RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself

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

Other Exercises

Objective Questions (MCQ)
Question 1.
Solution:
(a) x² – 3√x + 2 = 0
It is not a quadratic equation, it has a fractional power of √x
(b) x + \(\frac { 1 }{ x }\) = x²
⇒ x² + 1 = x3
It is not a quadratic equation.
(c) x² + \(\frac { 1 }{ { x }^{ 2 } }\) = 5
⇒ x4 + 1 + 5x²
It is not a quadratic equation.
(d) 2x² – 5x = (x – 1)²
⇒ 2x² – 5x = x² – 2x + 1
⇒ x² – 3x – 1 = 0
It is a quadratic equation. (d)

Question 2.
Solution:
(a) (x² + 1) = (2 – x)² + 3
⇒ x² + 1 = 4 + x² – 4x + 3 is not a quadratic equation.
(b) x3 – x² = (x – 1)3
⇒ x3 – x² = x3 – 3x² + 3x – 1
⇒ 3x² – x² – 3x + 1 = 0
⇒ 2x² – 3x + 1 = 0
It is a quadratic equation.
(c) 2x² + 3 = 10x – 15 + 2x² – 3x
⇒ 3x – 15 – 3 = 0
It is not a quadratic equation. (b)

Question 3.
Solution:
(a) It is a quadratic equation.
(b) (x + 2)² = 2(x² – 5)
⇒ x² + 4x + 4 = 2x² – 10
⇒ x² – 4x – 14 = 0
It is a quadratic equation.
(c) (√2 x + 3)² = 2x² + 6
⇒ 2x² + 3√2 x + 9 = 2x² + 6
⇒ 3√2 + 3 = 0
It is not a quadratic equation.
(d) (x – 1)² = 3x² + x – 2
⇒ x² – 2x +1 = 3x² + x – 2
⇒ 2x² + 3x – 3 = 0
It is a quadratic equation. (c)

Question 4.
Solution:
x = 3 is solution of 3x² + (k – 1)x + 9 = 0
It will satisfy it
3(3)² + (k – 1)(3) + 9 = 0
⇒ 27 + 3k – 3 + 9 = 0
⇒ 3k + 33 = 0
⇒ k = -11 (b)

Question 5.
Solution:
2 is one root of equation 2x² + ax + 6 = 0
It will satisfy it
2(2)² + a(2) + 6 = 0
⇒ 8 + 2a + 6 = 0
⇒ 2a = -14
⇒ a = -7
a = -7 (b)

Question 6.
Solution:
In equation x² – 6x + 2 = 0
Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -(-6) }{ 1 }\) = 6 (c)

Question 7.
Solution:
In equation x² – 3x + k = 10
x² – 3x + (k – 10) = 0
Product of roots = \(\frac { c }{ a }\) = \(\frac { k – 10 }{ 1 }\) = k – 10
k – 10 = -2 then k = 10 – 2 = 8 (c)

Question 8.
Solution:
In equation 7x² – 12x + 18 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 1

Question 9.
Solution:
In equation 3x² – 10x + 3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 2

Question 10.
Solution:
In equation 5x² + 13x + k = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 3

Question 11.
Solution:
In equation kx² + 2x + 3k = 0
Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -2 }{ k }\)
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 4

Question 12.
Solution:
Roots of an equation are 5, -2
Sum of roots (S) = 5 – 2 = 3
and product (P) = 5 x (-2) = -10
Equation will be
x² – (S)x + (P) = 0
⇒ x² – 3x – 10 = 0 (b)

Question 13.
Solution:
Sum of roots (S) = 6
Product of roots (P) = 6
Equation will be x² – (S)x + (P) = 0
x² – 6x + 6 = 0 (a)

Question 14.
Solution:
α and β are the roots of the equation 3x² + 8x + 2 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 5
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 6

Question 15.
Solution:
In equation ax² + bx + c = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 7

Question 16.
Solution:
In equation ax² + bx + c = 0
Let α and β are the roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 8

Question 17.
Solution:
In equation 9x² + 6kx + 4 = 0, roots are equal
Let roots be α, α then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 9

Question 18.
Solution:
In equation x² + 2 (k + 2) x + 9k = 0
Roots are equal
Let α, α be the roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 10

Question 19.
Solution:
In the equation
4x² – 3kx + 1 = 0 roots are equal
Let α, α be the roots
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 11

Question 20.
Solution:
Roots of ax² + bx + c = 0, a ≠ 0 are real and unequal if D > 0
⇒ b² – 4ac > 0 (a)

