Quadratic Equations

The given equation is a quadratic equation.

Explanation - It is of degree 2, it is in the form (a ≠ 0, a, b, c are real numbers)

where a = 1, b = - 1, c = 3.

The given equation equation is a quadratic equation.

Explanation - It is of degree 2, it is in the form (a ≠ 0, a, b, c are real numbers)

where a = 2, b = , c =

The given equation is a quadratic equation.

Explanation - It is of degree 2, it is in the form (a ≠ 0, a, b, c are real numbers)

where a = √2, b = 7, c = 5√2.

The given equation is a quadratic equation.

Explanation - It is of degree 2, it is in the form (a ≠ 0, a, b, c are real numbers)

where a = 1/3 , b = 1/5, c = - 2.

The given equation is not a quadratic equation.

Explanation - It is not in the form of because it has an extra term - √x with power 1/2

The given equation is a quadratic equation.

Explanation - Given

On solving the equation it gets reduced to; ; It is of degree 2 and it is in the form (a ≠ 0, a, b, c are real numbers) where a = 1, b = - 3, c = - 6.

The given equation is not a quadratic equation.

Explanation - Given

On getting reduced it becomes, it has degree = 3, it is not in the form

(a ≠ 0, a, b, c are real numbers).

The given equation is not a quadratic equation.

Explanation - Given

On getting reduced it becomes ;

It is not in the form ax² + bx + c = 0 (a ≠ 0, a, b, c are real numbers)

The given equation is a quadratic equation.

Explanation Given

On getting reduced it becomes

=

Now, using

where a = 6, b = 12, c = 16

It is in the form (a ≠ 0, a, b, c are real numbers)

The given (2x + 3)(3x + 2) = 6(x - 1)(x - 2)equation is not a quadratic equation.

Explanation - Given (2x + 3)(3x + 2) = 6(x - 1)(x - 2)

On getting reduced it becomes

It is not in the form (a ≠ 0, a, b, c are real numbers)

The given equation is not a quadratic equation.

Explanation - Given

On getting reduced it becomes -

It is not in the form (a ≠ 0, a, b, c are real numbers)

(i) - 1 is the root of given equation.

_{Explanation - Substituting value - 1 in LHS}

=

= 3 - 2 - 1

= 3 - 3 = 0 = RHS

Value satisfies the equation or LHS = RHS.

(ii) is the root of the given euation

_{Explanation - Substituting value in LHS}

=

=

= 1 - 1 = 0 = RHS

Value satisfies the equation or LHS = RHS.

(iii) is not the root of given equation

_{Explanation - Substituting value in LHS}

=

=

= ≠ 0 ≠ RHS

Value does not satisfy the equation or LHS ≠ RHS.

Given x = 1 is a root of the equation x² + kx + 3 = 0 it means it satisfies the equation.

Substituting x = 1 in equation -

1² + k(1) + 3 = 0

Putting the value of k in the given equation : x² + kx + 3 = 0

This reduced to the quadratic equation x² - 4x + 3 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation

= 1.3

= 3

And either of their sum or difference = b

= - 4

Thus the two terms are - 1 and - 3

Sum = - 1 - 3 = - 4

Product = - 1. - 3 = 3

x(x-1)-3(x-1) = 0

(x-1)(x-3) = 0

x = 1 or x = 3

Thus other root is 3.

Given x = 3/4 or x = - 2 are the roots of the equation

Putting in the equation gives -

;

9a + 12b-96 = 0

3a + 4b-32 = 0 - - - - - - - - - - - - - - - (1)

putting x = - 2 in equation gives

4a-2b-6 = 0

2a-b-3 = 0

2a-3 = b - - - - - - - - - - - (2)

Substituting (2) in (1)

3a + 4(2a-3)-32 = 0

⇒ 11a-44 = 0

⇒ a = 4

⇒ b = 2(4)-3 = 5

Thus for a = 4 or b = 5; or x = - 2 are the roots of the equation

(2x-3) (3x + 1) = 0

2x(3x + 1)-3(3x + 1) = 0 taking common from first two terms and last two terms

(2x-3)(3x + 1) = 0

(2x-3) = 0 or (3x + 1) = 0

x = 3/2 or x = (-1)/3

Roots of equation are 3/2, (-1)/3

x(4x + 5) = 0 (On taking x common)

