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 
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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) 2x^{2} – 7x + 6 = 0
(ii) 3x^{2} – 2x + 8 = 0
(iii) 2x^{2}  5√x + 4 = 0
(iv) √3x^{2} + 2√2x  2√3 = 0
(v) (x – 1)(2x – 1) = 0
(vi) 1 – x = 2x^{2}
Solution
(i) 2x^{2} – 7x + 6 = 0
Here,
a = 2,
b =  7,
c = 6
Discriminant D is given by
D = b^{2} – 4ac
= (7)^{2} – 4 × 2 × 6
= 49 – 48
= 1
(ii) 3x^{2} – 2x + 8 = 0
Here,
a = 3,
b = 2,
c = 8
Discriminant D is given by:
D = b^{2} – 4ac
= (2)^{2} – 4 × 3 × 8
= 4 – 96
=  92
(iii) 2x^{2}  5√x + 4 = 0
Here,
a = 2,
b = 5√2,
c = 4
Discriminant D is given by:
D = b^{2 }– 4ac
= (5√2)^{2} – 4 × 2 × 4
= (25 × 2) – 32
= 50 – 32
= 18
(iv) √3x^{2} + 2√2x  2√3 = 0
Here,
a = √3
b = 2√2,
c = 2√3
Discriminant D is given by:
D = b^{2} – 4ac
= (2√2)^{2} – 4 × √3 × (2√3)
= (4 × 2) + (8 × 3)
= 8 + 24
= 32
(v) (x – 1)(2x – 1) = 0
⇒ 2x^{2} – 3x + 1 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 2, b =  3 and c = 1
∴ Discriminant, D = b^{2} – 4ac
= (3)^{2 }– 4 × 2 × 1
= 9 – 8
= 1
(vi) 1 – x = 2x^{2}
⇒ 2x^{2} + x – 1 = 0
Here,
a = 2,
b = 1,
c =  1,
Discriminant D is given by:
D = b^{2} – 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) x^{2} – 4x – 1 = 0
Solution
x^{2} – 4x – 1 = 0
On comparing it with ax^{2} + bx + c = 0, we get:
a = 1, b = 4 and c = 1
Discriminant D is given by:
D = (b^{2} – 4ac)
= (4)^{2} – 4 × 1 × (1)
= 16 + 4
= 20
= 20 > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
(ii) x^{2} – 6x + 4 = 0
Solution
Given:
x^{2} – 6x + 4 = 0
On comparing it with ax^{2} + bx + c = 0, we get:
a = 1, b =  6 and c = 4
Discriminant D is given by:
D = (b^{2} – 4ac)
= (6)^{2} – 4 × 1 × 4
= 36 – 16
= 20 > 0
Hence, the roots of the equation are real.
Root Î± and Î² are given by:
Thus, the roots of the equation are (3 + 2√5) and (3  2√5).
(iii) 2x^{2} + x – 4 = 0
Solution
The given equation is 2x^{2} + x – 4 = 0.
Comparing it with ax^{2} + bx + c = 0, we get
a = 2, b = 1 and c =  4
∴ Discriminant, D = b^{2} – 4ac = (1)^{2}  4 × 2 × (4)
= 1 + 32
= 33 > 0
So, the given equation has real roots.
Given:
25x^{2} + 30x + 7 = 0
On comparing it with ax^{2} + bx + x = 0, we get;
a = 25, b = 30 and c = 7
Discriminant D is given by:
D = (b^{2} – 4ac)
= (30^{2} – 4 × 25 × 7)
= 900 – 700
= 200
= 200 > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
Thus, the roots of the equation are (3 + √2)/5 and (3 – √2)/5.
(v) 16x^{2} + 24x + 1
Solution
16x^{2 }+ 24x + 1
⇒ 16x^{2} – 24x – 1 = 0
On comparing it with ax^{2 }+ bx + x = 0, we get;
a = 16, b =  24 and c =  1
Discriminant D is given by:
D = (b^{2} – 4ac)
= (24)^{2} – 4 × 16 × (1)
= 576 + (64)
= 640 > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
Solution
15x^{2} – 28 = x
⇒ 15x^{2 }– x – 28 = 0
On comparing it with ax^{2} + bx + c = 0, we get:
a = 25, b =  1 and c =  28
Discriminant D is given by:
D = (b^{2} – 4ac)
= (1)^{2} – 4 × 15 × (28)
= 1 – (1680)
= 1 + 1680
= 1681
= 1681 > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
Thus, the roots of the equation are 7/5 and (4)/3.
