# RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations MCQs and VSAQs

**Answer each of the following questions either in one word or one sentence or as per requirement of the 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 – 4×1×4 = 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

**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 – 4×4×(-9)

= 144 + 144 = 288

D > 0

Roots are real and distinct.

**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.

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

**Solution**

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

Solution

Solution

**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 – 4×1×(-1) = a² + 4

Roots are real

D ≥ 0 ⇒ a² + 4 ≥ 0

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

**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.

**8. Write the value of Î», for which x² + 4x + Î» is a perfect square.Solution**

In x² + 4x + Î»

a = 1, b = 4, c = Î»

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

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

D = 0

⇒ 16 – 4Î» = 0

⇒ 16 = 4A

⇒ Î» = 4

Hence Î» = 4

**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

Roots are equal.

4ac = 4b^{2}⇒ b^{2} = ac

∴ The required condition is b2 = ac.

**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 – 4×2×(-8) = k² + 64

The roots are real

D ≥ 0

k² + 64 ≥ 0

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

**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

**12. Show that x = –3 is a solution of x² + 6x + 9 = 0.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)

**13. Show that x = –2 is a solution of 3x² + 13x + 14 = 0.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 = 3×4 – 26 + 14

= 12 – 26 + 14 = 26 – 26 = 0 = R.H.S.

Hence x = – 2 is its solution

**14. Find the discriminant of the quadratic equation 3√3 x² + 10x + √3 =0.**

**Solution**

= 100 - 12×3 = 100 - 36 = 64

**15. If x = −1/2, is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k.**

Solution

Solution

**Mark the correct alternative in each of the following :**

**1. If the equation x² + 4x + k = 0 has real and distinct roots, then**

(a) k < 4

(b) k > 4

(c) k ≥ 4

(d) k ≤ 4**Solution****(a)** In the equation x² + 4x + k = 0

a = 1, b = 4, c = k

D = b² – 4ac = (4)² – 4×1×k = 16 – 4k

Roots are real and distinct

D > 0

⇒ 16 – 4k > 0

⇒ 16 > 4k

⇒ 4 > k

⇒ k < 4

**2. If the equation x² – ax + 1 = 0 has two distinct roots, then**

(a) |a| = 2

(b) |a| < 2

(c) |a| > 2

(d) None of these**Solution****(c)** In the equation x² – ax + 1 = 0

a = 1, b = – a, c = 1

D = b² – 4ac = (-a)² – 4×1×1 = a² – 4

Roots are distinct

D > 0

⇒ a² – 4 > 0

⇒ a² > 4

⇒ a² > (2)²

⇒ |a| > 2

**3. If the equation 9x**^{2}** + 6kx + 4 = 0, has equal roots, then the roots are both equal to**

(a) ± 2/3

(b) ± 3/2

(c) 0

(d) ± 3**Solution****(a)**

**4. If ax**

(a) -b/2a

^{2}+ bx + c = 0 has equal roots, then c =(b) b/2a

**Solution**

**(d)**In the equation ax

^{2}+ bx + c = 0

D = b

^{2}– 4ac

Roots are equal

D = 0

⇒ b

^{2}– 4ac = 0

⇒ 4ac = b

^{2}

⇒ c = b

^{2}/4a

**5. If the equation ax ^{2} + 2x + a = 0 has two distinct roots, if**(a) a = ±1

(b) a = 0

(c) a = 0, 1

(d) a = -1, 0

**Solution**

**(a)**In the equation ax

^{2}+ 2x + a = 0

D = b

^{2}– 4ac = (2)

^{2}– 4×a× a = 4 – 4a

^{2}

Roots are real and equal

D = 0

⇒ 4 – 4a

^{2}= 0

⇒ 4 = 4a

^{2}

⇒ 1 = a

^{2}

⇒ a

^{2}= 1

⇒ a

^{2}= (±1)

^{2}

⇒ a = ±1

**6. The positive value of k for which the equation x ^{2} + kx + 64 = 0 and x^{2} – 8x + k = 0 will both have real roots, is**(a) 4

(b) 8

(c) 12

(d) 16

**Solution**

**(d)**In the equation x

^{2}+ kx + 64 = 0

a = 1, b = k, c = 64

D = b

^{2}- 4ac = k

^{2}- 4×1×64

⇒ k

^{2}- 256

∵ The roots are real.