x = 0 or (4x + 5) = 0

x = (-5)/4

Roots of equation are 0, (-5)/4

x = √81

Roots of equation are 9, - 9

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2; b = 1; c = -6

= 2. - 6

= - 12

And either of their sum or difference = b

= 1

Thus the two terms are 4 and - 3

Difference = 4 - 3 = 1

Product = 4. - 3 = - 12

2x(x + 2)-3(x + 2) = 0

(2x-3)(x + 2) = 0

(2x-3) = 0 or (x + 2) = 0

x = 3/2, x = -2

Roots of equation are 3/2, - 2

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1, b = 6, c = 5

= 1.5 = 5

And either of their sum or difference = b

= 6

Thus the two terms are 1 and 5

Sum = 5 + 1 = 6

Product = 5.1 = 5

x(x + 1) + 5(x + 1) = 0

(x + 1)(x + 5) = 0

(x + 1) = 0 or (x + 5) = 0

x = -1, x = -5

Roots of equation are - 1, - 5

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9; b = -3; c = -2

= 9. - 2 = - 18

And either of their sum or difference = b

= - 3

Thus the two terms are - 6 and 3

Sum = - 6 + 3 = - 3

Product = - 6.3 = - 18

3x(3x-2) + 1(3x-2) = 0

(3x + 1)(3x-2) = 0

(3x + 1) = 0 or (3x-2) = 0

x = (-1)/3 or x = 2/3

Roots of equation are (-1)/3, 2/3

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = 12; c = 35

= 1.35 = 35

And either of their sum or difference = b

= 12

Thus the two terms are 7 and 5

Sum = 7 + 5 = 12

Product = 7.5 = 35

x(x + 7) + 5(x + 7) = 0

(x + 5)(x + 7) = 0

(x + 5) = 0 or (x + 7) = 0

x = -5 or x = -7

Roots of equation are - 5, - 7

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = -18; c = 77

= 1.77 = 77

And either of their sum or difference = b

= - 18

Thus the two terms are - 7 and - 11

Sum = - 7 - 11 = - 18

Product = - 7. - 11 = 77

x(x-7)-11(x-7) = 0

(x-7)(x-11) = 0

(x-7) = 0 or (x-11) = 0

x = 7 or x = 11

Roots of equation are 7, 11

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 6; b = 11; c = 3

= 6.3 = 18

And either of their sum or difference = b

= 11

Thus the two terms are 9 and 2

Sum = 9 + 2 = 11

Product = 9.2 = 18

3x(2x + 3) + 1(2x + 3) = 0

(3x + 1)(2x + 3) = 0

(3x + 1) = 0 or (2x + 3) = 0

x = (-1)/3 or x = (-3)/2

Roots of equation are ,

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 6; b = 1; c = -12

= 6. - 12 = - 72

And either of their sum or difference = b

= 1

Thus the two terms are 9 and - 8

Difference = 9 - 8 = 1

Product = 9. - 8 = - 72

3x(2x + 3)-4(2x + 3) = 0

(2x + 3)(3x-4) = 0

(2x + 3) = 0 or (3x-4) = 0

x = (-3)/2 or x = 4/3

Roots of equation are ,

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3; b = -2; c = -1

= 3. - 1 = - 3

And either of their sum or difference = b

= - 2

Thus the two terms are - 3 and 1

Difference = - 3 + 1 = - 2

Product = - 3.1 = - 3

3x(x-1) + 1(x-1) = 0

(x-1)(3x + 1) = 0

(x-1) = 0 or (3x + 1) = 0

x = 1 or x = (-1)/3

Roots of equation are 1, (-1)/3

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4; b = -9 ; c = -100

= 4. - 100 = - 400

And either of their sum or difference = b

= - 9

Thus the two terms are - 25 and 16

Difference = - 25 + 16 = - 9

Product = - 25.16 = - 400

x(4x-25) + 4(4x-25) = 0

(4x-25)(x + 4) = 0

(4x-25) = 0 or (x + 4) = 0

x = 25/4 or x = -4

Roots of equation are 25/4, - 4

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 15 ; b = -1; c = -28

= 15. - 28 = - 420

And either of their sum or difference = b

= - 1

Thus the two terms are - 21 and 20

Difference = - 21 + 20 = - 1

Product = - 21.20 = - 420

3x(5x-7) + 4(5x-7) = 0

(5x-7)(3x + 4) = 0

(5x-7) = 0 or (3x + 4) = 0

x = 7/5 or x = (-4)/3

Roots of equation are 7/5, - 4/3

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3; b = 11 ;c = -4

= 3. - 4 = - 12

And either of their sum or difference = b

= 11

Thus the two terms are 12 and - 1

Difference = 12 - 1 = 11

Product = 12. - 1 = - 12

3x(x + 4)-1(x + 4) = 0

(x + 4)(3x-1) = 0

(x + 4) = 0 or (3x-1) = 0

x = -4 or x = 1/3

Roots of equation are - 4, 1/3

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 48; b = -13 ;c = -1

= 48× - 1 = - 48

And either of their sum or difference = b

= - 13

Thus the two terms are - 16 and 3

Difference = - 16 + 3 = - 13

Product = - 16.3 = - 48

16x(3x-1) + 1(3x-1) = 0

(16x + 1)(3x-1) = 0

(16x + 1) = 0 or (3x-1) = 0

x = (-1)/6 or x = 1/3

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = 2√2; c = -6

= 1. - 6 = - 6

And either of their sum or difference = b

= 2√2

Thus the two terms are 3√2 and -√2

Difference = 3√2-√2 = 2√2

Product = 3√2.-√2 = 3.-2 = -6

using

x(x + 3√2)-√2(x + 3√2) = 0

(x-√2)(x + 3√2) = 0

(x-√2) = 0 or (x + 3√2) = 0

x = √2 or x = -3√2

Roots of equation are √2 or -3√2

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = √3 ;b = 10; c = 7√3

= √3.7√3 = 21

(using 3 = √3×√3)

And either of their sum or difference = b

= 10

Thus, the two terms are 7 and 3

Sum = 7 + 3 = 10

Product = 7.3 = 21

(using )

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = √(3;) b = 11; c = 6√3

= √3.6√3 = 3.6 = 18

(using 3 = √3.√3

And either of their sum or difference = b

= 11

Thus the two terms are 9 and 2

Sum = 9 + 2 = 11

Product = 9.2 = 18

√3 x(x + 3√3) + 2(x + 3√3) = 0

(using 9 = 3.3 = 3√3 √3)

(√3 x + 2)(x + 3√3) = 0

(√3 x + 2)(x + 3√3) = 0

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3√7 ;b = 4 ;c = -√7

= 3√7.-√7 = 3.-7 = -21

(using 7 = √7.√7)

And either of their sum or difference = b

= 4

Thus the two terms are 7 and - 3

Difference = 7 - 3 = 4

Product = 7× - 3 = - 21

( using 7 = √7.√7)

(√7 x-1)(3x + √7) = 0

(√7 x-1) = 0 or (3x + √7) = 0

x = 1/√7 or x = (-7)/√3

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation

= √7.-13√7 = -13.7 = -91

And either of their sum or difference = b

= - 6

Thus the two terms are 7 and - 13

Difference = - 13 + 7 = - 6

Product = 7. - 13 = - 91

x(√7 x-13) + √7 (√7 x-13) = 0

(x + √7)(√7 x-13) = 0

(x + √7) = 0 or (√7 x-13) = 0

x = -√7 or x = 13/√7

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4√6 ;b = -13 ;c = -2√6

= 4√6.-2√6 = -48

And either of their sum or difference = b

= - 13

Thus the two terms are - 16 and 3

Difference = - 16 + 3 = - 13

Product = - 16.3 = - 48

(On using √6 = √3 √2 and 16 = 4.2.√2 √2 )

⇒ (4√2 x + √3)(√3 x-2√2) = 0

⇒ (4√2 x + √3) = 0 or (√3 x-2√2) = 0

x = (-√3)/(4√2 ) or x = (2√2)/√3

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3; b = -2√(6;) c = 2

= 3.2 = 6

And either of their sum or difference = b

= -2√6

Thus the two terms are -√6 and -√6

Sum = -√6 -√6 = -2√6

Product = -√6. -√6 = -6 6 = √6. √6

(On using 3 = √3.√3 and √6 = √3 √2)

Equation has repeated roots

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = √3 b = -2√2 c = -2√3

= √3.-2√3 = -2.3 = -6

And either of their sum or difference = b

= -2√2

Thus the two terms are -3√2 and √2

Difference = -3√2 + √2 = -2√2

Product = -3√2 × √2 = -3.2 = -6

(On using3√2 = √3 √3 √2 = √3.√6)

(∵2√3 = √2 √2 √3 = √2.√6)

Roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 ;b = -3√5 ;c = 10

= 1.10 = 10

And either of their sum or difference = b

= -3√5

Thus the two terms are -2√5 and -√5

Sum = -2√5-√5 = -3√5

Product = -2√5. -√5 = 2.5 = 10 using 5 = √5. √5

(On using: 10 = 2.5 = 2.√5 √5)

x(x-2√5)-√5 (x-2√5) = 0

(x-√5)(x-2√5) = 0

(x-√5) = 0 or (x-2√5) = 0

x = √5 or x = 2√5

Hence the roots of equation are √5 or 2√5

On taking x common from first two terms and - 1 from last two

x(x-√3)-1(x-√3) = 0

(x-√3)(x-1) = 0

(x-√3) = 0 or (x-1) = 0

x = √3 or x = 1

Roots of equation are √3 or 1

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = 3√3 ;c = -30

= 1. - 30 = - 30

And either of their sum or difference = b

= 3√3

Thus, the two terms are

Difference =

Product =

Hence the roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation

=

And either of their sum or difference = b

= 7

Thus the two terms are 5 and 2

Sum = 5 + 2 = 7

Product = 5.2 = 10

Hence the roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 5; b = 13; c = 8

= 5.8 = 40

And either of their sum or difference = b

= 13

Thus the two terms are 5 and 8

Sum = 5 + 8 = 13

Product = 5.8 = 40

Hence the roots of equation are

On taking x common from first two terms and - 1 from last two

Hence the roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9; b = 6; c = 1

= 9.1 = 9

And either of their sum or difference = b

= 6

Thus the two terms are 3 and 3

Sum = 3 + 3 = 6

Product = 3.3 = 9

Hence the equation has repeated roots

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 100 ;b = - 20 ;c = 1

= 100.1 = 100

And either of their sum or difference = b

= - 20

Thus the two terms are - 10 and - 10

Sum = - 10 - 10 = - 20

Product = - 10. - 10 = 100

10x(10x-1)-1(10x-1) = 0

(10x-1)(10x-1) = 0

(10x-1) = 0 or (10x-1) = 0

Roots of equation are repeated

(taking LCM)

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 16; b = - 8 ;c = 1

= 16.1 = 16

And either of their sum or difference = b

= - 8

Thus the two terms are - 4 and - 4

Sum = - 4 - 4 = - 8

Product = - 4. - 4 = 16

4x(4x-1)-1(4x-1) = 0

(4x-1)(4x-1) = 0

(4x-1) = 0 or (4x-1) = 0

The equation has repeated roots

taking LCM

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 10 ;b = - 3; c = - 1

= 10. - 1 = - 10

And either of their sum or difference = b

= - 3

Thus the two terms are - 5 and 2

Difference = - 5 + 2 = - 3

Product = - 5.2 = - 10

5x(2x-1) + 1(2x-1) = 0

(5x + 1)(2x-1) = 0

(5x + 1) = 0 or (2x-1) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2; b = - 5; c = 2

= 2.2 = 4

And either of their sum or difference = b

= - 5

Thus the two terms are - 4 and - 1

Difference = - 4 - 1 = - 5

Product = - 4. - 1 = 4

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

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

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

Hence the roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2; b = a; c = - a²

= - 2.a² = - 2 a²

And either of their sum or difference = b

= a

Thus the two terms are 2a and - a

Difference = 2a - a = a

Product = 2a. - a = - 2a²

2x (x + a)-a (x + a) = 0

(2x-a) (x + a) = 0

(2x-a) = 0 or (x + a) = 0

Hence the roots of equation are

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4 b = 4b c = - (a² - b²)

= 4. - (a² - b²)

= - 4a² + 4b²

And either of their sum or difference = b

= 4b

Thus the two terms are 2(a + b) and - 2(a - b)

Difference = 2a + 2b - 2a + 2b = 4b

Product = 2(a + b). - 2(a - b) = - 4(a² - b²)

using

⇒ 2x[2x + (a + b)]-(a-b) [2x + (a + b)] = 0

⇒ [2x + (a + b)] [2x-(a-b)] = 0

⇒ [2x + (a + b)] = 0 or [2x-(a-b)] = 0

Hence the roots of equation are

4x² - 4a²x + (a⁴ - b⁴) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4 ;b = - 4a² ; c = (a⁴ - b⁴)

= 4. (a⁴ - b⁴)

= 4a⁴ - 4b⁴

And either of their sum or difference = b

= - 4a²

Thus the two terms are - 2(a² + b²) and - 2(a² - b²)

Difference = - 2(a² + b²) - 2(a² - b²)

= - 2a² - 2b² - 2a² + 2b²

= - 4a²

Product = - 2(a² + b²). - 2(a² - b²)

= 4(a² + b²)(a² - b²)

= 4. (a⁴ - b⁴)

(∵ using a² - b² = (a + b) (a - b))

⇒ 4x² - 4a²x + (a⁴ - b⁴) = 0

⇒ 4x² - 4a²x + ((a²)² – (b²)²) = 0

(∵ using a² - b² = (a + b) (a - b))

⇒ 4x² - 2(a² + b²) x - 2(a² - b²) x + (a² + b²) (a² - b²) = 0

⇒ 2x [2x - (a² + b²)] - (a² - b²) [2x - (a² + b²)] = 0

⇒ [2x - (a² + b²)] [2x - (a² - b²)] = 0

⇒ [2x - (a² + b²)] = 0 or [2x - (a² - b²)] = 0

Hence the roots of given equation are

x² + 5x - (a² + a - 6) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = 5; c = - (a² + a - 6)

= 1. - (a² + a - 6)

= - (a² + a - 6)

And either of their sum or difference = b

= 5

Thus the two terms are (a + 3) and - (a - 2)

Difference = a + 3 –a + 2

= 5

Product = (a + 3). - (a - 2)

= - [(a + 3)(a - 2)]

= - (a² + a - 6)

x² + 5x - (a² + a - 6) = 0

⇒ x ² + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0

⇒ x[x + (a + 3)] - (a - 2) [x + (a + 3)] = 0

⇒ [x + (a + 3)] [x - (a - 2)] = 0

⇒ [x + (a + 3)] = 0 or [x - (a - 2)] = 0

⇒ x = - (a + 3) or x = (a - 2)

Hence the roots of given equation are - (a + 3) or (a - 2)

x² - 2ax - (4b² - a²) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = - 2a c = - (4b² - a²)

= 1. - (4b² - a²)

= - (4b² - a²)

And either of their sum or difference = b

= - 2a

Thus the two terms are (2b - a) and - (2b + a)

Difference = 2b - a - 2b - a

= - 2a

Product = (2b - a) - (2b + a)

(∵ using a² - b² = (a + b) (a - b))

