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 |
Other Exercises |
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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) 2x2 – 8x + 5 = 0
(ii) 3x2 - 2√6x + 2 = 0
(iii) 5x2 – 4x + 1 = 0
(iv) 5x(x – 2) + 6 = 0
(iv) 5x(x – 2) + 6 = 0
(v) 12x2 + 4√15 + 5 = 0
(vi) x2 – x + 2 = 0
Solution
(i) The given equation is 2x2 – 8x + 5 = 0
This is of the form ax2 + bx + c = 0, where a = 2, b = - 8 and c = 5.
∴ Discriminant, D = b2 – 4ac = (-8)2 – 4 × 2 × 5 = 64 – 40
= 24 > 0
Hence, the given equation has real and unequal roots.
(ii) The given equation is 3x2 - 2√6x + 2 = 0
This is of the form ax2 + bx + c = 0, where a = 3, b = -2√6 and c = 2
∴ Discriminant, D = b2 – 4ac
= (-2√6)2 – 4 × 3 × 2
= 24 – 24
= 0
Hence, the given equation has real and equal roots.
(iii) The given equation is 5x2 – 4x + 51 = 0.
This is of the form ax2 + bx + c = 0, where a = 5, b = - 4 and c = 1.
∴ Discriminant,
D = b2 – 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
⇒ 5x2 – 10x + 6 = 0
This is of the form ax2 + bx + c = 0, where a = 5, b = - 10 and c = 6.
∴ Discriminant. D = b2 – 4ac
= (-10)2 – 4 × 5 × 6
= 100 – 120
= -20 < 0
Hence, the given equation has no real roots.
(v)
(vi) The given equation is x2 – 2x + 2 = 0
This is of the form ax2 + bx + c = 0, where a = 1, b = - 1 and c = 2.
∴ Discriminant, D = b2 – 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(a2 + b2)x2 + 2(a + b)x + 1 = 0. has no real roots.
Solution
The given equation is 2(a2 + b2)x2 + 2(a + b)x + 1 = 0.
∴ D = [2(a + b)]2 – 4 × 2(a2 + b2) × 1
= 4(a2 + 2ab + b2) – 8(a2 + b2)
= 4a2 + 8ab + 4b2 – 8a2 – 8b2
= -4a2 + 8ab – 4b2
= -4(a2 – 2ab + b2)
= -4(a – b)2 < 0
Hence, the given equation has no real roots.
3. Show that the roots of the equation x2 + px – q2 = 0 are real for all real values of p and q.
Solution
x2 + px – q2 = 0
Here,
a = 1, b = p and c = - q2
Discriminant D is given by:
D = (b2 – 4ac)
= p2 - 4 × 1 × (-q)2
= (p2 + 4q2) > 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 3x2 + 2kx + 27 = 0 real and equal ?
Solution
3x2 + 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
⇒ 4k2 – 324 = 0
⇒ 4k2 = 324
⇒ k2 = 81
⇒ k2 = ± 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
⇒ kx2 - 2√5kx + 10 = 0
This is of the form ax2 + bx + c = 0, where a = k, b = -2√5k and c = 10.
∴ D = b2 – 4ac
= (-2√5k)2 – 4 × k × 10
= 20k2 – 40k
The given equation will have real and equal roots if D = 0.
∴ 20k2 – 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 4x2 + px + 3 = 0. real and equal?
Solution
The given equation is 4x2 + px + 3 = 0
This is of the form ax2 + bx + c = 0, where a = 4, b = p and c = 3.
∴ D = b2 – 4ac
= p2 – 4 × 4 × 3
= p2 – 48
The given equation will have real and equal roots if D = 0.
∴ p – 48 = 0
⇒ p2 = 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 9x2 – 3kx + k = 0.
Solution
The given equation is 9x2 – 3kx + k = 0.
This is of the form ax2 + bx + c = 0, where a = 9, b = - 3k and c = k.
∴ D = b2 – 4ac
= (-3k)2 – 4 × 9 × k
= 9k2 – 36k
The given equation will have real and equal roots is D = 0.
∴ 9k2 – 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)x2 + 2(k + 1)x + 1 = 0. has real and equal roots.
Solution
The given equation is (3k + 1)x2 + 2(k + 1)x + 1 = 0.
This is of the form ax2 + bx + c = 0, where a = 3k + 1, b = 2(k + 1) and c = 1.