Question 21.
Solution:
In the equation ax² + bx + c = 0
D = b² – 4ac > 0, then roots are real and unequal. (b)

Question 22.
Solution:
In the equation 2x² – 6x + 7 = 0
D = b² – 4ac = (-6)² – 4 x 2 x 7 = 36 – 56 = -20 < 0
Roots are imaginary (not real) (d)

Question 23.
Solution:
In equation 2x² – 6x + 3 = 0
D = b² – 4ac = (-6)² – 4 x 2 x 3 = 36 – 24 = 12 > 0
Roots are real, unequal and irrational, (b)

Question 24.
Solution:
In equation 5x² – kx + 1 = 0
D = b² – 4ac = (-k)² – 4 x 5 x 1 = k² – 20
Roots are real and distinct
D > 0
⇒ k² – 20 > 0
⇒ k² > 20
⇒ k > √±20
⇒ k > ±2√5
⇒ k > 2√5 or k < -2√5 (d)

Question 25.
Solution:
In equation x² + 5kx + 16 = 0
D = b² – 4ac = (5k)² – 4 x 1 x 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 12

Question 26.
Solution:
The equation x² – kx + 1 = 0
D = b2 – 4ac = (-k)² – 4 x 1 x 1 ⇒ k² – 4
Roots are not real
D < 0
⇒ k² – 4 < 0
⇒ k² < 4
⇒k < (±2)²
⇒ k < ±2
-2 < k < 2 (c)

Question 27.
Solution:
In the equation kx² – 6x – 2 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 13

Question 28.
Solution:
Let the number be = x
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 14

Question 29.
Solution:
Perimeter of a rectangle = 82 m
and Area = 400
Let breadth (b) = x, then
Length = \(\frac { P }{ 2 }\) – x = \(\frac { 82 }{ 2 }\) – x = 41 – x
Area = lb
400 = x (41 – x) = 41x – x²
⇒ x² – 41x + 400 = 0
⇒ x² – 25x – 16x + 400 = 0
⇒ x (x – 25) – 16(x – 25) = 0
⇒ (x – 25) (x – 16) = 0
Either, x – 16 = 0, then x = 16
or x – 25 = 0, then x = 25
Breadth = 16 m (c)

Question 30.
Solution:
Let breadth of a rectangular field = x m
Then length = (x + 8) m
and area = 240 m²
x (x + 8) = 240
⇒ x² + 8x – 240 = 0
⇒ x² + 20x – 12x – 240 = 0
⇒ x (x + 20) – 12 (x + 20) = 0
⇒ (x + 20) (x – 12) = 0
Either, x + 20 = 0, then x = -20 which is not possible being negative,
or x – 12 = 0, then x = 12
Breadth = 12 m (c)

Question 31.
Solution:
2x² – x – 6 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 15

Very-Short-Answer Questions
Question 32.
Solution:
Sum of two natural numbers = 8
Let first number – x
Then second number = 8 – x
According to the condition,
x (8 – x) = 15
⇒ 8x – x² = 15
⇒ x² – 8x + 15 = 0
⇒ x² – 3x – 5x + 15 = 0
⇒ x(x – 3) – 5(x – 3) = 0
⇒ (x – 3)(x – 5) = 0
Either, x – 3 = 0, then x = 3
or x – 5 = 0, then x = 5
Natural numbers are 3, 5

Question 33.
Solution:
x = -3 is a solution of equation x² + 6x + 9 = 0 Then it will satisfy it
LHS = x² + 6x + 9 = (-3)² + 6(-3) + 9 = 9 – 18 + 9 = 0 = RHS

Question 34.
Solution:
3x² + 13x + 14 = 0
If x = -2 is its root then it will satisfy it
LHS = 3(-2)² + 13(-2) + 14 = 3 x 4 – 26 + 14 = 12 – 26 + 14 = 26 – 26 = 0 = RHS

Question 35.
Solution:
x = y is a solution of equation 3x² + 2kx – 3 = 0, then it will satisfy it
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 17

Question 36.
Solution:
2x² – x – 6 = 0
⇒ 2x² – 4x + 3x – 6 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 18

Question 37.
Solution:
3√3 x² + 10x + √3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 19
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 20

Question 38.
Solution:
Roots of the quadratic equation 2x² + 8x + k = 0 are equal
Let α, α be its roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 21