= - (4b² - a²)

x² - 2ax - (4b² - a²) = 0

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

⇒ x[x + (2b - a)] - (2b + a)[x + (2b - a)] = 0

⇒ [x + (2b - a)] [x - (2b + a)] = 0

⇒ [x + (2b - a)] = 0 or [x - (2b + a)] = 0

⇒ x = (a - 2b) or x = (a + 2b)

Hence the roots of given equation are (a - 2b) or x = (a + 2b)

x² - (2b - 1)x + (b² - b - 20) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = - (2b - 1); c = b² - b - 20

= 1(b² - b - 20)

= (b² - b - 20)

And either of their sum or difference = b

= - (2b - 1)

Thus the two terms are - (b - 5) and - (b + 4)

Sum = - (b - 5) - (b + 4)

= - b + 5 - b - 4

= - 2b + 1

= - (2b - 1)

Product = - (b - 5) - (b + 4)

= (b - 5) (b + 4)

= b² - b - 20

x² - (2b - 1)x + (b² - b - 20) = 0

⇒ x² - (b - 5)x - (b + 4)x + (b - 5)(b + 4) = 0

⇒ x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0

⇒ [x - (b - 5)] [x - (b + 4)] = 0

⇒ [x - (b - 5)] = 0 or [x - (b + 4)] = 0

⇒ x = (b - 5) or x = (b + 4)

Hence the roots of equation are (b - 5) or (b + 4)

x² + 6x - (a² + 2a - 8) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1; b = 6 ;c = - (a² + 2a - 8)

= 1. - (a² + 2a - 8)

= - (a² + 2a - 8)

And either of their sum or difference = b

= 6

Thus the two terms are (a + 4) and - (a - 2)

Difference = a + 4 - a + 2

= 6

Product = (a + 4) - (a - 2)

= - (a² + 2a - 8)

⇒ x² + 6x - (a² + 2a - 8) = 0

⇒ x² + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0

⇒ x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0

⇒ [x + (a + 4)] [x - (a - 2)] = 0

⇒ [x + (a + 4)] = 0 or [x - (a - 2)] = 0

x = - (a + 4) or x = (a - 2)

Hence the roots of equation are - (a + 4) or (a - 2)

abx² + (b² - ac)x - bc = 0

abx² + (b² - ac)x - bc = 0

abx² + b²x - acx - bc = 0

bx (ax + b) - c (ax + b) = 0 taking bx common from first two terms and –c from last two

(ax + b) (bx - c) = 0

(ax + b) = 0 or (bx - c) = 0

Hence the roots of equation are

x² - 4ax - b² + 4a² = 0

x² - 4ax – ((b)² – (2a)²) = 0

{using a² - b² = (a + b)(a - b)}

x² - (b + 2a)x + (b - 2a)x - (b + 2a)(b - 2a) = 0

⇒ x [x - (b + 2a)] + (b - 2a) [x - (b + 2a)] = 0

⇒ [x - (b + 2a)] [x + (b - 2a)] = 0

⇒ [x - (b + 2a)] = 0 or [x + (b - 2a)] = 0

⇒ x = (b + 2a) or x = - (b - 2a)

⇒ x = (2a + b) or x = (2a - b)

Hence the roots of equation are (2a + b) or (2a - b)

4x² - 2a²x - 2 b²x + a²b² = 0

2x (2x - a²) - b²(2x - a²) = 0

(On taking 2x common from first two terms and –b² from last two)

⇒ (2x - a²) (2x - b²) = 0

⇒ (2x - a²) = 0 or (2x - b²) = 0

⇒

Hence the roots of equation are

12abx² - (9a² - 8b²)x - 6ab = 0

12abx² - 9a² x + 8b²x - 6ab = 0

3ax(4bx - 3a) + 2b(4bx - 3a) = 0 taking 3ax common from first two terms and 2b from last two

(4bx - 3a) (3ax + 2b) = 0

(4bx - 3a) = 0 or (3ax + 2b) = 0

Hence the roots of equation are

a²b²x² + b²x - a²x - 1 = 0

b²x(a²x + 1) - 1(a²x + 1) = 0 taking b²x common from first two terms and - 1 from last two

(a²x + 1) (b²x - 1) = 0

(a²x + 1) = 0 or (b²x - 1) = 0

Hence the roots of equation are

9x² - 9(a + b)x + (2 a² + 5ab + 2b²) = 0

Using the splitting middle term - the middle term of the general equation Ax² + Bx + C is divided in two such values that:

Product = AC

For the given equation A = 9, B = - 9(a + b), C = 2a² + 5ab + 2b²

= 9(2a² + 5ab + 2b²)
= 9(2a² + 4ab + ab + 2b²)
= 9[2a(a + 2b) + b(a + 2b)]
= 9(a + 2b)(2a + b)
= 3(a + 2b)3(2a + b)

Also,
3(a + 2b) + 3(2a + b) = 9(a + b)

Therefore,

9x² - 9 (a + b) x + (2 a² + 5ab + 2b²) = 0

9x² - 3(2a + b)x - 3(a + 2b)x + (a + 2b) (2a + b) = 0

3x[3x - (2a + b)] - (a + 2b)[3x - (2a + b)] = 0

[3x - (2a + b)] [3x - (a + 2b)] = 0

[3x - (a + 2b)] = 0 or [3x - (2a + b)] = 0

Hence the roots of equation are

taking LCM

x² + x = x + 16 cross multiplying

x² - 16 = 0

x² - (4)² = 0 using a² - b² = (a + b)(a - b)

(x + 4) (x - 4) = 0

(x + 4) = 0 or (x - 4) = 0

x = 4 or x = - 4

Hence the roots of equation are 4, - 4.

taking LCM

x + 4 = 2x² + 3x cross multiplying

2x² + 2x - 4 = 0 taking 2 common

x² + x - 2 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 1 c = - 2

= 1. - 2 = - 2

And either of their sum or difference = b

= 1

Thus the two terms are 2 and - 1

Difference = 2 - 1 = 1

Product = 2. - 1 = - 2

x² + x - 2 = 0

x² + 2x - x - 2 = 0

x(x + 2) - (x + 2) = 0

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

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

x = - 2 or x = 1

Hence the roots of equation are - 2 or 1.

taking LCM

3x² + 2x - 1 = 14x - 10 cross multiplying

3x² - 12x + 9 = 0 taking 3 common

x² - 4x + 3 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = - 4 c = 3

= 1.3 = 3

And either of their sum or difference = b

= - 4

Thus the two terms are - 3 and - 1

Sum = - 3 - 1 = - 4

Product = - 3. - 1 = 3

x² - 4x + 3 = 0

x² - 3x - x + 3 = 0

x(x - 3) - 1(x - 3) = 0

(x - 3) (x - 1) = 0

(x - 3) = 0 or (x - 1) = 0

x = 3 or x = 1

Hence the roots of equation are 3 or1.

