RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10D 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 10D Solutions
1. Find the nature of the roots of the following quadratic equations:
(i) 2x^{2} – 8x + 5 = 0
(ii) 3x^{2}  2√6x + 2 = 0
(iii) 5x^{2} – 4x + 1 = 0
(iv) 5x(x – 2) + 6 = 0
(iv) 5x(x – 2) + 6 = 0
(v) 12x^{2} + 4√15 + 5 = 0
(vi) x^{2} – x + 2 = 0
Solution
(i) The given equation is 2x^{2} – 8x + 5 = 0
This is of the form ax^{2} + bx + c = 0, where a = 2, b =  8 and c = 5.
∴ Discriminant, D = b^{2} – 4ac = (8)^{2} – 4 × 2 × 5 = 64 – 40
= 24 > 0
Hence, the given equation has real and unequal roots.
(ii) The given equation is 3x^{2}  2√6x + 2 = 0
This is of the form ax^{2 }+ bx + c = 0, where a = 3, b = 2√6 and c = 2
∴ Discriminant, D = b^{2} – 4ac
= (2√6)^{2} – 4 × 3 × 2
= 24 – 24
= 0
Hence, the given equation has real and equal roots.
(iii) The given equation is 5x^{2} – 4x + 51 = 0.
This is of the form ax^{2} + bx + c = 0, where a = 5, b =  4 and c = 1.
∴ Discriminant,
D = b^{2} – 4ac
= (4)^{2} – 4 × 5 × 1’
= 16 – 20
=  4 < 0
Hence, the given equation has no real roots.
(iv) The given equation is
5x(x – 2) + 6 = 0
⇒ 5x^{2} – 10x + 6 = 0
This is of the form ax^{2} + bx + c = 0, where a = 5, b =  10 and c = 6.
∴ Discriminant. D = b^{2} – 4ac
= (10)^{2} – 4 × 5 × 6
= 100 – 120
= 20 < 0
Hence, the given equation has no real roots.
(v)
(vi) The given equation is x^{2} – 2x + 2 = 0
This is of the form ax^{2 }+ bx + c = 0, where 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.
2. If a and b are distinct real numbers, show that the quadratic equations 2(a^{2} + b^{2})x^{2} + 2(a + b)x + 1 = 0. has no real roots.
Solution
The given equation is 2(a^{2} + b^{2})x^{2} + 2(a + b)x + 1 = 0.
∴ D = [2(a + b)]^{2} – 4 × 2(a^{2} + b^{2}) × 1
= 4(a^{2} + 2ab + b^{2}) – 8(a^{2} + b^{2})
= 4a^{2} + 8ab + 4b^{2} – 8a^{2 }– 8b^{2}
= 4a^{2} + 8ab – 4b^{2 }
= 4(a^{2} – 2ab + b^{2})
= 4(a – b)^{2} < 0
Hence, the given equation has no real roots.
3. Show that the roots of the equation x^{2} + px – q^{2} = 0 are real for all real values of p and q.
Solution
x^{2} + px – q^{2} = 0
Here,
a = 1, b = p and c =  q^{2 }
Discriminant D is given by:
D = (b^{2} – 4ac)
= p^{2}  4 × 1 × (q)^{2 }
= (p^{2} + 4q^{2}) > 0
D > 0 for all real values of p and q.
Thus, the roots of the equation are real.
4. For what values of k are the roots of the quadratic equation 3x^{2 }+ 2kx + 27 = 0 real and equal ?
Solution
3x^{2} + 2kx + 27 = 0
Here,
a = 3, b = 2k and c = 27
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (2k)^{2} – 4 × 3 × 27 = 0
⇒ 4k^{2} – 324 = 0
⇒ 4k^{2} = 324
⇒ k^{2} = 81
⇒ k^{2} = ± 9
∴ k = 9 or k = 9
5. For what value of k are the roots of the quadratic equation kx(x  2√5) + 10 = 0 real and equal.
Solution
The given equation is
kx(x  2√5) + 10 = 0
⇒ kx^{2}  2√5kx + 10 = 0
This is of the form ax^{2} + bx + c = 0, where a = k, b = 2√5k and c = 10.
∴ D = b^{2} – 4ac
= (2√5k)^{2} – 4 × k × 10
= 20k^{2} – 40k
The given equation will have real and equal roots if D = 0.