Question 39.
Solution:
px² – 2√5 px + 15 = 0
Here, a = p, b = 2√5 p, c = 15
D = b² – 4ac = (-2√5 p)² – 4 x p x 15 = 20p² – 60p
Roots are equal.
D = 0
⇒ 20p² – 60p = 0
⇒ p² – 3p = 0
⇒ p (p – 3) = 0
p – 3 = 0, then p = 3

Question 40.
Solution:
1 is a root of equation
ay² + ay + 3 = 0 and y² + y + b = 0
Then a(1)² + a(1) + 3 = 0
⇒ a + a + 3 = 0
⇒ 2a + 3 = 0
⇒ a = \(\frac { -3 }{ 2 }\)
and 1 + 1 + b = 0
⇒ 2 + b = 0
⇒ b = -2
ab = \(\frac { -3 }{ 2 }\) x (-2) = 3
Hence, ab = 3

Question 41.
Solution:
The polynomial is x² – 4x + 1
Here, a = 1, b = -4, c = 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 22

Question 42.
Solution:
In the quadratic equation 3x² – 10x + k = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 23

Question 43.
Solution:
The quadratic equation is
px (x – 2) + 6 = 0
⇒ px² – 2px + 6 = 0
D = b² – 4ac = (-2p)² – 4 x p x 6 = 4p² – 24p
Roots are equal
D = 0
Then 4p² – 24p = 0
⇒ 4p (p – 6) = 0
⇒ p – 6 = 0
⇒ p = 6

Question 44.
Solution:
x² – 4kx + k = 0
D = b² – 4ac = (-4k)² – 4 x 1 x k = 16k² – 4k
Roots are equal
D = 0
16k² – 4k = 0
⇒ 4k (4k – 1) = 0
⇒ 4k – 1 = 0
⇒ k = \(\frac { 1 }{ 4 }\)

Question 45.
Solution:
9x² – 3kx + k = 0
D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k
Roots are equal
D = 0
9k² – 36k = 0
9k (k – 4) = 0
Either, 9k = 0, then k = 0
or (k – 4) = 0 ⇒ k = 4
k = 0, 4

Short-Answer Questions
Question 46.
Solution:
x² – (√3 + 1) x + √3 = 0
D = b² – 4ac
= [-(√3 + 1)]² – 4 x 1 x √3
= 3 + 1 + 2√3 – 4√3
= 4 + 2√3 – 4√3
= 4 – 2√3
= 3 + 1 – 2√3
= (√3 – 1)²
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 24

Question 47.
Solution:
2x² + ax – a² = 0
D = B² – 4AC = a² – 4 x 2(-a)² = a² + 8a² = 9a²
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 25

Question 48.
Solution:
3x² + 5√5 x – 10 = 0
D = b² – 4ac = (5√5)² – 4 x 3 x (-10)
= 125 + 120 = 245 = 49 x 5 = (7√5)²
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 26
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 27

Question 49.
Solution:
√3 x² + 10x – 8√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 28

Question 50.
Solution:
√3 x² – 2√2 x – 2√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 29
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 30

Question 51.
Solution:
4√3 x² + 5x – 2√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 31
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 32

Question 52.
Solution:
4x² + 4bx – (a² – b²) = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 33

Question 53.
Solution:
x² + 5x – (a² + a – 6) = 0
a² + a – 6 = a² + 3a – 2a – 6 = a(a + 3) – 2(a + 3) = (a + 3)(a – 2)
and 6 = (a + 3) – (a – 2)
x² + (a + 3)x – (a – 2)x – (a + 3) (a – 2) = 0
x (x + a + 3) – (a – 2) (x + a + 3) = 0
(x + a + 3)(x – a + 2) = 0
Either, x + a + 3 = 0, then x = -(a + 3)
or x – a + 2 = -0 then x = (a – 2)
x = -(a + 3) or (a – 2)

Question 54.
Solution:
x² + 6x – (a² + 2a – 8) = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 34
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 35

Question 55.
Solution:
x² – 4ax + 4a² – b² = 0
4a² – b² = (2a)² – (b)² = (2a + b)(2a – b) – 4ax = (2a + b)x + (2a – b)x
x² – 4ax + 4a² – b² = x² – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0
⇒ x (x – 2a – b) – (2a – b)(x – 2a – b) = 0
⇒ (x – 2a – b)(x – 2a + b)
Either, x – 2a – b = 0, then x = 2a + b
or x – 2a + b = 0, then x = 2a – b
Hence, x = (2a + b) or (2a – b)

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Test Yourself are helpful to complete your math homework.

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