taking LCM

x² + 4x - 5 = 7 cross multiplying

x² + 4x - 12 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 4 c = - 12

= 1. - 12 = - 12

And either of their sum or difference = b

= 4

Thus the two terms are 6 and - 2

Difference = 6 - 2 = 4

Product = 6. - 2 = - 12

x² + 4x - 12 = 0

x² + 6x - 2x - 12 = 0

x(x + 6) - 2(x + 6) = 0

(x + 6)(x - 2) = 0

(x + 6) = 0 or (x - 2) = 0

x = - 6 or x = 2

Hence the roots of equation are - 6 or 2.

taking LCM

4x² + 4ax + 2bx = - 2ab cross multiplying

4x² + 4ax + 2bx + 2ab = 0

4x(x + a) + 2b(x + a) = 0 taking 4x common from first two terms and 2b from last two

(x + a) (4x + 2b) = 0

(x + a) = 0 or (4x + 2b) = 0

Hence the roots of equation are

taking LCM

8x² + 8 = 17x² - 34x cross multiplying

- 9x² + 34x + 8 = 0

9x² - 34x - 8 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9 b = - 34 c = - 8

= 9. - 8 = - 72

And either of their sum or difference = b

= - 34

Thus the two terms are - 36 and 2

Difference = - 36 + 2 = - 34

Product = - 36.2 = - 72

9x² - 34x - 8 = 0

9x² - 36x + 2x - 8 = 0

9x(x - 4) + 2 (x - 4) = 0

(9x + 2)(x - 4) = 0

Hence the roots of equation are

Given:

taking LCM

using (a - b)² = a² + b² - 2ab

cross multiplying

18x² - 48x + 130 = 105x - 140

18x² - 153x + 270 = 0 taking 9 common

2x² - 17x + 30 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = - 17 c = 30

= 2.30 = 60

And either of their sum or difference = b

= - 17

Thus the two terms are - 12 and - 5

Sum = - 12 - 5 = - 17

Product = - 12. - 5 = 60

2x² - 17x + 30 = 0

2x² - 12x - 5x + 30 = 0

2x(x - 6) - 5(x - 6) = 0

(x - 6) (2x - 5) = 0

(x - 6) = 0 or (2x - 5) = 0

Hence the roots of equation are

Given:

taking LCM

using (a - b)² = a² + b² - 2ab

8x² - 8x + 4 = 17x² - 17x cross multiplying

9x² - 9x - 4 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9 b = - 9 c = - 4

= 9. - 4 = - 36

And either of their sum or difference = b

= - 9

Thus the two terms are - 12 and 3

Sum = - 12 + 3 = - 9

Product = - 12.3 = - 36

9x² - 9x - 4 = 0

9x² - 12x + 3x - 4 = 0

3x(3x - 4) + 1(3x - 4) = 0

(3x - 4) (3x + 1) = 0

(3x - 4) = 0 or (3x + 1) = 0

Hence the roots of equation are

Given: taking LCM

30x² + 30x + 15 = 34x² + 34x cross multiplying

4x² + 4x - 15 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4 b = 4 c = - 15

= 4. - 15 = - 60

And either of their sum or difference = b

= 4

Thus the two terms are 10 and - 6

Difference = 10 - 6 = 4

Product = 10. - 6 = - 60

4x² + 4x - 15 = 0

4x² + 10x - 6x - 15 = 0

2x(2x + 5) - 3(2x + 5) = 0

(2x + 5) (2x - 3) = 0

(2x + 5) = 0 or (2x - 3) = 0

Hence the roots of equation are

Given:

taking LCM

3x² - 33x + 87 = 5x² - 60x + 175 cross multiplying

2x² - 27x + 88 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = - 27 c = 88

= 2.88 = 176

And either of their sum or difference = b

= - 27

Thus the two terms are - 16 and - 11

Sum = - 16 - 11 = - 27

Product = - 16. - 11 = 176

2x² - 27x + 88 = 0

2x² - 16x - 11x + 88 = 0

2x(x - 8) - 11(x - 8) = 0

(x - 8) (2x - 11) = 0

(x - 8) = 0 or (2x - 11) = 0

Hence the roots of equation are

Given:

taking LCM

cross multiplying

3x² - 15x + 15 = 5x² - 30x + 40

2x² - 15x + 25 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = - 15 c = 25

= 2.25 = 50

And either of their sum or difference = b

= - 15

Thus the two terms are - 10 and - 5

Sum = - 10 - 5 = - 15

Product = - 10. - 5 = 50

2x² - 15x + 25 = 0

2x² - 10x - 5x + 25 = 0

2x(x - 5) - 5(x - 5) = 0

(x - 5)(2x - 5) = 0

(x - 5) = 0 or (2x - 5) = 0

Hence the roots of equation are

Given:

taking LCM

cross multiplying

3x² - 5x = 6x² - 18x + 12

3x² - 13x + 12 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3 b = - 13 c = 12

= 3.12 = 36

And either of their sum or difference = b

= - 13

Thus the two terms are - 9 and - 4

Sum = - 9 - 4 = - 13

Product = - 9. - 4 = 36

3x² - 13x + 12 = 0

3 x² - 9x - 4x + 12 = 0

3x (x - 3) - 4(x - 3) = 0

(x - 3) (3x - 4) = 0

Hence the roots of equation are

Given:

taking LCM

(3x + 4)(x + 4) = 5x² + 15x + 10 cross multiplying

3x² + 16x + 16 = 5x² + 15x + 10

2x² - x - 6 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = - 1 c = - 6

= 2. - 6 = - 12

And either of their sum or difference = b

= - 1

Thus the two terms are - 4 and 3

Difference = - 4 + 3 = - 1

Product = - 4.3 = 12

2x² - x - 6 = 0

2x² - 4x + 3x - 6 = 0

2x(x - 2) + 3(x - 2) = 0

(x - 2) (2x + 3) = 0

(x - 2) = 0 or (2x + 3) = 0

Hence the roots of equation are

Given:

taking LCM

using (a + b)² = a² + b² + 2ab; (a - b)² = a² + b² - 2ab

19x² - 42x - 15 = 30x² + 35x - 15 cross multiplying

11 x² + 77x = 0

11x(x + 7) = 0 taking 11x common

11x = 0 or (x + 7) = 0

x = 0 or x = - 7

Hence the roots of equation are 0, - 7

Given:

taking LCM; using (a + b)² = a² + b² + 2ab

47x² + 162x - 33 = 385x² - 176x - 33 cross multiplying

338x² - 338x = 0

338x(x - 1) = 0 taking 338x common

338x = 0 or (x - 1) = 0

x = 1 or x = 0

Hence the roots of equation are 1, 0

Given:

taking LCM; using (a + b)² = a² + b² + 2ab

- 24x² - 64x - 1 = 3(8x² - 2x - 3) cross multiplying

- 24x² - 64x - 1 = 24x² - 6x - 9

48 x² + 58x - 8 = 0 taking 2 common

24 x² + 29x - 4 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 24 b = 29 c = - 4

