24

⇒ 3x² – 24x - 3840 = 0

⇒ x² – 8x - 1280 = 0 (required quadratic form)

(i) X² – 3x – 10

= x² – 5x +2x – 10

= x(x – 5 ) + 2(x – 5 )

= (x – 5 )(x+2)

Roots of this equation are the values for which (x – 5 )(x+2) = 0

∴ x - 5 = 0 or x + 2 = 0

i.e.,

x = 5 or x = -2

(ii) 2 x² + x – 6

= 2x² + 4x – 3x – 6

= 2x (x+2) – 3(x+2)

= (x+2)(2x – 3 )

Roots of this equation are the values for which (x+2)(2x – 3) = 0

∴ x+ 2= 0 or 2x – 3 = 0

x = −2 or x =

(iii)

Roots of this equation are the values for which = 0

∴ = 0 or = 0

(iv)

Roots of this equation are the values for which (4x – 1 )² = 0

Therefore,

(4x – 1 ) = 0 or (4x – 1) = 0

i.e.,

(v) 100x² – 20x +1

= 100x² – 10x – 10x +1

=10x(10x – 1) – 1 (10x – 1)

=(10x – 1)²

Roots of this equation are the values for which (10x – 1)² = 0

Therefore,

(10x – 1) = 0 or (10x – 1) = 0

i.e.,

(i) Let the number of John’s marbles be x.

Therefore, number of Jivanti’s marble = 45 − x

After losing 5 marbles,

Number of John’s marbles = x − 5

Number of Jivanti’s marbles = 45 − x − 5 = 40 − x

Given that the product of their marbles is 124.

∴ (x – 5) (40 – x) = 124

⇒ x² – 45x +324 = 0

Now, to factorize this equation, we need to take numbers such that their product is 324 and sum is 45

⇒ x² – 36x – 9x +324 = 0

⇒ x(x – 36 ) – 9 (x – 36 ) = 0

⇒ (x – 36) (x – 9) = 0

Either x – 36 = 0 or x − 9 = 0

i.e.,

x = 36 or x = 9

If the number of John’s marbles = 36,

Then, number of Jivanti’s marbles = 45 − 36 = 9

If number of John’s marbles = 9,

Then, number of Jivanti’s marbles = 45 − 9 = 36

(ii) Let the number of toys produced be x.

∴ Cost of production of each toy = Rs (55 − x)

It is given that, total production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x² – 55x + 750 = 0

⇒ x² – 25x – 30x + 750 = 0

⇒ x(x – 25 ) – 30(x – 25 ) = 0

⇒ (x – 25)(x – 30) = 0

Either x – 25 = 0 or x − 30 = 0

i.e., x = 25 or x = 30

Hence, the number of toys will be either 25 or 30.

Let the first number be x and the second number is 27 − x.
[As the sum of both the numbers is 27]

Therefore, their product = x (27 − x)

It is given that the product of these numbers is 182.

x (27 – x) = 182

- x² + 27x - 182 = 0
Changing the signs on both sides we get,
x² - 27x + 182 = 0
Factorizing we get , 13 and 14 are the numbers whose sum is 27 and product is 182

x² – 13x – 14x + 182 = 0

= x(x – 13) – 14 (x – 13)= 0

= (x – 13) (x – 14) = 0

Either x – 13 = 0 or x − 14 = 0

i.e., x = 13 or x = 14

If first number = 13, then

Other number = 27 − 13 = 14

If first number = 14, then

Other number = 27 − 14 = 13

Therefore, the numbers are 13 and 14.

To Find: Consecutive integers sum of whose square is 365
Consecutive integers mean that the difference between the integers is of 1
Let the consecutive positive integers be x and x + 1.

