# RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10C Class 10 Maths

 Chapter Name RS Aggarwal Chapter 10 Quadratic Equation Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 10AExercise 10BExercise 10DExercise 10EExercise 10F Related Study NCERT Solutions for Class 10 Maths

### Exercise 10C Solutions

1. Find the roots of the each of the following equations, if they exist, by applying the quadratic formula:

(i) 2x2 – 7x + 6 = 0

(ii) 3x2 – 2x + 8 = 0

(iii) 2x2 - 5√x + 4 = 0

(iv) √3x2 + 2√2x - 2√3 = 0

(v) (x – 1)(2x – 1) = 0

(vi) 1 – x = 2x2

Solution

(i) 2x2 – 7x + 6 = 0

Here,

a = 2,

b = - 7,

c = 6

Discriminant D is given by

D = b2 – 4ac

= (-7)2 – 4 × 2 × 6

= 49 – 48

= 1

(ii) 3x2 – 2x + 8 = 0

Here,

a = 3,

b = -2,

c = 8

Discriminant D is given by:

D = b2 – 4ac

= (-2)2 – 4 × 3 × 8

= 4 – 96

= - 92

(iii) 2x2 - 5√x + 4 = 0

Here,

a = 2,

b = -5√2,

c = 4

Discriminant D is given by:

D = b2 – 4ac

= (-5√2)2 – 4 × 2 × 4

= (25 × 2) – 32

= 50 – 32

= 18

(iv) √3x2 + 2√2x - 2√3 = 0

Here,

a = √3

b = 2√2,

c = -2√3

Discriminant D is given by:

D = b2 – 4ac

= (2√2)2 – 4 × √3 × (-2√3)

= (4 × 2) + (8 × 3)

= 8 + 24

= 32

(v) (x – 1)(2x – 1) = 0

⇒ 2x2 – 3x + 1 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 2, b = - 3 and c = 1

∴ Discriminant, D = b2 – 4ac

= (-3)2 – 4 × 2 × 1

= 9 – 8

= 1

(vi) 1 – x = 2x2

⇒ 2x2 + x – 1 = 0

Here,

a = 2,

b = 1,

c = - 1,

Discriminant D is given by:

D = b2 – 4ac

= 12 – 4 × 2(-1)

= 1 + 8

= 9

2. Find the roots of the each of the following equations, if they exist, by applying the quadratic formula:

(i) x2 – 4x – 1 = 0

Solution

x2 – 4x – 1 = 0

On comparing it with ax2 + bx + c = 0, we get:

a = 1, b = -4 and c =- 1

Discriminant D is given by:

D = (b2 – 4ac)

= (-4)2 – 4 × 1 × (-1)

= 16 + 4

= 20

= 20 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Thus, the roots of the equation are (2 + √5) and (2 - √5).

(ii) x2 – 6x + 4 = 0

Solution

Given:

x2 – 6x + 4 = 0

On comparing it with ax2 + bx + c = 0, we get:

a = 1, b = - 6 and c = 4

Discriminant D is given by:

D = (b2 – 4ac)

= (-6)2 – 4 × 1 × 4

= 36 – 16

= 20 > 0

Hence, the roots of the equation are real.

Root α and β are given by:

Thus, the roots of the equation are (3 + 2√5) and (3 - 2√5).

(iii) 2x2 + x – 4 = 0

Solution

The given equation is 2x2 + x – 4 = 0.

Comparing it with ax2 + bx + c = 0, we get

a = 2, b = 1 and c = - 4

∴ Discriminant, D = b2 – 4ac = (1)2 - 4 × 2 × (-4)

= 1 + 32

= 33 > 0

So, the given equation has real roots.

(iv) 25x2 + 30x + 7 = 0

Solution

Given:

25x2 + 30x + 7 = 0

On comparing it with ax2 + bx + x = 0, we get;

a = 25, b = 30 and c = 7

Discriminant D is given by:

D = (b2 – 4ac)

= (302 – 4 × 25 × 7)

= 900 – 700

= 200

= 200 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Thus, the roots of the equation are (-3 + √2)/5 and (-3 – √2)/5.