= 24. - 4 = - 96

And either of their sum or difference = b

= 29

Thus the two terms are 32 and - 3

Difference = 32 - 3 = 29

Product = 32. - 3 = - 96

24 x² + 29x - 4 = 0

24 x² + 32x - 3x - 4 = 0

8x(3x + 4) - 1(3x + 4) = 0

(3x + 4)(8x - 1) = 0

(3x + 4) = 0 or (8x - 1) = 0

Hence the roots of equation are

Given: - - - - - - - - (1)

Let

y² - 5y + 6 = 0 substituting value for y in (1)

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = - 5 c = 6

= 1.6 = 6

And either of their sum or difference = b

= - 5

Thus the two terms are - 3 and - 2

Difference = - 3 - 2 = - 5

Product = - 3. - 2 = 6

y² - 5y + 6 = 0

y² - 3y - 2y + 6 = 0

y(y - 3) - 2 (y - 3) = 0

(y - 3)(y - 2) = 0

(y - 3) = 0 or (y - 2) = 0

y = 3 or y = 2

Case I: if y = 3

x = 3x + 3

2x + 3 = 0

x = - 3/2

Case II: if y = 2

x = 2x + 2

x = - 2

Hence the roots of equation are

Given:

taking - 1 with both terms

taking LCM

taking common (a - x - b)

taking LCM

(a - x + b)[2x - (a + b)] = 0

(a - x + b) = 0 or [2x - (a + b)] = 0

Hence the roots of equation are

Given:

taking LCM

taking common (a + b - abx)

taking LCM

(a + b - abx)[(a + b)x - 2] = 0

(a + b - abx) = 0 or [(a + b)x - 2] = 0

Hence the roots of equation are

Given: 3^{(x + 2)} + 3 ^{- x} = 10

- - - - - - - - (1)

Let 3^x = y - - - - - - - - - - (2)

substituting for y in (1)

9y² - 10y + 1 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9 b = - 10 c = 1

= 9.1 = 9

And either of their sum or difference = b

= - 10

Thus the two terms are - 9 and - 1

Sum = - 9 - 1 = - 10

Product = - 9. - 1 = 9

9y² - 9y - 1y + 1 = 0

9y(y - 1) - 1(y - 1) = 0

(y - 1) (9y - 1) = 0

(y - 1) = 0 or (9y - 1) = 0

y = 1 or y = 1/9

3^x = 1 or 3^x = 1/9

On putting value of y in equation (2)

3^x = 3⁰ or 3^x = 3 ^{- 2}

x = 0 or x = - 2

Hence the roots of equation are 0, - 2

Given: 4^{(x + 1)} + 4^{(1 - x)} = 10

- - - - - - - (1)

Let 4^x = y - - - - - - - - - - (2)

substituting for y in (1)

4y² - 10y + 4 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4 b = - 10 c = 4

= 4.4 = 16

And either of their sum or difference = b

= - 10

Thus the two terms are - 8 and - 2

Sum = - 8 - 2 = - 10

Product = - 8. - 2 = 16

4y² - 10y + 4 = 0

4y² - 8y - 2y + 4 = 0

4y(y - 2) - 2(y - 2) = 0

(y - 2) (4y - 2) = 0

(y - 2) = 0 or (4y - 2) = 0

y = 2 or y = 1/2

substituting the value of y in (2)

4^x = 2 or 4^x = 2 ^{- 1}

2^2x = 2¹ or 2^2x = 2 ^{- 1}

2x = 1 or 2x = - 1

Hence the roots of equation are

Given: 2^2x - 3.2^{(x + 2)} + 32 = 0

(2^x)² - 3.2^x.2² + 32 = 0 - - - - (1)

Let 2^x = y - - - - - - (2)

substituting for y in (1)

y² - 12y + 32 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = - 12 c = 32

= 1.32 = 32

And either of their sum or difference = b

= - 12

Thus the two terms are - 8 and - 4

Sum = - 8 - 4 = - 12

Product = - 8. - 4 = 32

y² - 8y - 4y + 32 = 0

y(y - 8) - 4(y - 8) = 0

(y - 8) (y - 4) = 0

(y - 8) = 0 or (y - 4) = 0

y = 8 or y = 4

2^x = 8 or 2^x = 4

substituting the value of y in (2)

2^x = 2³ or 2^x = 2²

x = 2 or x = 3

Hence the roots of equation are 2, 3

Given: x² - 6x + 3 = 0

x² - 6x = - 3

x² - 2.x.3 + 3² = - 3 + 3² (adding 3² on both sides)

(x - 3)² = - 3 + 9 = 6 using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given: x² - 4x + 1 = 0

x² - 4x = - 1

x² - 2.x.2 + 2² = - 1 + 2² (adding 2² on both sides)

(x - 2)² = - 1 + 4 = 3 using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given: x² + 8x - 2 = 0

x² + 8x = 2

x² + 2.x.4 + 4² = 2 + 4² (adding 4² on both sides)

(x + 4)² = 2 + 16 = 18 using a² + 2ab + b² = (a + b)²

(taking square root on both sides)

Hence the roots of equation are

Given:

(adding on both sides)

using a² + 2ab + b² = (a + b)²

Hence the equation has repeated roots

Given: 2x² + 5x - 3 = 0

4x² + 10x - 6 = 0 (multiplying both sides by 2)

4x² + 10x = 6

(adding on both sides)

using a² + 2ab + b² = (a + b)²

(taking square root on both sides)

Hence the roots of equation are

Given: 3x² - x - 2 = 0

9x² - 3x - 6 = 0 (multiplying both sides by 3)

9x² - 3x = 6

(adding on both sides)

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given: 8x² - 14x - 15 = 0

16x² - 28x - 30 = 0 (multiplying both sides by 2)

16x² - 28x = 30

(adding on both sides)

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given: 7x² + 3x - 4 = 0

49x² + 21x - 28 = 0 (multiplying both sides by 7)