Given that x² + (x + 1)² = 365

⇒ x² + x² + 1 + 2x = 365

⇒ 2x² + 2x – 364 = 0

⇒ x² + x – 182 = 0

⇒ x² + 14x – 13x – 182 = 0

⇒ x (x + 14) – 13 (x + 14) = 0

(x+14)(x – 13) = 0

Either x + 14 = 0 or x − 13 = 0, i.e., x = −14 or x = 13

Since the question ask for positive integers, x can only be 13.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Consider a right angle ΔABC

Let the base of the Δ be x

⇒ Altitude = x - 7

In the right-angled triangle

According to the Pythagoras theorem

(Hypotenuse)² = (base)² + (altitude)²

⇒ 13² = x² + (x - 7)²

⇒ 169 – x² – x² - 49 + 14x = 0

⇒ 2x² - 14x – 120 = 0

⇒ x² – 7x - 60 = 0

⇒ x² – 12x + 5x - 60 = 0

⇒ x( x - 12) + 5( x – 12) = 0

⇒ (x + 5) (x - 12) = 0

⇒ x = - 5 or 12

But base cannot be negative, so x = 12 cm

⇒ altitude = 12 - 7 = 5 cm

Hence 5 cm and 12 cm are the two sides of the given triangle.

To find: Number of articles produced and cost of each article.

Let the number of articles produced be x.

Therefore, cost of production of each article = Rs (2x + 3)

It is given that the total cost of production is Rs 90.

∴ x (2x + 3) = 90

⇒ 2x² + 3x – 90 = 0

⇒ 2x² +15x - 12x-90 =0

⇒ x (2x + 15) - 6(2x + 15) = 0

⇒(2x + 15) (x - 6) = 0

Either 2x + 15 = 0 or x − 6 = 0,

i.e., x = or x = 6

As the number of articles produced can only be a positive integer,

Therefore, x can only be 6.

Hence, the number of articles produced = 6

Cost of each article = 2 × 6 + 3
= 12 + 3
= Rs 15

_i) 2x² – 7x +3 =0

Dividing by coefficient of x², we get

Adding and subtracting the square of

Use the formula (a-b)² = a² + b² - 2ab to get,

So,

When,

When,

Hence the value of x is 3 and

ii) 2x² + x – 4 = 0

Dividing by coefficient of

Adding and subtracting the square of

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

iii)
Divide the whole equation by 4 to get,

Add and subtract the square of half of the coefficient of x,


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






∴ Roots of equation are = x =

iv) 2x² + x + 4

Divide by coefficient of

Adding and subtracting the square of coefficient of x we get,

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

Square of a number cannot be negative, Hence roots of the given equation do not exist.

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

The roots of equation is given by formula



i) 2x² – 7x + 3 = 0

According to the quadratic formula;

∴ The roots are =

=

=

∴ The roots are 3 and

ii) 2x² + x – 4 = 0

According to the quadratic formula;

=

∴ The roots are =

= =

∴ The roots are =

iii) 4x² + 4√3x + 3 = 0

According to the quadratic formula;

=

∴ The roots are =

=

∴ The roots are

iv) 2x² + x + 4 = 0

According to the quadratic formula;

=

∴ The roots are =

=

This is not possible, Hence the roots do not exists.

(i)

Taking L.C.M and then cross multiplying we get

According to the quadratic formula, the roots of the equation: a x² + bx + c = 0 is given by the expression



For the given equation,

So, the roots are,

∴ The roots are =

_(ii)

Factorizing the quadratic equation, such that the product of two numbers is 2 and their sum is 3 ( 2 x 1 and 2 + 1)

x(x - 2) - 1(x - 2) = 0

(x - 2)(x - 1) = 0
i.e x - 2 = 0 or x - 1 = 0
therefore,

x = 1 and x = 2 are the roots of the given equation.

Let the present age of Rehman be x years.

Three years ago, his age was (x − 3) years.

Five years hence, his age will be (x + 5) years.

It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is .

∴

⇒

⇒

⇒ 3(2 x + 2) = (x - 3)(x + 5)

⇒ 6 x + 6 = x² + 2 x – 15

⇒ x² – 4x - 21 = 0

⇒ x² – 7 x + 3 x – 21 = 0

⇒ x (x - 7) +3 (x - 7) = 0

⇒ x = 7, -3

However, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

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

⇒ 27a -a² + 54 - 2a = 210

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

⇒a² – 25a + 156 = 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.