(v) 16x2 + 24x + 1

Solution

16x2 + 24x + 1

⇒ 16x2 – 24x – 1 = 0

On comparing it with ax2 + bx + x = 0, we get;

a = 16, b = - 24 and c = - 1

Discriminant D is given by:

D = (b2 – 4ac)

= (24)2 – 4 × 16 × (-1)

= 576 + (64)

= 640 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

(vi) 15x2 – 28 = x

Solution

15x2 – 28 = x

⇒ 15x2 – x – 28 = 0

On comparing it with ax2 + bx + c = 0, we get:

a = 25, b = - 1 and c = - 28

Discriminant D is given by:

D = (b2 – 4ac)

= (1)2 – 4 × 15 × (-28)

= 1 – (-1680)

= 1 + 1680

= 1681

= 1681 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Thus, the roots of the equation are 7/5 and (-4)/3.

(vii) 2x2 - 2√2x + 1 = 0

Solution

The given equation is 2x2 - 2√2x + 1 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 2, b = -2√2 and c = 1

∴ Discriminant, D = b2 – 4ac

= (-2√2)2 – 4 × 2 × 1

= 8 – 8

= 0

So, the given equation has real roots.

Now, √D = 0

∴ α = (-b + √D)/2a = -(-2√2)/(2 × 2) = (2√2)/4 = √2/2

β = (-b - √D)/2a = {-(-2√2) - √0}/(2 × 2) = 2√2/4 = √2/2

Hence, √2/2 is the repeated root of the given equation.

(viii) √2x2 + 7 + 5√2 = 0

Solution

The given equation is √2x2 + 7 + 5√2 = 0.

Comparing it with ax2 + bx + c = 0, we get

a = √2, b = 7 and c = 5√2

∴ Discriminant, D = b2 – 4ac

= (7)2 – 4 × √2 × 5√2 = 49 – 40

= 9 > 0

So, the given equation has real roots.

Now, √D = √9 = 3

∴ α = (-b + √D)/2a = (-7 + 3)/(2 × √2)

= -4/(2√2)

= -√2

β = (-b - √D)/2a

= (-7 – 3)/(2 × √2)

= -10/(2√2)

= -(5√2)/2

Hence, -√2 and –(5√2/2) are the root of the given equation.

(ix) √3x2 + 10x - 8√3 = 0

Solution

Given:

√3x2 + 10x - 8√3 = 0

On comparing it with ax2 + bx + x = 0, we get;

a = √3, b = 10 and c = -8√3

Discriminant D is given by:

D = (b2 – 4ac)

= (10)2 – 4 × √3 × (-8√3)

= 100 + 96

= 196 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Thus, the roots of the equation are (2√3)/3 and –(4√3).

(x) √3x2 - 2√2x - 2√3 = 0.

Solution

The given equation is √3x2 - 2√2x - 2√3 = 0.

Comparing it with ax2 + bx + c = 0, we get

a = √3, b = -2√2 and c = -2√3

∴ Discriminant, D = b2 – 4ac

= (-2√2)2 – 4 × √3 × (-2√3)

= 8 + 24

= 32 > 0

So, the given equation has real roots.

Now, √D = √32 = 4√2

∴ α = (-b + √D)/2a = -(-2√2) + 4√2)/(2 × √3) = 6√2/2√3 = √6

β = (-b - √D)/2a = {-(-2√2) - 4√2)}/(2 × √3) = (-2√2)/(2√3) = -(√6/3)

Hence, √6 and –(√6/3) are the root of the given equation.

(xi) 2x2 + 6√3x – 60 = 0.

Solution

The given equation is 2x2 + 6√3x – 60 = 0.

Comparing it with ax2 + bx + c = 0, we get

a = 2, b = 6√3 and c = - 60

∴ Discriminant, D = b2 – 4ac

= (6√3)2 – 4 × 2 × (-60)

= 180 + 480

= 588 > 0

So, the given equation has real roots.

Hence, 2√3 and -5√3 are the roots of the given equation.