∴ 20k^{2} – 40k = 0
⇒ 20k(k – 2) = 0
⇒ k = 0 or k – 2 = 0
⇒ k = 0 or k = 2
But, for k = 0, we get 10 = 0, which is not true
Hence, 2 is the required value of k.
6. For what values of p are the roots of the equation 4x^{2} + px + 3 = 0. real and equal?
Solution
The given equation is 4x^{2} + px + 3 = 0
This is of the form ax^{2} + bx + c = 0, where a = 4, b = p and c = 3.
∴ D = b^{2} – 4ac
= p^{2} – 4 × 4 × 3
= p^{2} – 48
The given equation will have real and equal roots if D = 0.
∴ p – 48 = 0
⇒ p^{2} = 48
Hence, 4√3 and 4√3 are the required values of p.
7. Find the nonzero value of k for which the roots of the quadratic equation 9x^{2} – 3kx + k = 0.
Solution
The given equation is 9x^{2} – 3kx + k = 0.
This is of the form ax^{2} + bx + c = 0, where a = 9, b =  3k and c = k.
∴ D = b^{2} – 4ac
= (3k)^{2} – 4 × 9 × k
= 9k^{2} – 36k
The given equation will have real and equal roots is D = 0.
∴ 9k^{2} – 36k = 0
⇒ 9k(k – 4) = 0
⇒ k = 0 or k – 4 = 0
⇒ k = 0 or k = 4
But, k ≠ 0 (Given)
Hence, the required values of k is 4.
8. Find the values of k for which the quadratic equation (3k + 1)x^{2} + 2(k + 1)x + 1 = 0. has real and equal roots.
Solution
The given equation is (3k + 1)x^{2} + 2(k + 1)x + 1 = 0.
This is of the form ax^{2} + bx + c = 0, where a = 3k + 1, b = 2(k + 1) and c = 1.
∴ D = b^{2} – 4ac
= [2(k + 1)]^{2} – 4 × (3k + 1) × 1
= 4(k^{2} + 2k + 1) – 4(3k +1)
= 4k^{2 }+ 8k + 4 – 12k – 4
= 4k^{2} – 4k
The given equation will have real and equal roots if D = 0.
∴ 4k^{2} – 4k = 0
⇒ 4k(k – 1) = 0
⇒ k = 0 or k – 1 = 0
⇒ k = 0 or k = 1
Hence, 0 and 1 are the required values of k.
9. Find the value of p for which the quadratic equation (2p + 1)x^{2} – (7p + 2)x + (7p – 3) = 0. has real and equal roots.
Solution
The given equation is (2p + 1)x^{2} – (7p + 2)x + (7p – 3) = 0
This is of the form ax^{2} + bx + c = 0, where a = 2p + 1, b = (7p + 2) and c = 7p – 3
∴ D = b^{2} – 4ac
= [(7p + 2)]^{2 }– 4 × (2p + 1) × (7p – 3)
= (49p^{2 }+ 28p + 4)  4(14p^{2} + p – 3)
= 49p^{2} + 28p + 4 – 56p^{2} – 4p + 12
= 7p^{2} + 24p + 16
The given equation will have real and equal roots if D = 0.
∴ 7p^{2} + 24p + 16 = 0
⇒ 7p^{2} – 24p – 16 = 0
⇒ 7p^{2} – 28p + 4p – 16 = 0
⇒ 7p(p – 4) + 4(p – 4) = 0
⇒ (p – 4)(7p + 4) = 0
⇒ p – 4 = 0 or 7p + 4 = 0
⇒ p = 4 or p = (4/7)
Hence, 4 and –(4/7) are the required values of p.
10. Find the values of p for which the quadratic equation (p + 1)x^{2} – 6(p + 1)x + 3(x + 9) = 0., p ≠  1has equal roots. hence, find the roots of the equation.
Solution
The given equation is (p + 1)x^{2} – 6(p + 1)x + 3(p + 9) = 0
This is of the form ax^{2} + bx + c = 0, where a = p + 1, b = 6(p + 1) and c = 3(p + 9).
∴ D = b^{2} – 4ac
= [6(p+ 1)]^{2} – 4 × (p + 1) × 3(p + 9)
= 12(p + 1)[3(p + 1) – (p + 9)]
= 12(p + 1)(2p – 6)
The given equation will have real and equal roots if D = 0.