⇒a² – 13a – 12a + 156 = 0

⇒a(a – 13) – 12(a – 13) = 0

⇒(a – 12)(a – 13) = 0

⇒a = 12 or 13

If marks in mathematics is 12
marks in english is 30 - a = 30 - 12 = 18

If marks in mathematics is 13
marks in english is 30 - a = 30 - 13 = 17

Hence

Marks in Mathematics = 12, Marks in English = 18

Or

Marks in Mathematics = 13, Marks in English = 17

To find: length and breadth of the field
Let the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of the rectangle =

It is given that the diagonal of the rectangle is 60 m more than the shorter side.

∴ = x+60

⇒ x² + (x+30)² = (x+60)²

⇒ x² +x² +900+60x = x²+3600+120x

⇒ x² – 60x – 2700 = 0

⇒ x(x – 90 )+30(x – 90 ) = 0

⇒ (x - 90)(x+30) = 0

⇒ x = 90, -30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.

Hence, length of the larger side will be (90 + 30) m = 120 m

Assumption: Let the larger and smaller number be x and y respectively.
Given:

According to the given question difference of squares of two numbers is 180. and the square of smaller number is 8 times square of the larger number.
So,

⇒ x² – y² = 180 eq(i)

⇒ x² – 8x = 180

⇒ x² – 8x – 180 = 0

⇒ x² – 18x +10x – 180 = 0

⇒ x(x – 18 ) +10(x – 18 ) = 0

⇒ (x – 18 )(x+10) = 0

⇒ x = 18, -10

The larger number cannot be negative as it makes the square of smaller number negative, which is not possible.

Therefore the larger number will be 18 only.

x = 18

∴ y² = 8x
= 8×18
= 144

=

∴ smaller number =

Therefore, the numbers are 18 and 12 or 18 and −12.
Note: As we need to solve this problem using one variable, so the variable y is written in terms of variable x.

To find: Speed of the train
Let the speed of the train be x km/hr.

Time taken to cover 360 km = hr, As

Now, given that if the speed would be 5 km/hr more, the same distance would be covered in 1 hour less, i.e.
if speed = x + 5, and

⇒ x² + 5 x – 1800 = 0
Now we have to factorize in such a way that the product of the two numbers is 1800 and the difference is 5
⇒ x² + 45 x – 40 x – 1800 = 0

⇒ x(x + 45) – 40(x+45) = 0

⇒ (x+45)(x – 40 ) = 0

⇒ x = - 45, 40

since, the speed of train can’t be negative,
so, speed will be 40 km/hour.

Let the time taken by the smaller pipe to fill the tank be x hr.

Time taken by the larger pipe = (x − 10) hr

Part of the tank filled by a smaller pipe in 1 hour =

Part of the tank filled by the larger pipe in 1 hour =

It is given that the tank can be filled in hours by both the pipes together.
So 75/8 hours, multiplied by the sum of parts filled with both pipes in one hour equal to complete work i.e 1.

⇒

⇒

⇒

⇒ 75(2x - 10) = 8x² – 80x

⇒ 150x – 750 = 8x² – 80x

⇒ 8x² – 230x + 750 = 0

⇒ 8x² – 200x – 30x +750 = 0

⇒ 8x(x – 25) – 30(x – 25) = 0
⇒ (x – 25 )(8x – 30) = 0

⇒ x = 25,

Time taken by the smaller pipe cannot be = 3.75 hours.

As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.

Let the average speed of passenger train be x km/h.

Average speed of express train = (x + 11) km/h

It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

⇒

⇒ 132×11 = x(x+11)

⇒ x² +11x – 1452 = 0

⇒ x² + 44x – 33x – 1452 = 0

⇒ x(x+44) – 33(x+44) = 0

⇒ (x+44)(x – 33) = 0

⇒ x = - 44, 33

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively and their areas will be x² and y² respectively.