∴ D ≥ 0

⇒ k

^{2}- 256 ≥ 0

⇒ k

^{2}≥ 256

⇒ k

^{2}≥ (±16)

^{2}

⇒ k ≥ 16

**...(1)**

Only positive value is taken.

Now in second equation,

x

^{2}- 8x + k = 0

D = (-8)2 - 4×1×k = 64 - 4k

∵ Roots are real

∴ D ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 64 ≥ 4k

⇒ 16 ≥ k

**...(2)**

From 1 and 2,

16 ≥ k ≥ 16

⇒ k = 16

**7. **

(a) 4

(b) 3

(c) -2

(d) 3.5**Solution(c)**

∴ x = 3 is correct.

**8. If 2 is a root of the equation x ^{2} + bx + 12 = 0 and the equation x^{2} + bx + q = 0 has equal roots, then q =**(a) 8

(b) – 8

(c) 16

(d) -16

**Solution**

**(c) 16**

**9. If the equation (a**

^{2}+ b^{2}) x^{2}– 2 (ac + bd) x + c^{2}+ d^{2}= 0 has equal roots, then(a) ab = cd

(b) ad = bc

(c) ad = √bc

(d) ab = √cd

**Solution****(b)**

In the equation,

(a^{2} + b^{2})x^{2} - 2(ac + bd)x + (c^{2} + d^{2}) = 0

D = B^{2 }- 4AC

= [-2(ac bd)]2 - 4 (a^{2} + b) (c^{2} + d^{2})

= 4[a^{2}c^{2} + b^{2}d^{2} + 2abcd] - 4[a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}]

= 4a^{2}c^{2} + 4b^{2}d^{2} + 8abcd - 4a^{2}c^{2} - 4a^{2}d^{2} - 4b^{2}c^{2} - 4b^{2}d^{2}= 8abcd - 4a^{2}d^{2} - 4b^{2}c^{2}= -4[a^{2}d^{2} + b^{2}c^{2} - 2abcd]

= -4 (ad - bc)^{2}

∵ Roots are equal

∴ D = 0

-4 (ad - bc)^{2} = 0

⇒ ad - bc = 0

⇒ ad = bc

**10. If the roots of the equation (a ^{2} + b^{2}) x^{2} – 2b (a + c) x + (b^{2} + c^{2}) = 0 are equal, then:**

(a) 2b = a + c

(b) b^{2} = ac

(c) b = 2ac/a+c

(d) b = ac

**Solution**

**(b)**

In the equation,

**11. If the equation x**

^{2}– bx + 1 = 0 does not possess real roots, then(a) -3 < b < 3

(b) -2 < b < 2

(c) b > 2

(d) b < -2

**Solution****(b)**

In the equation,

∵ b < 2 and b > -2 or -2 < b

∴ -2 < b < 2

**12. If x = 1 is a common root of the equations ax ^{2} + ax + 3 = 0 and x^{2} + x + b = 0, then ab =**

(a) 3

(b) 3.5

(c) 6

(d) -3

**Solution**

**(a)** In the equation

ax^{2} + ax + 3 = 0 and x^{2} + x + b = 0

Substituting the value of x = 1, then in ax^{2} + ax + 3 = 0

**13. If p and q are the roots of the equation x**

^{2}– px + q + 0, then(a) p = 1, q = -2

(b) p = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1

**Solution**

**(a)**p and q are the roots of the equation

x

^{2}- px + q = 0,

Sum of roots = -(-p) = p

Product of roots = q

(a) If p = 1, q = -2, then equation will be

x^{2} -(s)x + p = 0

⇒ x^{2} -(1-2)x + 1×(-2) = 0

⇒ x^{2} + x - 2 = 0

(b) If p = 0, q = 1, then equation will be

x^{2} - (0 +1)x + 0×1 = 0

⇒ x^{2} - x + 0 = 0

(c) If p = -2, q = 0, then equation will be

x^{2} - (-2+0)x +(-2×0)

⇒ x^{2} + 2x + 0 = 0

(d) p = -2, q = 1, then equation will be

x^{2} - (-2+1)x + (-2×1) = 0

⇒ x^{2} + x - 2 = 0

We see that only (a) is correct.

When p = 1, q = -2

**14. If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax ^{2} + bx + 1 = 0 having real roots is**

(a) 10

(b) 7

(c) 6

(d) 12

**Solution****(b)**

ax^{2} + bx + 1 = 0

D = b^{2} – 4a = b^{2} – 4a

Roots are real.