(xii) 4√3x2 + 5x - 2√3 = 0

Solution

The given equation is 4√3x2 + 5x - 2√3 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 4√3, b = 5 and c = -2√3

∴ Discriminant, D = b2 – 4ac

= 52 – 4 × 4√3 × (-2√3)

= 25 + 96

= 121 > 0

So, the given equation has real roots.

∴ α = (-b + √D)/2a = (-5 – 11)/(2 × 4√3) = 6/(8√3) = √3/4.

β = (-b - √D)/2a = (-5 – 11)/(2 × 4√3) = -16/(8√3)

= -(2√3)/3

Hence, √3/4 and –(2√3)/3 are the root of the given equation.

(xiii) 3x2 - 2√6x + 2 = 0

Solution

The given equation is 3x2 - 2√6x + 2 = 0

Comparing it with a2 + bx + c = 0, we get

a = 3, b = -2√6 and c = 2

∴ Discriminant, D = b2 – 4ac

= (-2√6)2 – 4 × 3 × 2

= 24 – 24

= 0

So, the given equation has real roots.

Now, √D = 0

∴ α = (- b + √D)/2a = -(-2√6) + 0/(2 × 3)

= (2√6)/6

= √6/3

β = (-b - √D)/2a = -(-2√6)/(2 × 3)

= (2√6)/6

= √6/3

Hence, √6/3 are the repeated of the given equation.

(xiv) 2√3x2 – 5x + √3 = 0

Solution

The given equation is 2√3x2 – 5x + √3 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 2√3, b = -5 and c = √3

∴ Discriminant, D = b2 – 4ac

= (-5)2 – 4 × 2√3 × √3

= 25 – 25

= 1 > 0

So, the given equation has real roots.

Now, √D = √1 = 1

∴ α = (-b + √D)/2a = -(-5 + 1) + 1/(2 × 2√3) = 6/(4√3) = √3/2

β = (-b - √D)/2a = -(-5) – 1/(2 × 2√3)

= 4/4√3

= √3/3

Hence, √3/2 and √3/3 are the roots of the given equation.

(xv) x2 + x + 2 = 0.

Solution

The given equation is x2 + x + 2 = 0

comparing it with ax2 + bx + c = 0, we get

a = 1, b = 1 and c = 2

∴ Discriminant D = b2 – 4ac = 12 – 4 × 1 × 2

= 1 – 8

= - 7 < 0

Hence, the given equation has no real roots (or real roots does not exist).

(xvi) 2x2 + ax – a2 = 0

Solution

The given equation is 2x2 + ax – a2 = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 2, B = a and C = - a2

∴ Discriminant, D = B2 – 4AC

= a2 – 4 × 2 × -a2

= a2 + 8a2

= 9a2  ≥  0

So, the given equation has real roots.

Now, √D = √9a2 = 3a

∴ α = (-B + √D)/2A = (-a + 3a)/(2 × 2) = 2a/4 = a/2

β = (-B - √D)/2A = (-a – 3a)/(2 × 2) = -4a/4 = - a

Hence, a/2 and – a are the roots of the given equation.

(xvii) (x2 - √3 + 1)x + √3 = 0.

Solution

The given equation is x2 – (√3 + 1)x + √3 = 0.

Comparing it with ax2 + bx + c = 0, we get

a = 1, b = -(√3 + 1) and c = √3

∴ Discriminant,

D = b2 – 4ac = [-(√3 + 1)]2 – 4 × 1 × √3

= 3 + 1 + 2√3 - 4√3

= 3 - 2√3 + 1

= (√3 – 1)2 > 0

So, the given equation has real roots.

Hence, √3 and 1 are the roots of the given equation.

(xviii) 2x2 + 5√3x + 6 = 0

Solution

The given equation is 2x2 + 5√3x + 6 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 2, b = 5√3 and c = 6

∴ Discriminant, D = b2 – 4ac = (5√3)2 – 4 × 2 × 6

= 75 – 48

= 27 > 0

So, the given equation has real roots.

Now, √D = √27 = 3√3

∴ α = (-b + √D)/2A = (-5√3 + 3√3)/(2 × 2) = (-2√3)/4 = - √3/2

β = (-b - √D)/2a = (-5√3 – 3√3)/(2 × 2) = (-8√3)/4 = - 2√3

Hence, -√3/2 and -2√3 are the roots of the given equation.

