## Online Education for RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself

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

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10B
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10C
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10E
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Test Yourself

**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 = x^{3}

It is not a quadratic equation.

(c) x² + \(\frac { 1 }{ { x }^{ 2 } }\) = 5

⇒ x^{4} + 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) x^{3} – x² = (x – 1)^{3}

⇒ x^{3} – x² = x^{3} – 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

**Question 9.**

**Solution:**

In equation 3x² – 10x + 3 = 0

**Question 10.**

**Solution:**

In equation 5x² + 13x + k = 0

**Question 11.**

**Solution:**

In equation kx² + 2x + 3k = 0

Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -2 }{ k }\)

**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

**Question 15.**

**Solution:**

In equation ax² + bx + c = 0

**Question 16.**

**Solution:**

In equation ax² + bx + c = 0

Let α and β are the roots, then

**Question 17.**

**Solution:**

In equation 9x² + 6kx + 4 = 0, roots are equal

Let roots be α, α then

**Question 18.**

**Solution:**

In equation x² + 2 (k + 2) x + 9k = 0

Roots are equal

Let α, α be the roots, then

**Question 19.**

**Solution:**

In the equation

4x² – 3kx + 1 = 0 roots are equal

Let α, α be the roots

**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

**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

**Question 28.**

**Solution:**

Let the number be = x

**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

**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

**Question 36.**

**Solution:**

2x² – x – 6 = 0

⇒ 2x² – 4x + 3x – 6 = 0

**Question 37.**

**Solution:**

3√3 x² + 10x + √3 = 0

**Question 38.**

**Solution:**

Roots of the quadratic equation 2x² + 8x + k = 0 are equal

Let α, α be its roots, then

**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

**Question 42.**

**Solution:**

In the quadratic equation 3x² – 10x + k = 0

**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

**Question 47.**

**Solution:**

2x² + ax – a² = 0

D = B² – 4AC = a² – 4 x 2(-a)² = a² + 8a² = 9a²

**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

**Question 49.**

**Solution:**

√3 x² + 10x – 8√3 = 0

**Question 50.**

**Solution:**

√3 x² – 2√2 x – 2√3 = 0

**Question 51.**

**Solution:**

4√3 x² + 5x – 2√3 = 0

**Question 52.**

**Solution:**

4x² + 4bx – (a² – b²) = 0

**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

**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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