D ≥ 0

⇒ b^{2} – 4a ≥ 0

⇒ b^{2} ≥ 4a

Here, value of b can be 2, 3 or 4

If b = 2, then a can be 1,

If b = 3, then a can be 1, 2

If b = 4, then a can be 1, 2, 3, 4

No. of equation can be 7

**15. The number of quadratic equations having real roots and which do not change by squaring their roots is**

(a) 4

(b) 3

(c) 2

(d) 1

**Solution****(c)** There can be two such quad, equations whose roots can be 1 and 0

The square of 1 and 0 remains same

No. of quad equation are 2

**16. If (a ^{2} + b^{2}) x^{2} + 2(ab + bd) x + c^{2} + d^{2} = 0 has no real roots, then**

(a) ad = bc

(b) ab = cd

(c) ac = bd

(d) ad ≠ bc

**Solution****(d)**(a

^{2}+b

^{2})x

^{2}+ 2(ab+bd)x + c

^{2}+ d

^{2}= 0

Here, A = a

^{2}+ b

^{2}, B = 2 (ab + bd), C = c

^{2}+ d

^{2}D = B

^{2}- 4AC = [2(ac + bd)]2 - 4(a

^{2}+ b

^{2}) (c

^{2}+ d

^{2})

= 4[a

^{2}c

^{2}+ b

^{2}d

^{2}+ 2abcd] - 4[a

^{2}c

^{2}+ a

^{2}d

^{2}+ b

^{2}c

^{2}+ b

^{2}d

^{2}]

= 4a

^{2}c

^{2}+ 4b

^{2}d

^{2}+ 8abcd - 4a

^{2}c

^{2}- 4a

^{2}d

^{2}- 4b

^{2}c

^{2}- 4b

^{2}d

^{2}= -4a

^{2}d

^{2}- 4b

^{2}c

^{2}+ 8abcd

= -4(a

^{2}d

^{2}+ b

^{2}c

^{2}- 2abcd)

= -4(ad - bc)

^{2}∵ Roots are not real.

∴ D < 0

∴ -4 (ad - bc)

^{2}< 0

= (ad - bc)

^{2}< 0

⇒ ad-bc < O or ad ≠ bc

**17. If the sum of the roots of the equation x ^{2} – x = Î»(2x – 1) is zero, then Î» =**

(a) -2

(b) 2

(c) –1/2

(d) 1/2

**Solution****(c)**

x^{2} – x = Î»(2x – 1)

⇒ x^{2} – x = 2Î»x – Î»

⇒ x^{2} – x - 2Î»x + Î» = 0

⇒ x^{2} – (1 + 2Î»)x + Î» = 0

**18. If x = 1 is a common root of ax ^{2} + ax + 2 = 0 and x^{2} + x + b = 0 then, ab =**

(a) 1

(b) 2

(c) 4

(d) 3

**Solution****(b)**ax

^{2}+ ax + 2 = 0

**...(i)**

x

^{2}+ x + b = 0

**...(ii)**

x = 1 is common root of equations (i) and (ii)

Then in (i),

a(1)^{2} + a×1 + 2 = 0

⇒ a + a + 2 = 0

⇒ 2a + 2 = 0

**19. The value of c for which the equation ax ^{2} + 2bx + c = 0 has equal roots is**

(a) b^{2}/a

(b) b^{2}/4a

(c) a^{2}/b

(d) a^{2}/4b

**Solution****(a)**ax

^{2}+ 2bx + c = 0

D = b

^{2 }- 4ac

**20. If x ^{2} + k (4x + k – 1) + 2 = 0 has equal roots, then k =**

(a) -2/3, 1

(b) 2/3, -1

(c) 3/2, 1/3

(d) 3/2, -1/3

**Solution**

**(b)**x

^{2}+ k (4x + k – 1) + 2 = 0

⇒ x

^{2}+ 4kx + k

^{2}– k + 2 = 0

Here, a =1, b = 4k, c = k

^{2}– k + 2

∵ D = b

^{2 }- 4ac

**21. If the sum and product of the roots of the equation kx ^{2} + 6x + 4k = 0 are equal, then k =**

(a) -3/2

(b) 3/2

(c) 2/3

(d) -2/3

**Solution**

**(b)**kx

^{2}+ 6x + 4k = 0

Here, a = k, b = 6 and c = 4k

**22. If sin Î± and cos Î± are the roots of the equations ax**

^{2}+ bx + c = 0, then b^{2}=(a) a^{2} – 2ac

(b) a^{2} + 2ac

(b) a^{2} – ac

(d) a^{2} + ac

**Solution****(b)**sin Î± and cos Î± are the roots of the equations ax

^{2}+ bx + c = 0

**23. If 2 is a root of the equation x**

^{2}+ ax + 12 = 0 and the quadratic equation x^{2}+ ax + q = 0 has equal roots, then q =(a) 12