∴ 12(p + 1)(2p – 6) = 0
⇒ p + 1 = 0 or 2p – 6 = 0
⇒ p = 1 or p = 3
But, p ≠ 1 (Given)
Thus, the value of p is 3
Putting p = 3, the given equation becomes 4x^{2} – 24x + 36 = 0
4x^{2 }– 24x + 36 = 0
⇒ 4(x^{2} – 6x + 9) = 0
⇒ (x – 3)^{2} = 0
⇒ x – 3 = 0
⇒ x = 3
Hence, 3 is the repeated root of this equation.
11. If – 5 is a root of the quadratic equation 2x^{2} + px – 15 = 0. and the quadratic equation p(x^{2} + x) + k = 0 has equal
Solution
It is given that 5 is a root of the quadratic equation 2x2 + px  15 = 0.
∴ 2(5)^{2 }+ p × (5) – 15 = 0
⇒  5p + 35 = 0
⇒ p = 7
The roots of the equation px^{2} + px + k = 0 = 0 are equal.
∴ D = 0
⇒ p^{2 }– 4pk = 0
⇒ (7)^{2} – 4 × 7 × k = 0
⇒ 49 – 28k = 0
⇒ k = 49/28 = 7/4
Thus, the value of k is 7/4.
12. If 3 is a root of the quadratic equation x^{2} – x + k = 0., find the value of p so that the roots of the equation x^{2} + 2kx + (k^{2} + 2k + p) = 0 are equal.
Solution
It is given that 3 is a root of the quadratic equation x^{2} – x + k = 0
∴ (3)^{2} – 3 + k = 0
⇒ k + 6 = 0
⇒ k =  6
The roots of the equation x^{2} + 2kx + (k^{2} + 2k + p) = 0 are equal.
∴ D = 0
⇒ (2k)^{2 }– 4 × 1 × (k^{2} + 2k + p) = 0
⇒ 4k^{2} – 4k^{2} – 8k – 4p = 0
⇒  8k – 4p = 0
⇒ p = 8k/4 =  2k
⇒ p = 8k/4 = 2k
⇒ p = 2 × (6) = 12
Hence, the value of p is 12.
13. If 4 is a root of the equation x^{2 }+ 2x + 4p = 0. find the value of k for the which the quadratic equation x^{2} + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.
Solution
It is given that 4 is a root of the quadratic equation x^{2} + 2x + 4op = 0
∴ (4)^{2} + 2 × (4) + 4p = 0
⇒ 16 – 8 + 4p = 0
⇒ 4p + 8 = 0
⇒ p =  2
The equation x^{2 }+ px(1 + 3k) + 7(3 + 2k) = 0 has real roots.
∴ D = 0
⇒ [p(1 + 3k)]^{2} – 4 × 1 × 7(3 + 2k) = 0
⇒ [2(1 + 3k)]^{2} – 28(3 + 2k) = 0
⇒ 4(1 + 6k + 9k^{2}) – 28(3 + 2k) = 0
⇒ 4(1 + 6k + 9k^{2} – 21 – 14k) = 0
⇒ 9k^{2} – 8k – 20 = 0
⇒ 9k^{2} – 18k + 10k – 20 = 0
⇒ 9k(k – 2) + 10(k – 2) = 0
⇒ (k – 2)(9k + 10) = 0
⇒ k – 2 = 0 or 9k + 10 = 0
⇒ k = 2 or k = (10/9)
Hence, the required value of k is 2 or –(10/9).
14. If the quadratic equation (1 + 4m2)x^{2}/ + 2mcx + (c^{2} – a^{2}) = 0 has equal roots, prove that c^{2} = a^{2} (1 + m^{2}).
Solution
Given:
(1 + m^{2})x^{2} + 2mcx + (c^{2} – a^{2}) = 0
Here,
a = (1 + m^{2}), b = 2mc and c = (c^{2} – a^{2})
It is given that the roots of the equation are equal; therefore, we have:
D = 0
⇒ (b^{2} – 4ac) = 0
⇒ (2mc)^{2} – 4 × (1 + m^{2}) × (c^{2} – a^{2}) = 0
⇒ 4m^{2}c^{2} – 4(c^{2} – a^{2} + m^{2}c^{2}  m^{2}a^{2}) = 0
⇒ 4m^{2}c^{2} – 4c^{2} + 4a^{2} – 4m^{2}c^{2} + 4m^{2}a^{2} = 0
⇒  4c^{2} + 4a^{2 }+ 4m^{2}a^{2} =0
⇒ a^{2} + m^{2}a^{2} = c^{2}
⇒ a^{2}(1 + m^{2}) = c^{2}
⇒ c^{2} = a^{2}(1 + m^{2})
Hence proved.