(vii) 2x^{2}  2√2x + 1 = 0
Solution
The given equation is 2x^{2}  2√2x + 1 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 2, b = 2√2 and c = 1
∴ Discriminant, D = b^{2} – 4ac
= (2√2)^{2} – 4 × 2 × 1
= 8 – 8
= 0
So, the given equation has real roots.
Now, √D = 0
∴ Î± = (b + √D)/2a = (2√2)/(2 × 2) = (2√2)/4 = √2/2
Î² = (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) √2x^{2} + 7 + 5√2 = 0
Solution
The given equation is √2x^{2} + 7 + 5√2 = 0.
Comparing it with ax^{2} + bx + c = 0, we get
a = √2, b = 7 and c = 5√2
∴ Discriminant, D = b^{2} – 4ac
= (7)^{2} – 4 × √2 × 5√2 = 49 – 40
= 9 > 0
So, the given equation has real roots.
Now, √D = √9 = 3
∴ Î± = (b + √D)/2a = (7 + 3)/(2 × √2)
= 4/(2√2)
= √2
Î² = (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) √3x^{2} + 10x  8√3 = 0
Solution
Given:
√3x^{2} + 10x  8√3 = 0
On comparing it with ax^{2} + bx + x = 0, we get;
a = √3, b = 10 and c = 8√3
Discriminant D is given by:
D = (b^{2} – 4ac)
= (10)^{2 }– 4 × √3 × (8√3)
= 100 + 96
= 196 > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
Thus, the roots of the equation are (2√3)/3 and –(4√3).
(x) √3x^{2}  2√2x  2√3 = 0.
Solution
The given equation is √3x^{2}  2√2x  2√3 = 0.
Comparing it with ax^{2} + bx + c = 0, we get
a = √3, b = 2√2 and c = 2√3
∴ Discriminant, D = b^{2} – 4ac
= (2√2)^{2} – 4 × √3 × (2√3)
= 8 + 24
= 32 > 0
So, the given equation has real roots.
Now, √D = √32 = 4√2
∴ Î± = (b + √D)/2a = (2√2) + 4√2)/(2 × √3) = 6√2/2√3 = √6
Î² = (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) 2x^{2} + 6√3x – 60 = 0.
Solution
The given equation is 2x^{2} + 6√3x – 60 = 0.
Comparing it with ax^{2} + bx + c = 0, we get
a = 2, b = 6√3 and c =  60
∴ Discriminant, D = b^{2 }– 4ac
= (6√3)^{2} – 4 × 2 × (60)
= 180 + 480
= 588 > 0
So, the given equation has real roots.
(xii) 4√3x^{2 }+ 5x  2√3 = 0
Solution
The given equation is 4√3x^{2} + 5x  2√3 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 4√3, b = 5 and c = 2√3
∴ Discriminant, D = b^{2} – 4ac
= 5^{2} – 4 × 4√3 × (2√3)
= 25 + 96
= 121 > 0
So, the given equation has real roots.
∴ Î± = (b + √D)/2a = (5 – 11)/(2 × 4√3) = 6/(8√3) = √3/4.
Î² = (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) 3x^{2}  2√6x + 2 = 0
Solution
The given equation is 3x^{2 } 2√6x + 2 = 0
Comparing it with a^{2} + bx + c = 0, we get
a = 3, b = 2√6 and c = 2
∴ Discriminant, D = b^{2} – 4ac
= (2√6)^{2 }– 4 × 3 × 2
= 24 – 24
= 0
So, the given equation has real roots.
Now, √D = 0
∴ Î± = ( b + √D)/2a = (2√6) + 0/(2 × 3)
= (2√6)/6
= √6/3
Î² = (b  √D)/2a = (2√6)/(2 × 3)
= (2√6)/6
= √6/3
Hence, √6/3 are the repeated of the given equation.
(xiv) 2√3x^{2} – 5x + √3 = 0
Solution
The given equation is 2√3x^{2} – 5x + √3 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 2√3, b = 5 and c = √3
∴ Discriminant, D = b^{2 }– 4ac
= (5)^{2} – 4 × 2√3 × √3
= 25 – 25
= 1 > 0
So, the given equation has real roots.
Now, √D = √1 = 1
∴ Î± = (b + √D)/2a = (5 + 1) + 1/(2 × 2√3) = 6/(4√3) = √3/2
Î² = (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) x^{2 }+ x + 2 = 0.