(xix) 3x2 – 2x + 2 = 0.b

Solution

The given equation is 3x2 – 2x + 2 = 0

Comparing it with ax2 + bx + c = 0, we get

a = 3, b = - 2 and c = 2

∴ Discriminant D = b2 – 4ac

= (2)2 – 4 × 3 × 2

= 4 – 24

= - 20 < 0

Hence, the given equation has no real roots (or real roots does not exist).

(xx) x + 1/x = 3, x ≠ 0

Solution

The given equation is

x + 1/x = 3, x ≠ 0

⇒ (x2 + 1)/x = 3

⇒ (x2 + 1)/x = 3

⇒ x2 + 1 = 3x

⇒ x2 – 3x +1 = 0

This equation is of from of ax2 + bx + c = 0, where, a = 1, b = - 3 and c = 1.

∴ Discriminant, D = b2 – 4ac

= (-3)2 – 4 × 1 × 1

= 9 – 4

= 5 > 0

So, the given equation has real roots.

Now, √D = √5

∴ α = (-b + √D)/2a

= {-(-3) + √5)/(2 × 1) = (3 + √5)/2

β = (-b - √D)/2a = {-(-3) - √5)}/(2 × 1) = (3 - √5)/2

Hence, (3 + √5)/2 and (3 - √5)/2 are -the roots of the given equation.

(xxi) 1/x – 1/(x – 2) = 3, x ≠ 0, 2

Solution

The given equation is

1/x – 1/(x – 2) = 3, x ≠ 0, 2

⇒ (x – 2 – x)/(x(x – 2) = 3

⇒ -2/(x2 – 2x) = 3

⇒ - 2 = 3x2 – 6x

⇒ 3x2 – 6x + 2 = 0

This equation is of form ax2 + bx + c = 0, where = 3, b = -6, and c = 2.

∴ Discriminant, D = b2 – 4ac = (-6)2 - 4 × 3 × 2

= 36 – 24

= 12 > 0

So, the given equation has real roots.

Hence, (3 + √3)/3 and (3 - √3)/3 are the roots of the given equation.

(xxii) x – 1/x = 3, x ≠ 0

Solution

The given equation is

x – 1/x = 3, x ≠ 0

⇒ (x2 – 1)/x = 3

⇒ x2 – 1 = 3x

⇒ x2 – 3x – 1 = 0

This equation is of the form ax2 + bx + c = 0, where a = 1, b = - 3 and c = -1.

∴ Discriminant, D = b2 – 4ac

= (-3)2 – 4 × 1 × (-1)

= 9 + 4

= 13 > 0

So, the given equation has real roots.

(xxiii) m/nx2. n/m = 1 – 2x

Solution

The given equation is

m/nx2.n/m = 1 – 2x

⇒ (m2x2 + n2)/mn = 1 – 2x

⇒ m2x2 + n2 = mn – 2mnx

⇒ m2x2 + 2mnx + n2 – mn = 0

This equation is of the form ax2 + bx + c = 0, where a = m2, b = 2mn and c = n2 – mn

∴ Discriminant,

D = b2 – 4ac = (2mn)2 – 4 × m2 × (n2 – mn)

= 4m2n2 – 4m2n2 + 4m3n2

= 4m3n > 0

So, the given equation has real roots.

(xxiv) 36x2 – 12ax + (a2 – b2) = 0

Solution

The given equation is 36x2 – 12ax + (a2 – b2) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 36, B = - 12a and C = a2 – b2

∴ Discriminant,

D = b2 – 4AC = (-12a)2 – 4 × 36 × (a2 – b2)

= 144a2 – 144a2 + 144b2

= 144b2

So, the given equation has real roots.

∴ α = (-B + √D)/2A

= {-(-12a) + 12b)/(2 × 36) = 12(a + b)/72 = (a + b)/0

β = (-B - √D)/2A = {-(-12a) – 12b)}/(2 × 36) = 12(a - b)/72

= (a – b)/6

Hence, (a + b)/6 and (a – b)/6 are the roots of the given equation.