∴ D = b2 – 4ac
= [2(k + 1)]2 – 4 × (3k + 1) × 1
= 4(k2 + 2k + 1) – 4(3k +1)
= 4k2 + 8k + 4 – 12k – 4
= 4k2 – 4k
The given equation will have real and equal roots if D = 0.
∴ 4k2 – 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)x2 – (7p + 2)x + (7p – 3) = 0. has real and equal roots.
Solution
The given equation is (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
This is of the form ax2 + bx + c = 0, where a = 2p + 1, b = -(7p + 2) and c = 7p – 3
∴ D = b2 – 4ac
= -[-(7p + 2)]2 – 4 × (2p + 1) × (7p – 3)
= (49p2 + 28p + 4) - 4(14p2 + p – 3)
= 49p2 + 28p + 4 – 56p2 – 4p + 12
= -7p2 + 24p + 16
The given equation will have real and equal roots if D = 0.
∴ -7p2 + 24p + 16 = 0
⇒ 7p2 – 24p – 16 = 0
⇒ 7p2 – 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)x2 – 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)x2 – 6(p + 1)x + 3(p + 9) = 0
This is of the form ax2 + bx + c = 0, where a = p + 1, b = -6(p + 1) and c = 3(p + 9).
∴ D = b2 – 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 4x2 – 24x + 36 = 0
4x2 – 24x + 36 = 0
⇒ 4(x2 – 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 2x2 + px – 15 = 0. and the quadratic equation p(x2 + 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 px2 + px + k = 0 = 0 are equal.
∴ D = 0
⇒ p2 – 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 x2 – x + k = 0., find the value of p so that the roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.
Solution
It is given that 3 is a root of the quadratic equation x2 – x + k = 0
∴ (3)2 – 3 + k = 0
⇒ k + 6 = 0
⇒ k = - 6
The roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.
∴ D = 0
⇒ (2k)2 – 4 × 1 × (k2 + 2k + p) = 0
⇒ 4k2 – 4k2 – 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 x2 + 2x + 4p = 0. find the value of k for the which the quadratic equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.
Solution
It is given that -4 is a root of the quadratic equation x2 + 2x + 4op = 0
∴ (-4)2 + 2 × (-4) + 4p = 0
⇒ 16 – 8 + 4p = 0
⇒ 4p + 8 = 0
⇒ p = - 2
The equation x2 + 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 + 9k2) – 28(3 + 2k) = 0
⇒ 4(1 + 6k + 9k2 – 21 – 14k) = 0
⇒ 9k2 – 8k – 20 = 0
⇒ 9k2 – 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)x2/ + 2mcx + (c2 – a2) = 0 has equal roots, prove that c2 = a2 (1 + m2).
Solution
Given:
(1 + m2)x2 + 2mcx + (c2 – a2) = 0
Here,
a = (1 + m2), b = 2mc and c = (c2 – a2)
It is given that the roots of the equation are equal; therefore, we have:
D = 0
⇒ (b2 – 4ac) = 0
⇒ (2mc)2 – 4 × (1 + m2) × (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 - m2a2) = 0
⇒ 4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0
⇒ - 4c2 + 4a2 + 4m2a2 =0
⇒ a2 + m2a2 = c2
⇒ a2(1 + m2) = c2
⇒ c2 = a2(1 + m2)
Hence proved.
15. If the roots of the quadratic equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are real and equal, show that either a = 0 or (a3 + b3 + c3 = 3abc)
Solution
Given:
(c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0
Here,
a = (c2 – ab), b = - 2(a2 – bc), c = (b2 – ac)
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (b2 – 4ac) = 0
⇒ {-2(a2 – bc)}2 – 4 × (c2 – ab) × (b2 – ac) = 0
⇒ 4(a4 – 2a2bc + b2c2) – 4(b2c2 – ac3 – ab3 + a2bc) = 0
⇒ a4 – 2a2bc + b2c2 – b2c2 + ac3 + ab3 – a2bc = 0
⇒ a4 – 3a2bc + ac3 + ab3 = 0
⇒ a(a3 – 3abc + c3 + b3) = 0
Now,
a = 0 or a3 – 3abc + c3 + b3 = 0
a = 0 or a3 + b3 + c3 = 3abc
16. Find the value of p for which the quadratic equation 2x2 + px + 8 = 0 has real roots.
Solution
Given:
2x2 + px + 8 = 0
Here,
a = 2, b = p and c = 8
Discriminant D is given by
D = (b2 – 4ac)
= p2 – 4 × 2 × 8
= (p2 – 64)
If D ≥ 0, the roots of the equation will be real
⇒ (p2 – 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 (α – 12)x2 + 2(α – 12)x + 2 = 0 has equal roots.