Solution
The given equation is x^{2} + x + 2 = 0
comparing it with ax^{2} + bx + c = 0, we get
a = 1, b = 1 and c = 2
∴ Discriminant D = b^{2} – 4ac = 1^{2} – 4 × 1 × 2
= 1 – 8
=  7 < 0
Hence, the given equation has no real roots (or real roots does not exist).
(xvi) 2x^{2} + ax – a^{2} = 0
Solution
The given equation is 2x^{2} + ax – a^{2} = 0
Comparing it with Ax^{2 }+ Bx + C = 0, we get
A = 2, B = a and C =  a^{2}
∴ Discriminant, D = B^{2} – 4AC
= a^{2} – 4 × 2 × a^{2}
= a^{2} + 8a^{2}
= 9a^{2} ≥ 0
So, the given equation has real roots.
Now, √D = √9a^{2} = 3a
∴ Î± = (B + √D)/2A = (a + 3a)/(2 × 2) = 2a/4 = a/2
Î² = (B  √D)/2A = (a – 3a)/(2 × 2) = 4a/4 =  a
Hence, a/2 and – a are the roots of the given equation.
(xvii) (x^{2 } √3 + 1)x + √3 = 0.
Solution
The given equation is x^{2} – (√3 + 1)x + √3 = 0.
Comparing it with ax^{2} + bx + c = 0, we get
a = 1, b = (√3 + 1) and c = √3
∴ Discriminant,
D = b^{2} – 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.
Solution
The given equation is 2x^{2 }+ 5√3x + 6 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 2, b = 5√3 and c = 6
∴ Discriminant, D = b^{2} – 4ac = (5√3)^{2} – 4 × 2 × 6
= 75 – 48
= 27 > 0
So, the given equation has real roots.
Now, √D = √27 = 3√3
∴ Î± = (b + √D)/2A = (5√3 + 3√3)/(2 × 2) = (2√3)/4 =  √3/2
Î² = (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) 3x^{2} – 2x + 2 = 0.b
Solution
The given equation is 3x^{2} – 2x + 2 = 0
Comparing it with ax^{2} + bx + c = 0, we get
a = 3, b =  2 and c = 2
∴ Discriminant D = b^{2} – 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
⇒ (x^{2} + 1)/x = 3
⇒ (x^{2} + 1)/x = 3
⇒ x^{2} + 1 = 3x
⇒ x^{2} – 3x +1 = 0
This equation is of from of ax^{2} + bx + c = 0, where, a = 1, b =  3 and c = 1.
∴ Discriminant, D = b^{2} – 4ac
= (3)^{2} – 4 × 1 × 1
= 9 – 4
= 5 > 0
So, the given equation has real roots.
Now, √D = √5
∴ Î± = (b + √D)/2a
= {(3) + √5)/(2 × 1) = (3 + √5)/2
Î² = (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/(x^{2} – 2x) = 3
⇒  2 = 3x^{2 }– 6x
⇒ 3x^{2} – 6x + 2 = 0
This equation is of form ax^{2} + bx + c = 0, where = 3, b = 6, and c = 2.
∴ Discriminant, D = b^{2} – 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
⇒ (x^{2 }– 1)/x = 3
⇒ x^{2} – 1 = 3x
⇒ x^{2} – 3x – 1 = 0
This equation is of the form ax^{2} + bx + c = 0, where a = 1, b =  3 and c = 1.
∴ Discriminant, D = b^{2} – 4ac
= (3)^{2} – 4 × 1 × (1)
= 9 + 4
= 13 > 0
So, the given equation has real roots.
Solution
The given equation is
m/nx^{2}.n/m = 1 – 2x
⇒ (m^{2}x^{2} + n^{2})/mn = 1 – 2x
⇒ m^{2}x^{2} + n^{2} = mn – 2mnx
⇒ m^{2}x^{2 }+ 2mnx + n^{2} – mn = 0
This equation is of the form ax^{2} + bx + c = 0, where a = m^{2}, b = 2mn and c = n^{2} – mn
∴ Discriminant,
D = b^{2 }– 4ac = (2mn)^{2} – 4 × m^{2} × (n^{2 }– mn)
= 4m^{2}n^{2} – 4m^{2}n^{2} + 4m^{3}n^{2}
= 4m^{3}n > 0
So, the given equation has real roots.
