RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 2
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 3

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 4

Question 3.
Solution:
x² + px – q² = 0
Comparing it with ax² + bx + c = 0
a = 1, b = p, c = -q²
Discriminant (D) = b² – 4ac
= (p)² – 4 x 1 x (-q²)
= p² + 4 q²
p and q are of two powers
p² + 4q² is always greater than 0
The roots are real for all real values of p and q.

Question 4.
Solution:
The quadratic equation is 3x² + 2kx + 27 = 0
Comparing it with ax² + bx + c = 0
a = 3, b = 2k, c = 27
Discriminant (D) = b² – 4ac
= (2k)² – 4 x 3 x 27
= (2k)² – 324
Roots are real and equal
(2k)² – 324 = 0
⇒ (2k)² – (18)² = 0
⇒ (k)² – (9)² = 0
⇒ (k + 9) (k – 9) = 0
Either k + 9 = 0, then k = -9
or k – 9 = 0, then k = 9
Hence, k = 9, -9

Question 5.
Solution:
The quadratic equation is
kx (x – 2√5) x + 10 = 0
kx² – 2√5 kx + 10 = 0
Comparing it with ax² + bx + c = 0
a = k, b = -2√5 k, c = 10
D = b² – 4ac = (-2 k)² – 4 x k x 10 = 20k² – 40k
Roots are real and equal.
D = 0
20k² – 40k = 0
⇒ k² – 2k = 0
⇒ k (k – 2) = 0
Either, k = 0 or k – 2 = 0, then k = 2
k = 0, k = 2

Question 6.
Solution:
The quadratic equation is 4x² + px + 3 = 0
Comparing it with ax² + bx + c = 0
a = 4, b = p, c = 3
D = b² – 4ac = p² – 4 x 4 x 3 = p²- 48
Roots are real and equal.
D = 0
⇒ p² – 48 = 0
⇒ p² = 48 = (±4√3)²
⇒ P = ± 4√3
P = 4√3, p = -4√3

Question 7.
Solution:
The quadratic equation is 9x² – 3kx + k = 0
Comparing it with ax? + bx + c = 0
a = 9, b = -3k, c = k
D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k
Roots are real and equal.
D = 0
9k² – 36k = 0
⇒ 9k (k – 4) = 0
Either, k = 0 or k – 4 = 0, then k = 4
The value of k is non-zero.
k = 4

Question 8.
Solution:
(i) The equation is (3k + 1) x² + 2(k + 1) x + 1 = 0
Comparing it with ax² + bx + c = 0
a = (3k + 1), b = 2(k + 1), c = 1
D = b² – 4 ac
= [2(k + 1)]² – 4(3k + 1) x 1
= 4k² + 4 + 8k – 12k – 4
= 4k² – 4k
= 4k (k – 1)
Roots are real and equal.
Either, k = 0 or k – 1 = 0, then k = 1
k = 0, k = 1
(ii) x² + k(2x + k – 1) + 2 = 0
⇒ x² + 2kx + (k² – k + 2) = 0
Here, a = 1, b = 2k, c = (k² – k + 2)
Discriminant (D) = b² – 4ac = (2k)² – 4 x 1 x (k² – k + 2)
= 4k² – 4k² + 4k – 8
= 4k – 8
Roots are real and equal.
D = 0
⇒ 4k – 8 = 0
⇒ k = 2
Hence, k = 2

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 5

Question 10.
Solution:
The given quadratic equation is
(p + 1) x² – 6(p + 1) x + 3(p + 9) = 0, p ≠ -1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 6

Question 11.
Solution:
-5 is a root of 2x² + px – 15 = 0
x = -5 will satisfy it
Now, substituting the value of x = -5
⇒ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35
⇒ P = 7
In quadratic equation p(x² + x) + k = 0
⇒ 7 (x² + x) + k = 0 (p = 7)
⇒ 7x² + 7x + k = 0
Comparing it with ax² + bx + c = 0
a = 7, b = 7, c = k
D = b² – 4ac = (7)² – 4 x 7 x k
= 49 – 28k
Roots are real and equal.
49 – 28k = 0
⇒ 28k = 49
k = \(\frac { 49 }{ 28 }\) = \(\frac { 7 }{ 4 }\)

Question 12.
Solution:
3 is a root of equation x² – x + k = 0
It will satisfy it
Now, substituting the value of x = 3 in it
(3)² – (3) + k = 0
⇒ 9 – 3 + k = 0
⇒ 6 + k = 0
⇒ k = -6
Now in the equation, x² + k (2x + k + 2) + p = 0
x² + (-6)(2x – 6 + 2) + p = 0
⇒ x² – 12x + 36 – 12 + p = 0
⇒ x² – 12x + (24 + p) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -12, c = 24 + p
D = b² – 4ac
= (-12)² – 4 x 1 x (24 + p)
= 144 – 96 – 4p = 48 – 4p
Roots are real and equal.
D = 0
48 – 4p = 0
⇒ 4p = 48
⇒ p = 12
Hence, p = 12

Question 13.
Solution:
-4 is a root of the equation x² + 2x + 4p = 0
Then it will satisfy the equation
Now, substituting the value of x = -4
(-4)² + 2(-4) + 4p = 0
16 – 8 + 4p = 0
⇒ 8 + 4p = 0
⇒ 4p = -8
⇒ p = -2
In the quadratic equation x² + px (1 + 3k) + 7(3 + 2k) = 0
⇒ x² – 2x (1 + 3k) + 7(3 + 2k) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -2 (1 + 3k), c = 7 (3 + 2k)
D = b² – 4ac
= [-2(1 + 3k)]² – 4 x 1 x 7(3 + 2k)
= 4(1 + 9k² + 6k) – 28(3 + 2k)
= 4 + 36k² + 24k – 84 – 56k
= 36k² – 32k – 80
Roots are equal.
D = 0
⇒ 36k² – 32k – 80 = 0
⇒ 9k² – 8k – 20 = 0
⇒ 9k² – 18k + 10k – 20 = 0
⇒ 9k (k – 2) + 10(k – 2) = 0
⇒ (k – 2) (9k + 10) = 0
Either, k – 2 = 0, then k = 2
or 9k + 10 = 0, then 9k = -10 ⇒ k = \(\frac { -10 }{ 9 }\)
k = 2, k = \(\frac { -10 }{ 9 }\)

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 7

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 8

Question 16.
Solution:
The quadratic equation is 2x² + px + 8 = 0
Comparing it with ax² + bx + c = 0
a = 2, b = p, c = 8
D = b2 – 4ac = p² – 4 x 2 x 8 = p² – 64
Roots are real.
D ≥ 0
p² – 64 ≥ 0
⇒ p² ≥ 64 ≥ (±8)²
p ≥ 8 or p ≤ -8

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 9
Roots are equal
D = 0
⇒ 4(α – 12) (α – 14) = 0
⇒ α – 14 = 0 {(α – 12) ≠ 0}
⇒ α = 14
Hence, α = 14

Question 18.
Solution:
9x² + 8kx + 16 = 0
Comparing it with ax2 + bx + c = 0
a = 9, b = 8k, c = 16
D = b² – 4ac
= (8k)² – 4 x 9 x 16 = 64k² – 576
Roots are real and equal.
D = 0
64k² – 576 = 0
64k² = 576
⇒ k² = 9 = (±3)²
k = 3, k = -3

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 10
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 11
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 12

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 13
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 14

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 15
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 17

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 18

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