15. If the roots of the quadratic equation (c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) = 0 are real and equal, show that either a = 0 or (a^{3} + b^{3 }+ c^{3} = 3abc)
Solution
Given:
(c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) = 0
Here,
a = (c^{2} – ab), b =  2(a^{2} – bc), c = (b2 – ac)
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (b^{2 }– 4ac) = 0
⇒ {2(a^{2} – bc)}^{2} – 4 × (c^{2} – ab) × (b^{2} – ac) = 0
⇒ 4(a^{4} – 2a^{2}bc + b^{2}c^{2}) – 4(b^{2}c^{2} – ac^{3} – ab^{3} + a^{2}bc) = 0
⇒ a^{4} – 2a^{2}bc + b^{2}c^{2} – b^{2}c^{2} + ac^{3} + ab^{3} – a^{2}bc = 0
⇒ a^{4} – 3a^{2}bc + ac^{3} + ab^{3} = 0
⇒ a(a^{3} – 3abc + c^{3} + b^{3}) = 0
Now,
a = 0 or a^{3 }– 3abc + c^{3} + b^{3} = 0
a = 0 or a^{3} + b^{3} + c^{3} = 3abc
16. Find the value of p for which the quadratic equation 2x^{2} + px + 8 = 0 has real roots.
Solution
Given:
2x^{2} + px + 8 = 0
Here,
a = 2, b = p and c = 8
Discriminant D is given by
D = (b^{2 }– 4ac)
= p^{2} – 4 × 2 × 8
= (p^{2} – 64)
If D ≥ 0, the roots of the equation will be real
⇒ (p^{2} – 64) ≥ 0
⇒ (p + 8)(p – 8) ≥ 0
⇒ p ≥ 8 and p ≤  8
Thus, the roots of the equation are real for p ≥ 8 and p ≤ 8.
17. Find the value of a for which the equation (Î± – 12)x^{2} + 2(Î± – 12)x + 2 = 0 has equal roots.
Solution
(Î± – 12)x^{2} + 2(Î± – 12x + 2 = 0
Here,
a = (Î± – 12), b = 2(Î± – 12) and c = 2
It is given that the roots of the equation are equal; therefore, we have
D = 0
⇒ (b^{2} – 4ac) = 0
⇒ {2(Î± – 12)}^{2} – 4 × (Î± – 12) × 2 = 0
⇒ 4(Î±^{2}  24Î± + 144)  8Î± + 96 = 0
⇒ 4Î±^{2}  96Î± + 576  8Î± + 96 = 0
⇒ 4Î±^{2}  104Î± + 672 = 0
⇒ Î±^{2}  26Î± + 168 = 0
⇒ Î±^{2}  14Î±  12Î± + 168 = 0
⇒ Î±(Î± – 14) – 12(Î± – 14) = 0
⇒ (Î± – 14)(Î± – 12) = 0
∴ Î± = 14 or Î± = 12
If the value of Î± is 12, the given equation becomes nonquadratic.
Therefore, the value of Î± will be 14 for the equation to have equal roots.
18. Find the value of k for which the roots of 9x^{2} + 8kx + 16 = 0 are real and equal
Solution
Given:
9x^{2} + 8kx + 16 = 0
Here,
a = 9, b = 8k and c = 16
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (b^{2} – 4ac) = 0
⇒ (8k)^{2} – 4 × 9 × 16 = 0
⇒ 64k^{2} – 576 = 0
⇒ 64k^{2} = 576
⇒ k^{2} = 9
⇒ k = ± 3
∴ k = 3 or k = 3
19. Find the values of k for which the given quadratic equation has real and distinct roots:
(i) kx^{2} + 6x + 1 = 0.
(ii) x^{2} – kx + 9 = 0
(iii) 9x^{2} + 3kx + 4 = 0
(iv) 5x^{2} – kx + 1 = 0
Solution
(i) The given equation is kx^{2} + 6x + 1 = 0.