(xxv) x2 – 2ax + (a2 – b2) = 0

Solution

x2 – 2ax + (a2 – b2) = 0

On comparing it with Ax2 + Bx + C = 0, we get;

A = 1, B = - 2a and C = (a2 – b2)

Discriminant D is given by:

D = b2 – 4AC

= (-2a)2 – 4 × 1 × (a2 – b2)

= 4a2 – 4a2 + 4b2

= 4b2 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Hence, the roots of the equation are (a + b) and (a – b).

(xxvi) x2 – 2ax – (4b2 – a2) = 0

Solution

The given equation is x2 – 2ax – (4b2 – a2) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = - 2a and C = -(4b2 – a2)

∴ Discriminant,

B2 – 4AC = (-2a)2 – 4 × 1 × [-(4b2 – a2)]

= 4a2 + 16b2 – 4a2

= 16b2 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A

= {-(-2a) + 4b)/(2 × 1) = 2(a + 2b)/2 = a + 2b

β = (-B - √D)/2A = {-(-2a) – 4b)}/(2 × 1) = 2(a - b)/2

= a – 2b

Hence, a + 2b and a – 2b are the roots of the given equation.

(xxvii) x2 + 6x – (a2 – 2a + 8) = 0.

Solution

The given equation is x2 + 6x – (a2 + 2a – 8) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = 6 and C = - (a2 + 2a – 8)

∴ Discriminant,

D = B2 – 4AC = 62 – 4 × 1 × [-(a2 + 2a – 8)]

= 36 + 4a2 + 8a – 32

= 4a2 + 8a – 32

= 4a2 + 8a + 4

= 4(a2 + 2a + 1)

= 4(a + 1)2 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A

= (-6 + 2(a + 1))/(2 × 1) = (2a - 4)/2 = a - 2

β = (-B - √D)/2A = (- 6 – 2(a + 1))/(2 × 1) = (-2a – 8)/2

= - a – 4

= - (a + 4)

Hence, (a – 2) and – (a + 4) are the roots of the given equation.

(xxviii) x2 + 5x – (a2 – a - 6) = 0.

Solution

The given equation is x2 + 5x – (a2 + a – 6) = 0.

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = 5 and C = - (a2 + a – 8)

∴ Discriminant,

D = B2 – 4AC

= 52 – 4 × 1 × [-(a2 + a – 6)]

= 25 + 4a2 + 4a - 24

= 4a2 + 4a2 + 4a + 1

= (2a + 1)2 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A

= (-5 + 2a + 1)/(2 × 1) = (2a - 4)/2 = a - 2

β = (-B - √D)/2A = (- 5 – 2(a + 1))/(2 × 1) = (-2a – 6)/2

= - a – 3

= - (a + 3)

Hence, (a – 2) and –(a + 3) are the roots of the given equation.

(xxix) x2 – 4ax – b2 + 4a2 = 0.

Solution

The given equation is x2 – 4ax – b2 + 4a2 = 0.

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = - 4a and C = - b2 + 4a2

∴ Discriminant, D = B2 – 4AC

= (-4a)2 – 4 × 1 × (-b2 + 4a2)

= 16a2 + 4b2 – 16a2

= 4b2 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A = (-(-4a) + 2b)/(2 × 1)

= (4a + 2b)/2 = 2a + b

β = (-B - √D)/2A = (-(-4a) – 2b)/(2 × 1) = (4a – 2b)/2

= 2a – b.

Hence, (2a + b) and (2a – b) are the roots of the given equation.

(xxx) 4x2 – 4a2x + (a4 – b4) = 0

Solution

The given equation is 4x2 – 4a2x + (a4 – b4) = 0.

Comparing it with Ax2 + Bx + C = 0, we get

A = 4, B = - 4a2 and C = a4 – b4

∴ Discriminant, B2 – 4AC

= (-4a2)2 – 4 × 4 × (a2 – b2)

= 16a4 – 16a4 + 16b4

= 16b4 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A = (-(-4a) + 4b2)/(2 × 4)

= 4(a2 + b2)/8

= (a2 + b2)/2

∴ β = (-B - √D)/2A

= -(-4a2) – 4b2/(2 × 4)

= 4(a2 – b2)/8

= (a2 – b2)/2

Hence, 1/2(a2 + b2) and 1/2(a2 – b2) are the roots of the given equation.