Solution
(α – 12)x2 + 2(α – 12x + 2 = 0
Here,
a = (α – 12), b = 2(α – 12) and c = 2
It is given that the roots of the equation are equal; therefore, we have
D = 0
⇒ (b2 – 4ac) = 0
⇒ {2(α – 12)}2 – 4 × (α – 12) × 2 = 0
⇒ 4(α2 - 24α + 144) - 8α + 96 = 0
⇒ 4α2 - 96α + 576 - 8α + 96 = 0
⇒ 4α2 - 104α + 672 = 0
⇒ α2 - 26α + 168 = 0
⇒ α2 - 14α - 12α + 168 = 0
⇒ α(α – 14) – 12(α – 14) = 0
⇒ (α – 14)(α – 12) = 0
∴ α = 14 or α = 12
If the value of α is 12, the given equation becomes non-quadratic.
Therefore, the value of α will be 14 for the equation to have equal roots.
18. Find the value of k for which the roots of 9x2 + 8kx + 16 = 0 are real and equal
Solution
Given:
9x2 + 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
⇒ (b2 – 4ac) = 0
⇒ (8k)2 – 4 × 9 × 16 = 0
⇒ 64k2 – 576 = 0
⇒ 64k2 = 576
⇒ k2 = 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) kx2 + 6x + 1 = 0.
(ii) x2 – kx + 9 = 0
(iii) 9x2 + 3kx + 4 = 0
(iv) 5x2 – kx + 1 = 0
Solution
(i) The given equation is kx2 + 6x + 1 = 0.
∴ D = 62 – 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 x2 – kx + 9 = 0
∴ D = (-k)2 – 4 × 1 × 9
= k2 – 36
The given equation has real and distinct roots if D > 0
∴ k2 – 36 > 0
⇒ (k – 6)(k + 6) > 0
⇒ k < -6 or k > 6
(iii) the given equation is 9x2 + 3kx + 4 = 0
∴ D = (3k)2 – 4 × 9 × 4
= 9k2 – 144
The given equation has real and distinct roots if D > 0.
∴ 9k2 – 144 > 0
⇒ 9(k2 – 16) > 0
⇒ (k – 4)(k + 4) > 0
⇒ k < - 4 or k > 4
(iv) The given equation is 5x2 – kx + 1 = 0.
∴ D = (-k)2 – 4 × 5 × 1
= k2 – 20
The given equation has real and distinct roots if D > 0.
∴ k2 – 20 > 0
⇒ k2 - (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)x2 + 5(a + b)x – 2(a – b) = 0 are equal and unequal.
Solution
The given equation is (a – b)x2 + 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 (a2 + b2)x2 – 2(ac + bd)x = (c2 + d2) = 0 are equal, prove that a/b = c/d.
Solution
It is given that the roots of the equation (a2 + b2)x2 - 2(ac + bd)x + (c2 + d2) = 0 are equal.
∴ D = 0
⇒ [-2(ac + bd)]2 – 4(a2 + b2)(c2 + d2) = 0
⇒ 4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2 + b2d2) = 0
⇒ 4(a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2) = 0
⇒ (-a2d2 + 2abcd – b2c2) = 0
⇒ -(a2d2 – 2abcd + b2c2) = 0
⇒ (ad – bc)2 = 0
⇒ ad – bc = 0
⇒ ad = bc
⇒ a/b = c/d
Hence proved.
22. If the roots of the equations ax2 + 2bx + c = 0 and bax2 - 2√acx + b = 0 are simultaneously real then prove that b2 = ac
Solution
It is given that the roots of the equation ax2 + 2bx + c = 0 are real.
∴ D1 = (2b)2 – 4 × a × c ≥ 0
⇒ 4(b2 – ac) ≥ 0
⇒ b2 – ac ≥ 0 ....(1)
⇒ 4(ac – b2) ≥ 0
⇒ - 4(b2 – ac) ≥ 0
⇒ b2 – ac ≥ 0 ...(2)
The roots of he given equations are simultaneously real if (1) and (2) holds true together.
this is possible if
b2 – ac = 0
⇒ b2 = ac