(xxiv) 36x^{2} – 12ax + (a^{2} – b^{2}) = 0
Solution
The given equation is 36x^{2} – 12ax + (a^{2 }– b^{2}) = 0
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 36, B =  12a and C = a^{2} – b^{2}
∴ Discriminant,
D = b^{2} – 4AC = (12a)^{2 }– 4 × 36 × (a^{2} – b^{2})
= 144a^{2 }– 144a^{2} + 144b^{2}
= 144b^{2}
So, the given equation has real roots.
∴ Î± = (B + √D)/2A
= {(12a) + 12b)/(2 × 36) = 12(a + b)/72 = (a + b)/0
Î² = (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) x^{2 }– 2ax + (a^{2} – b^{2}) = 0
Solution
x^{2} – 2ax + (a^{2} – b^{2}) = 0
On comparing it with Ax^{2} + Bx + C = 0, we get;
A = 1, B =  2a and C = (a^{2 }– b^{2})
Discriminant D is given by:
D = b^{2} – 4AC
= (2a)^{2} – 4 × 1 × (a^{2} – b^{2})
= 4a^{2} – 4a^{2} + 4b^{2}
= 4b^{2} > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
(xxvi) x^{2} – 2ax – (4b^{2} – a^{2}) = 0
Solution
The given equation is x^{2} – 2ax – (4b^{2} – a^{2}) = 0
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 1, B =  2a and C = (4b^{2} – a^{2})
∴ Discriminant,
B^{2} – 4AC = (2a)^{2} – 4 × 1 × [(4b^{2} – a^{2})]
= 4a^{2} + 16b^{2} – 4a^{2}
= 16b^{2} > 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A
= {(2a) + 4b)/(2 × 1) = 2(a + 2b)/2 = a + 2b
Î² = (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) x^{2} + 6x – (a^{2} – 2a + 8) = 0.
Solution
The given equation is x^{2} + 6x – (a^{2} + 2a – 8) = 0
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 1, B = 6 and C =  (a^{2} + 2a – 8)
∴ Discriminant,
D = B^{2} – 4AC = 6^{2 }– 4 × 1 × [(a^{2} + 2a – 8)]
= 36 + 4a^{2} + 8a – 32
= 4a^{2} + 8a – 32
= 4a^{2} + 8a + 4
= 4(a^{2 }+ 2a + 1)
= 4(a + 1)^{2} > 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A
= (6 + 2(a + 1))/(2 × 1) = (2a  4)/2 = a  2
Î² = (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) x^{2 }+ 5x – (a^{2} – a  6) = 0.
Solution
The given equation is x^{2} + 5x – (a^{2 }+ a – 6) = 0.
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 1, B = 5 and C =  (a^{2} + a – 8)
∴ Discriminant,
D = B^{2} – 4AC
= 5^{2} – 4 × 1 × [(a^{2} + a – 6)]
= 25 + 4a^{2} + 4a  24
= 4a^{2} + 4a^{2} + 4a + 1
= (2a + 1)^{2} > 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A
= (5 + 2a + 1)/(2 × 1) = (2a  4)/2 = a  2
Î² = (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) x^{2} – 4ax – b^{2} + 4a^{2} = 0.
Solution
The given equation is x^{2} – 4ax – b^{2} + 4a^{2} = 0.
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 1, B =  4a and C =  b^{2} + 4a^{2 }
∴ Discriminant, D = B^{2} – 4AC
= (4a)^{2 }– 4 × 1 × (b^{2} + 4a^{2})
= 16a^{2} + 4b^{2} – 16a^{2}
= 4b^{2} > 0
So, the given equation has real roots
= (4a + 2b)/2 = 2a + b
Î² = (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) 4x^{2} – 4a^{2}x + (a^{4} – b^{4}) = 0
Solution
The given equation is 4x^{2} – 4a^{2}x + (a4 – b4) = 0.
Comparing it with Ax^{2 }+ Bx + C = 0, we get
A = 4, B =  4a^{2} and C = a^{4 }– b^{4}
∴ Discriminant, B^{2} – 4AC
= (4a^{2})^{2} – 4 × 4 × (a^{2} – b^{2})
= 16a^{4 }– 16a^{4} + 16b^{4 }
= 16b^{4} > 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A = ((4a) + 4b^{2})/(2 × 4)
= 4(a^{2} + b^{2})/8
= (a^{2} + b^{2})/2
∴ Î² = (B  √D)/2A
= (4a^{2}) – 4b^{2}/(2 × 4)
= 4(a^{2} – b^{2})/8
= (a^{2} – b^{2})/2
Hence, 1/2(a^{2} + b^{2}) and 1/2(a^{2} – b^{2}) are the roots of the given equation.