(xxxi) 4x2 – 4bx – (a2 – b2) = 0.

Solution

The given equation is 4x2 – 4bx – (a2 – b2) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 4, B = 4b and C = -(a2 – b2)

∴ Discriminant,

D = B2 – 4AC

= (4b)2 – 4 × 4 × [-(a2 – b2)]

= 16b2 + 16a2 – 16b2

= 16a2 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A = (-4b + 4a)/(2 × 4)

= 4(a – b)/8 = (a – b)/2

∴ β = (-B - √D)/2A

= -4(a + b)/8

= -(a + b)/2

Hence, 1/2(a – b) and – 1/2(a + b) are the roots of the given equation.

(xxxii) 33. x2 – (2b – 1)x + (b2 – b – 20) = 0

Solution

The given equation is x2 – (2b – 1)x + (b2 – b – 20) = 0.

Comparing it with Ax2 + Bx + C = 0, we get

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

∴ Discriminant,

D = B2 – 4AC

= [-(2b – 1)]2 – 4 × 1 × (b2 – b – 20)

= 4b2 – 4b + 1 – 4b2 + 4b + 80

= 81 > 0

So, the given equation has real roots

∴ α = (-B + √D)/2A = -[-(2b - 1) + 9]/(2 × 1)

= (2b + 8)/2

= b + 4

β = (-B - √D)/2A

= -[(-2b - 1)] - 9/(2 × 1)

= (2b – 10)/2

= b – 5

Hence, (b + 4) and (b – 5) are the roots of the given equation.

(xxxiii) 3a2x2 + 8abx + 4b2 = 0

Solution

3a2x2 + 8abx + 4b2 = 0

On comparing it with Ax2 + Bx + C = 0, we get:

A = 3a2, B = 8ab and C = 4b2

Discriminant D is given by:

D = (B2 – 4AC)

= (8ab)2 – 4 × 3a2 × 4b2

= 16a2b2 > 0

Hence, the roots of the equation are real.

Roots α and β are given by:

Thus, the roots of the equation are (-2b/3a) and (-2b/a).

(xxxiv) a2b2x2 – (4b4 – 3a4)x – 12a2b2 = 0, a ≠ 0 and b ≠ 0

Solution

The given equation is a2b2x2 – (4b4 – 3a4)x – 12a2b2 = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = a2b2, B = -(4b4 – 3a4) and c = - 12a2b2

∴ Discriminant,

B2 – 4AC

= [-(4b4 – 3a4)]2 - 4 × a2b2 × (-12a2b2)

= 16b8 – 24a4b4 + 9a8 + 48a4b4

= 16b8 + 24a4b4 + 9a8

= (4b4 + 3a4)2 > 0

So, the given equation has real roots.

∴ α = (- B + √D)/2A = -[-(4b4 – 3a4)] + (4b4 + 3a4)]/(2 × a2b2) = 8b4/(2a2b2) = 4b2/a2

β = (- B - √D)/2A = [-[-(4b4 – 3a4)] – (4b4 + 3a4)]/(2 × a2b2)

= -6a4/2a2b2

= -(3a2/b2)

Hence, 4b2/a2 and – (3a2/b) are the roots of the given equation.

(xxxv) 12abx2 – (9a2 – 8b2)x – 6ab = 0, where a ≠ 0 and b ≠ 0

Solution

Given:

12abx2 – (9a2 – 8b2)x – 6ab = 0

On comparing it with Ax2 + Bx + C = 0, we get:

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

Discriminant D is given by:

D = B2 – 4AC

= [-(9a2 – 8b2)]2 – 4 × 12ab × (-6ab)

= 81a4 – 144a2b2 + 64b4 + 288a2b2

= 81a4 + 144a2b2 + 64b4

= (9a2 + 8b2)2 > 0

Hence, the roots of the equation are equal.

Roots α and β are given by:

Thus, the roots of the equation are 3a/4b and (-2b)/3a.