∴ D = 6^{2} – 4 × k × 1 = 36 – 4k
The given equation has real and distinct roots if D > 0.
∴ 36 – 4k > 0
⇒ 4k < 36
⇒ k < 9
(ii) The given equation is x^{2} – kx + 9 = 0
∴ D = (k)^{2} – 4 × 1 × 9
= k^{2} – 36
The given equation has real and distinct roots if D > 0
∴ k^{2} – 36 > 0
⇒ (k – 6)(k + 6) > 0
⇒ k < 6 or k > 6
(iii) the given equation is 9x^{2} + 3kx + 4 = 0
∴ D = (3k)^{2} – 4 × 9 × 4
= 9k^{2} – 144
The given equation has real and distinct roots if D > 0.
∴ 9k^{2} – 144 > 0
⇒ 9(k^{2} – 16) > 0
⇒ (k – 4)(k + 4) > 0
⇒ k <  4 or k > 4
(iv) The given equation is 5x^{2 }– kx + 1 = 0.
∴ D = (k)^{2 }– 4 × 5 × 1
= k^{2} – 20
The given equation has real and distinct roots if D > 0.
∴ k^{2} – 20 > 0
⇒ k^{2}  (2√5)^{2} > 0
⇒ (k  2√5)(k + 2√5) > 0
⇒ k = 2√5 or k > 2√5
20. If a and b are real and a ≠ b then show that the roots of the equation
(a – b)x^{2} + 5(a + b)x – 2(a – b) = 0 are equal and unequal.
Solution
The given equation is (a – b)x^{2} + 5(a + b)x – 2(a – b) = 0.
∴ D = [5(a + b)]^{2} – 4 × (a – b) × [2(a – b)]
= 25(a + b)^{2} + 8(a – b)^{2 }
Since a and b are real and a ≠ b, so (a – b)^{2} > 0 and (a + b)^{2} > 0
∴ 8(a – b)^{2} > 0 ...(1) (Product of two positive numbers is always positive)
Also, 25(a + b)^{2} > 0 ...(2) (Product of two positive numbers is always positive)
Adding (1) and (2), we get
25(a + b)^{2} + 8(a – b)^{2} > 0 (Sum of two positive numbers is always positive)
⇒ D > 0
Hence, the roots of the given equation are real and unequal.
21. If the roots of the equation (a^{2} + b2)x^{2} – 2(ac + bd)x = (c^{2} + d^{2}) = 0 are equal, prove that a/b = c/d.
Solution
It is given that the roots of the equation (a^{2} + b^{2})x^{2}  2(ac + bd)x + (c^{2} + d^{2}) = 0 are equal.
∴ D = 0
⇒ [2(ac + bd)]^{2} – 4(a^{2} + b^{2})(c^{2} + d^{2}) = 0
⇒ 4(a^{2}c^{2} + b^{2}d^{2} + 2abcd) – 4(a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}) = 0
⇒ 4(a^{2}c^{2} + b^{2}d^{2} + 2abcd – a^{2}c^{2} – a^{2}d^{2} – b^{2}c^{2} – b^{2}d^{2}) = 0
⇒ (a^{2}d^{2} + 2abcd – b^{2}c^{2}) = 0
⇒ (a^{2}d^{2} – 2abcd + b^{2}c^{2}) = 0
⇒ (ad – bc)^{2} = 0
⇒ ad – bc = 0
⇒ ad = bc
⇒ a/b = c/d
Hence proved.
22. If the roots of the equations ax^{2} + 2bx + c = 0 and bax^{2}  2√acx + b = 0 are simultaneously real then prove that b^{2} = ac
Solution
It is given that the roots of the equation ax^{2} + 2bx + c = 0 are real.
∴ D_{1} = (2b)^{2} – 4 × a × c ≥ 0
⇒ 4(b^{2 }– ac) ≥ 0
⇒ b^{2} – ac ≥ 0 ....(1)
⇒ 4(ac – b^{2}) ≥ 0
⇒  4(b^{2} – ac) ≥ 0
⇒ b^{2} – ac ≥ 0 ...(2)
The roots of he given equations are simultaneously real if (1) and (2) holds true together.
this is possible if
b^{2} – ac = 0
⇒ b^{2} = ac