(xxxi) 4x^{2} – 4bx – (a^{2} – b^{2}) = 0.
Solution
The given equation is 4x^{2} – 4bx – (a^{2} – b^{2}) = 0
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 4, B = 4b and C = (a^{2} – b^{2})
∴ Discriminant,
D = B^{2} – 4AC
= (4b)^{2} – 4 × 4 × [(a^{2} – b^{2})]
= 16b^{2} + 16a^{2} – 16b^{2}
= 16a^{2 }> 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A = (4b + 4a)/(2 × 4)
= 4(a – b)/8 = (a – b)/2
∴ Î² = (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. x^{2} – (2b – 1)x + (b^{2} – b – 20) = 0
Solution
The given equation is x^{2} – (2b – 1)x + (b^{2} – b – 20) = 0.
Comparing it with Ax^{2} + Bx + C = 0, we get
A = 1, B = (2b – 1) and C = b^{2} – b – 20
∴ Discriminant,
D = B^{2} – 4AC
= [(2b – 1)]^{2} – 4 × 1 × (b^{2} – b – 20)
= 4b^{2} – 4b + 1 – 4b^{2} + 4b + 80
= 81 > 0
So, the given equation has real roots
∴ Î± = (B + √D)/2A = [(2b  1) + 9]/(2 × 1)
= (2b + 8)/2
= b + 4
Î² = (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) 3a^{2}x^{2} + 8abx + 4b^{2} = 0
Solution
3a^{2}x^{2} + 8abx + 4b^{2} = 0
On comparing it with Ax^{2} + Bx + C = 0, we get:
A = 3a^{2}, B = 8ab and C = 4b^{2}
Discriminant D is given by:
D = (B^{2} – 4AC)
= (8ab)^{2} – 4 × 3a^{2} × 4b^{2 }
= 16a^{2}b^{2} > 0
Hence, the roots of the equation are real.
Roots Î± and Î² are given by:
Thus, the roots of the equation are (2b/3a) and (2b/a).
(xxxiv) a^{2}b^{2}x^{2 }– (4b^{4} – 3a^{4})x – 12a^{2}b^{2} = 0, a ≠ 0 and b ≠ 0
Solution
The given equation is a^{2}b^{2}x^{2 }– (4b^{4 }– 3a^{4})x – 12a^{2}b^{2} = 0
Comparing it with Ax^{2 }+ Bx + C = 0, we get
A = a^{2}b^{2}, B = (4b^{4} – 3a^{4}) and c =  12a^{2}b^{2}
∴ Discriminant,
B^{2} – 4AC
= [(4b^{4} – 3a^{4})]^{2}  4 × a^{2}b^{2} × (12a^{2}b^{2})
= 16b^{8} – 24a^{4}b^{4} + 9a^{8} + 48a^{4}b^{4}
= 16b^{8} + 24a^{4}b^{4} + 9a^{8}
= (4b^{4} + 3a^{4})^{2} > 0
So, the given equation has real roots.
Î² = ( B  √D)/2A = [[(4b^{4} – 3a^{4})] – (4b^{4} + 3a^{4})]/(2 × a^{2}b^{2})
= 6a^{4}/2a^{2}b^{2}
= (3a^{2}/b^{2})
Hence, 4b^{2}/a^{2} and – (3a^{2}/b) are the roots of the given equation.
(xxxv) 12abx^{2} – (9a^{2 }– 8b^{2})x – 6ab = 0, where a ≠ 0 and b ≠ 0
Solution
Given:
12abx^{2} – (9a^{2} – 8b^{2})x – 6ab = 0
On comparing it with Ax^{2} + Bx + C = 0, we get:
A = 12ab, B = (9a^{2} – 8b^{2}) and C =  6ab
Discriminant D is given by:
D = B^{2} – 4AC
= [(9a^{2} – 8b^{2})]^{2} – 4 × 12ab × (6ab)
= 81a^{4} – 144a^{2}b^{2} + 64b^{4} + 288a^{2}b^{2}
= 81a4 + 144a^{2}b^{2} + 64b^{4}
= (9a^{2} + 8b^{2})^{2} > 0
Hence, the roots of the equation are equal.
Roots Î± and Î² are given by:
Thus, the roots of the equation are 3a/4b and (2b)/3a.