(adding on both sides)

using a² + 2ab + b² = (a + b)²

(taking square root on both sides)

Hence the roots of equation are

Given: 3x² - 2x - 1 = 0

9x² - 6x = 3 (multiplying both sides by 3)

(adding on both sides)

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

3x - 1 = 2 or 3x - 1 = - 2

3x = 3 or 3x = - 1

Hence the roots of equation are

Given: 5x² - 6x - 2 = 0

25x² - 30x - 10 = 0 (multiplying both sides by 5)

25x² - 30x = 10

(adding on both sides)

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given:

2x² - 5x + 2 = 0

4x² - 10x + 4 = 0

4x² - 10x = - 4 (multiplying both sides by 2)

(adding on both sides)

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

4x² + 4bx = (a² - b²)

(adding on both sides)

using a² + 2ab + b² = (a + b)²

(taking square root on both sides)

2x + b = - a or 2x + b = a

Hence the roots of equation are

Given :

(adding on both sides)

using a² - 2ab + b² = (a - b)²

taking square root on both sides

Hence the roots of equation are

Given:

(multiplying both sides by )

[Adding on both sides]

using a² - 2ab + b² = (a - b)²

(taking square root on both sides)

Hence the roots of equation are

Given:

(multiplying both sides with )

[Adding 5²on both sides]

using a² + 2ab + b² = (a + b)²

(taking square root on both sides)

Hence the roots of equation are

2x² + x + 4 = 0

4x² + 2x + 8 = 0 (multiplying both sides by 2)

4x² + 2x = - 8

[Adding on both sides]

using a² + 2ab + b² = (a + b)²

But cannot be negative for any real value of x

So there is no real value of x satisfying the given equation.

Hence the given equation has no real roots.

Given: 2x² – 7x + 6 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 2, b = – 7, c = 6

Discriminant D = b² – 4ac

= (– 7)² – 4.2.6

= 49 – 48 = 1

Given: 3x² – 2x + 8 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 3, b = – 2, c = 8

Discriminant D = b² – 4ac

= (– 2)² – 4.3.8

= 4 – 96 = – 92

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 25.2 – 32

= 50 – 32 = 18

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 8 + 24 = 32

Given: (x – 1)(2x – 1) = 0

2x² – 3x + 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = , b = , c = – 1

Discriminant D = b² – 4ac

= (– 3)² – 4.2.1

= 9 – 8 = 1

Given: 1 – x = 2x²

2x² + x – 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = , b = 1, c = – 1

Discriminant D = b² – 4ac

= (1)² – 4.2. – 1

= 1 + 8 = 9

Given: x² – 4x – 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = – 4, c = – 1

Discriminant D = b² – 4ac

= (– 4)² – 4.1. – 1

= 16 + 4 = 20 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: x² – 6x + 4 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = – 6, c = 4

Discriminant D = b² – 4ac

= (6)² – 4.1.4

= 36 – 16 = 20 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 2x² + x – 4 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 2, b = 1, c = – 4

Discriminant D = b² – 4ac

= (1)² – 4.2. – 4

= 1 + 32 = 33 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 25x² + 30x + 7 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 25, b = 30, c = 7

Discriminant D = b² – 4ac

= (30)² – 4.25.7

= 900 – 700 = 200 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 16x² = 24x + 1

16x² – 24x – 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 16, b = – 24, c = – 1

Discriminant D = b² – 4ac

= (– 24)² – 4.16. – 1

= 576 + 64 = 640 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 15x² – 28 = x

15x² – x – 28 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 15, b = – 1, c = – 28

Discriminant D = b² – 4ac

= (– 1)² – 4.15. – 28

= 1 + 1680 = 1681 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 8 – 8 = 0

Hence the roots of equation are real.

Roots are given by

Hence these are the repeated roots of the equation

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

= 49 – 40 = 9 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

= 100 + 96 = 196 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 8 + 24 = 32 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 180 + 480 = 588 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 25 + 96 = 121 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

= 24 – 24 = 0

Hence the roots of equation are real and repeated.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 25 – 24 = 10

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: x² + x + 2 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = 1, c = 2

Discriminant D = b² – 4ac

= (1)² – 4.1.2

= 1 – 8 = – 7 < 0

Hence the roots of equation do not exist

Given: 2x² + ax – a² = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 2, B = a, C = – a²

Discriminant D = B² – 4AC

= (a)² – 4.2. – a²

= a² + 8 a² = 9a² ≥0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

Using a² – 2ab + b² = (a – b)²

Thus the roots of given equation are real.

Roots are given by

Hence the roots of equation are

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

=

= 75 – 48 = 27 >

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 3x² – 2x + 2 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 3, b = – 2, c = 2

Discriminant D = b² – 4ac

= (– 2)² – 4.3.2

= 4 – 24 = – 20 < 0

Hence the roots of equation do not exist

Given:

taking LCM

cross multiplying

x² + 1 = 3x

x² – 3x + 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = – 3, c = 1

Discriminant D = b² – 4ac

= (– 3)² – 4.1.1

= 9 – 4 = 5 >

Hence the roots of equation are real.

√D = √5

Roots α and β are given by

Hence the roots of equation are

Given:

taking LCM

3x² – 6x + 2 = 0 cross multiplying

Comparing with standard quadratic equation ax² + bx + c = 0

a = 3, b = – 6, c = 2

Discriminant D = b² – 4ac

= (– 6)² – 4.3.2

= 36 – 24 = 12 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

taking LCM

x² – 3x – 1 = 0 cross multiplying

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = – 3, c = – 1

Discriminant D = b² – 4ac

= (– 3)² – 4.1. – 1

= 9 + 4 = 13 > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given:

taking LCM m²x+ n² = mn – 2mnx

On cross multiplying

m²x+ 2mnx + n² – mn = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = m², b = 2mn, c = n² – mn

Discriminant D = b² – 4ac

= (2mn)² – 4.m². (n² – mn)

= 4m²n² – 4m²n² + 4m³n > 0

Hence the roots of equation are real.

Roots α and β are given by

Hence the roots of equation are

Given: 36x² – 12ax + (a² – b²) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 36, B = – 12a, C = a² – b²

Discriminant D = B² – 4AC

= (– 12a)² – 4.36.(a² – b²)

= 144a² – 144a² + 144 b² = 144 b² > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: x² – 2ax + (a² – b²) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = – 2a, C = a² – b²

Discriminant D = B² – 4AC

= (– 2a)² – 4.1.(a² – b²)

= 4a² – 4a² + 4 b² = 4 b² > 0

Hence the roots of equation are real.