(b) 8

(c) 20

(d) 16

**Solution****(d)**2 is a root of the equation x

^{2}+ ax + 12 = 0

**24. If the sum of the roots of the equation x**

^{2}– (k + 6) x + 2 (2k – 1) = 0 is equal to half of their product, then k =(a) 6

(b) 7

(c) 1

(d) 5

**Solution****(b)** In the quadratic equation

x^{2} – (k + 6) x + 2 (2k – 1) = 0

Here, a = 1, b = – (k + 6), c = 2 (2k – 1)

**25. If a and b are roots of the equation x ^{2} + ax + b = 0, then a + b =**

(a) 1

(b) 2

(c) -2

(d) -1

**Solution****(d)** a and b are the roots of the equation x^{2} + ax + b = 0

Sum of roots = – a and product of roots = b

Now a + b = –a

and ab = b

⇒ a = 1 **…(i)**2a + b = 0

⇒ 2×1 + b = 0

⇒ b = -2

Now, a + b = 1 – 2 = -1

**26. A quadratic equation whose one root is 2 and the sum of whose roots is zero, is**

(a) x^{2} + 4 = 0

(b) x^{2} – 4 = 0

(c) 4x^{2} – 1 = 0

(d) x^{2} – 2 = 0

**Solution****(b)** Sum of roots of a quad, equation = 0

One root = 2

Second root = 0 – 2 = – 2

and product of roots = 2×(-2) = – 4

Equation will be

x^{2} + (sum of roots) x + product of roots = 0

x^{2} + 0x + (-4) = 0

⇒ x^{2} – 4 = 0

**27. If one root of the equation ax ^{2} + bx + c = 0 is three times the other, then b^{2} : ac =**

(a) 3 : 1

(b) 3 : 16

(c) 16 : 3

(d) 16 : 1

**Solution****(c)**

Equation is ax^{2} + bx + c = 0

Let first root = É‘ then,

Second rot = 3É‘

**28. If one root of the equation 2x ^{2} + kx + 4 = 0 is 2, then the other root is**

(a) 6

(b) -6

(c) -1

(d) 1

**Solution****(d)** The given quadratic equation 2x^{2} + kx + 4 = 0

One root is 2

Product of roots = c/a = 4/2 = 2

Second root = 2/2 = 1

**29. If one root of the equation x ^{2} + ax + 3 = 0 is 1, then its other root is**(a) 3

(b) -3

(c) 2

(d) -2

**Solution****(a)** The quad, equation is x^{2} + ax + 3 = 0

One root =1

and product of roots = c/a = 3/1 = 3

Second root = 3/1 = 3

**30. If one root of the equation 4x ^{2} – 2x + (Î» – 4) = 0 be the reciprocal of the other, then Î» =**(a) 8

(b) -8

(c) 4

(d) -4

**Solution****(a)**The quadratic equation is 4x

^{2}– 2x + (Î» – 4) = 0

Let first root be É‘

**31. If y = 1 is a common root of the equations ay ^{2} + ay + 3 = 0 and y^{2} + y + b = 0, then ab equals**

(a) 3

(b) – 1/2

(c) 6

(d) -3

**Solution****(a)**y = 1

ay

^{2}+ ay + 3 = 0

**32. The values of k for which the quadratic equation 16x ^{2} + 4kx + 9 = 0 has real and equal roots are**

(a) 6, –1/6

(b) 36, -36

(c) 6, -6

(d) 3/4 , –3/4

**Solution****(c)** 16x^{2} + 4kx + 9 = 0

Here a = 16, b = 4k, c = 9

Now D = b^{2} – 4ac = (4k)^{2} – 4×16×9 = 16k^{2} – 576

Roots are real and equal

D = 0 or b^{2} – 4ac = 0

⇒ 16k^{2} – 576 = 0

⇒ k^{2} – 36 = 0

⇒ k^{2} = 36 = (± 6)^{2}

k = ± 6

k = 6, -6