It is given that 4x − 4y = 24 [Difference of perimeter]
or x − y = 6

x = y + 6

Also, x² + y² = 468 [sum of squares is 468]

(6+y)² +y² = 468

36 + y² +12y + y² = 468

2y² +12y – 432 = 0

y² + 6y – 216 = 0

y² +18y – 12y – 216 = 0

y(y+18)(y – 12) = 0

y = - 18 or 12.

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

There are three types of roots that are possible for a quadratic equation:

For a quadratic equation: a x² + b x + c = 0, we know that its roots is given by the formula




(i) Comparing this equation with ax² + bx + c = 0, we obtain,

a = 2, b = −3, c = 5

Discriminant D = b² − 4ac = (− 3)² − 4 (2) (5) = 9 − 40 = −31

As, b² − 4ac < 0,

Therefore, no real root is possible for the given equation.

_(ii) Comparing this equation with ax² + bx + c = 0,

we obtain,

a = 3

b = -4√3

c = 4

Discriminant = 48 − 48 = 0

As, b² − 4ac = 0,

Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be and .

Therefore, the roots are and .

(iii) Comparing this equation with ax²+bx + c = 0,

we obtain,

a = 2, b = −6, c = 3

Discriminant = b² − 4ac = (− 6)² − 4 (2) (3) = 36 − 24 = 12

As, b² − 4ac > 0,

Therefore, distinct real roots exist for this equation as follows.

So, the roots are or .

We know that if an equation ax² + bx + c = 0 has two equal roots,

its discriminant

(b² − 4ac) will be 0.

(i) 2x² + kx + 3 = 0

Comparing equation with ax² + bx + c = 0, we obtain,

a = 2, b = k, c = 3

Discriminate = b² − 4ac = (k)²− 4(2) (3) = k² − 24

For equal roots,

Discriminant = 0

k² − 24 = 0

k² = 24

=

(ii) kx (x − 2) + 6 = 0

or kx²− 2kx + 6 = 0

Comparing this equation with ax² + bx + c = 0, we obtain,

a = k, b = −2 k, c = 6

Discriminant = b² − 4ac = (− 2k)² − 4 (k) (6) = 4k² − 24k

For equal roots, b² − 4ac = 0

= 4k² − 24k = 0

= 4k (k − 6) = 0

Either 4k = 0 or k = 6

= k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x²’ and ‘x’.

Therefore, if this equation has two equal roots, k should be 6 only.

Let the breadth of mango grove be l.

Length of mango grove will be 2l.

Area of mango grove = (2l) (l) = 2l²

2l² = 800

=

⇒

⇒ l² = 400

However, length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

Let the age of one friend be x years.

Age of the other friend will be (20 − x) years.

4 years ago,

age of 1^st friend = (x − 4) years

And, age of 2^nd friend = (20 − x − 4) = (16 − x) years

Given that,

(x − 4) (16 − x) = 48

16x − 64 − x² + 4x = 48

x² − 20x + 112 = 0

Comparing this equation with ax² + bx + c = 0, we obtain

a = 1, b = −20, c = 112

Discriminant = b² − 4ac = (− 20)² − 4 (1) (112) = 400 − 448 = −48

As b² − 4ac < 0,

Therefore, no real root is possible for this equation and hence, this situation is not possible.

Let the length and breadth of the park be l and b.

Perimeter = 2 (l + b) = 80

l + b = 40 Or, b = 40 – l

Area = l × b = l (40 − l)

= 40l − l²= 400 Given

l² − 40l + 400 = 0

Comparing this equation with al² + bl + c = 0, we obtain

a = 1, b = −40, c = 400

Discriminant D = b² − 4ac = (− 40)² −4 (1) (400) = 1600 − 1600 = 0

As b² − 4ac = 0,

Therefore, this equation has equal real roots and hence, this situation is possible.

Root of this equation,

Therefore, length of park, l = 20 m

And breadth of park, b = 40 − l = 40 − 20 = 20 m