Roots are given by

x = (a + b) or x = (a – b)

Hence the roots of equation are (a + b) or (a – b)

Given: x² – 2ax – (4b² – a²) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = – 2a, C = – (4b² – a²)

Discriminant D = B² – 4AC

= (– 2a)² – 4.1. – (4b² – a²)

= 4a² – 4a² + 16 b² = 16b² > 0

Hence the roots of equation are real.

Roots are given by

x = (a + 2b) or x = (a – 2b)

Hence the roots of equation are (a + 2b) or (a – 2b)

Given: x² + 6x – (a² + b² – 8) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = 6, C = – (a² + b² – 8)

Discriminant D = B² – 4AC

= (6)² – 4.1. – (a² + b² – 8)

= 36+ 4a² + 8a – 32 = 4a² + 8a + 4

= 4(a² + 2a + 1)

= 4(a + 1)² > 0 Using a² + 2ab + b² = (a + b)²

Hence the roots of equation are real.

= 2(a + 1)

Roots are given by

x = (a – 2) or x = – (4 + a)

Hence the roots of equation are (a – 2) or – (4 + a)

Given: x² + 5x – (a² + a – 6) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = 5, C = – (a² + a – 6)

Discriminant D = B² – 4AC

= (5)² – 4.1. – (a² + a – 6)

= 25+ 4a² + 4a – 24 = 4a² + 4a + 1

= (2a + 1)² > 0 Using a² + 2ab + b² = (a + b)²

Hence the roots of equation are real.

= (2a + 1)

Roots are given by

x = (a – 2) or x = – (a + 3)

Hence the roots of equation are (a – 2) or x = – (a + 3)

Given: x² – 4ax – b² + 4a² = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = – 4a, C = – b² + 4a²

Discriminant D = B² – 4AC

= (– 4a)²– 4.1. (– b² + 4a²)

= 16a² + 4b² – 16a² = 4 b² > 0

Hence the roots of equation are real.

Roots are given by

x = (2a – b) or x = (2a + b)

Hence the roots of equation are (2a – b) or (2a + b)

Given: 4x² – 4a²x + (a⁴ – b⁴) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 4, B = – 4a², C = (a⁴ – b⁴)

Discriminant D = B² – 4AC

= (– 4a²)² – 4.4. (a⁴ – b⁴)

= 16a⁴ + 16b⁴ – 16a⁴ = 16 b⁴ > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: 4x² + 4bx – (a² – b²) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 4, B = 4b, C = – (a² – b²)

Discriminant D = B² – 4AC

= (4b)² – 4.4. – (a² – b²)

= 16b² + 16a² – 16b² = 16 a² > 0

Hence the roots of equation are real.

Roots are given by

Hence the roots of equation are

Given: x² – (2b – 1)x + (b² – b – 20) = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 1, B = – (2b – 1), C = (b² – b – 20)

Discriminant D = B² – 4AC

= [ – (2b – 1)²] – 4.1. (b² – b – 20) Using a² – 2ab + b² = (a – b)²

= 4b² – 4b + 1 – 4b² + 4b + 80 = 81 > 0

Hence the roots of equation are real.

Roots are given by

x = (b + 4) or x = (b – 5)

Hence the roots of equation are (b + 4) or (b – 5)

Given: 3a²x² + 8abx + 4b² = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 3a², B = 8ab, C = 4b²

Discriminant D = B² – 4AC

= (8ab)² – 4.3a^2. 4b²

= 64 a²b² – 48a²b² = 16a²b² > 0

Hence the roots of equation are real.

= 4ab

Roots are given by

Hence the roots of equation are

Given: a²b²x² – (4b⁴ – 3a⁴)x – 12a²b² = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = a²b², B = – (4b⁴ – 3a⁴), C = – 12a²b²

Discriminant D = B² – 4AC

= [ – (4b⁴ – 3a⁴)]² – 4a²b². – 12a²b²

= 16b⁸ – 24a⁴b⁴ + 9a⁸ + 48 a⁴b⁴

= 16b⁸ + 24a⁴b⁴ + 9a⁸

= (4b⁴ + 3a⁴)² > 0 Using a² + 2ab + b² = (a + b)²

Hence the roots of equation are real.

=

Roots are given by

Hence the roots of equation are

Given: 12abx² – (9a² – 8b²)x – 6ab = 0

Comparing with standard quadratic equation Ax² + Bx + C = 0

A = 12ab, B = – (9a² – 8b²), C = – 6ab

Discriminant D = B² – 4AC

= [ – (9a² – 8b²)]² – 4.12ab. – 6ab

= 81a⁴ – 144a²b² + 64b⁴ + 288 a²b²

= 81a⁴ + 144a²b² + 64b⁴

= (9a² + 8b²)² > 0 Using a² + 2ab + b² = (a + b)²

Hence the roots of equation are real.

= 9a² + 8b²

Roots are given by

Hence the roots of equation are

Given: 2x² – 8x + 5 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 2, b = – 8, c = 5

Discriminant D = b² – 4ac

= (– 8)² – 4.2.5

= 64 – 40 = 24 > 0

Hence the roots of equation are real and unequal.

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

= 24 – 24 = 0

Hence the roots of equation are real and equal.

Given: 5x² – 4x + 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 5, b = – 4, c = 1

Discriminant D = b² – 4ac

= (– 4)² – 4.5.1

= 16 – 20 = – 4 < 0

Hence the equation has no real roots.

Given: 5x (x – 2) + 6 = 0

5x² – 10x + 6 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 5, b = – 10, c = 6

Discriminant D = b² – 4ac

= (– 10)² – 4.5.6

= 100 – 120 = – 20 < 0

Hence the equation has no real roots.

Given:

Comparing with standard quadratic equation ax² + bx + c = 0

Discriminant D = b² – 4ac

= 240 – 240 = 0

Hence the equation has real and equal roots.

Given: x² – x + 2 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 1, b = – 1, c = 2

Discriminant D = b² – 4ac

= (– 1)² – 4.1.2

= 1 – 8 = – 7 < 0

Hence the equation has no real roots.

Given: 2 (a² + b²)x² + 2(a + b)x + 1 = 0

Comparing with standard quadratic equation ax² + bx + c = 0

a = 2 (a² + b²), b = 2(a + b), c = 1

Discriminant D = b² – 4ac

= [2(a + b)]² – 4. 2 (a² + b²).1

= 4(a² + b² + 2ab) – 8 a² – 8b²

= 4a² + 4b² + 8ab – 8a² – 8b²

= – 4a² – 4b² + 8ab

= – 4(a² + b² – 2ab)

= – 4(a – b)² < 0

Hence the equation has no real roots.

Given equation x² + px – q² = 0