## RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
- RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

**Answer each of the following questions either in one word or one sentence or as per requirement of the question :**

**Question 1.**

Write the value of k for which the quadratic equation x² – kx + 4 = 0 has equal roots.

**Solution:**

x² – kx + 4 = 0

Here a = 1, b = – k, c = 4

Discriminant (D) = b² – 4ac

= (-k)² – 4 x 1 x 4 = k² – 16

The roots are equal

D = 0 => k² – 16 = 0

=> (k + 4) (k – 4) = 0.

Either k + 4 = 0, then k = – 4

or k – 4 = 0, then k = 4

k = 4, -4

**Question 2.**

What is the nature of roots of the quadratic equation 4x² – 12x – 9 = 0 ?

**Solution:**

4x² – 12x – 9 = 0

Here a = 4, b = -12, c = – 9

Discriminant (D) = b² – 4ac = (-12)² – 4 x 4 x (-9)

= 144 + 144 = 288

D > 0

Roots are real and distinct

**Question 3.**

If 1 + √2 is a root of a quadratic equation with rational co-efficients, write its other root.

**Solution:**

The roots of the quadratic equation with rational co-efficients are conjugate

The other root will be 1 – √2

**Question 4.**

Write the number of real roots of the equation x² + 3 |x| + 2 = 0.

**Solution:**

**Question 5.**

Write the sum of the real roots of the equation x² + |x| – 6 = 0.

**Solution:**

**Question 6.**

Write the set of values of ‘a’ for which the equation x² + ax – 1 = 0, has real roots.

**Solution:**

x² + ax – 1=0

Here a = 1, b = a, c = -1

D = b² – 4ac = (a)² – 4 x 1 x (-1) = a² + 4

Roots are real

D ≥ 0 => a² + 4 ≥ 0

For all real values of a, the equation has real roots.

**Question 7.**

In there any real value of ‘a’ for which the equation x² + 2x + (a² + 1) = 0 has real roots ?

**Solution:**

x² + 2x + (a² + 1) = 0

D = (-b)² – 4ac = (2)² – 4 x 1 (a² + 1) = 4 – 4a² – 4 = – 4a²

For real value of x, D ≥ 0

But – 4a² ≤ 0

So it is not possible

There is no real value of a

**Question 8.**

Write the value of λ, for which x² + 4x + λ is a perfect square.

**Solution:**

In x² + 4x + λ

a = 1, b = 4, c = λ

x² + 4x + λ will be a perfect square if x² + 4x + λ = 0 has equal roots

D = b² – 4ac = (4)² – 4 x 1 x λ = 16 – 4λ

D = 0

=> 16 – 4λ = 0

=> 16 = 4A

=> λ = 4

Hence λ = 4

**Question 9.**

Write the condition to be satisfied for which equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 have equal roots.

**Solution:**

In ax² + 2bx + c = 0

**Question 10.**

Write the set of values of k for which the quadratic equation has 2x² + kx – 8 = 0 has real roots.

**Solution:**

In 2x² + kx – 8 = 0

D = b²- 4ac = (k)² – 4 x 2 x (-8) = k² + 64

The roots are real

D ≥ 0

k² + 64 ≥ 0

For all real values of k, the equation has real roots.

**Question 11.**

Write a quadratic polynomial, sum of whose zeros is 2√3 and their product is 2.

**Solution:**

Sum of zeros = 2√3

and product of zeros = 2

The required polynomial will be

**Question 12.**

Show that x = – 3 is a solution of x² + 6x + 9 = 0 **(C.B.S.E. 2008)**

**Solution:**

The given equation is x² + 6x + 9 = 0

If x = -3 is its solution then it will satisfy it

L.H.S. = (-3)² + 6 (-3) + 9 = 9 – 18 + 9 = 18 – 18 = 0 = R.H.S.

Hence x = – 3 is its one root (solution)

**Question 13.**

Show that x = – 2 is a solution of 3x² + 13x + 14 = 0. **(C.B.S.E. 2008)**

**Solution:**

The given equation is 3x² + 13x + 14 = 0

If x = – 2 is its solution, then it will satisfy it

L.H.S. = 3(-2)² + 13 (- 2) + 14 =3 x 4 – 26 + 14

= 12 – 26 + 14 = 26 – 26 = 0 = R.H.S.

Hence x = – 2 is its solution

**Question 14.**

Find the discriminant of the quadratic equation 3√3 x² + 10x + √3 =0. **(C.B.S.E. 2009)**

**Solution:**

**Question 15.**

If x = , is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k. **[CBSE 2015]**

**Solution:**

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS are helpful to complete your math homework.

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