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

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

**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

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

**Question 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 x 1 x k = 16 – 4k

Roots are real and distinct

D > 0

=> 16 – 4k > 0

=> 16 > 4k

=> 4 > k

=> k < 4

**Question 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 x 1 x 1 = a² – 4

Roots are distinct

D > 0

=> a² – 4 > 0

=> a² > 4

=> a² > (2)²

=> |a| > 2

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**Question 3.**

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

(a) ± \(\frac { 2 }{ 3 }\)

(b) ± \(\frac { 3 }{ 2 }\)

(c) 0

(d) ± 3

**Solution:**

**(a)**

**Question 4.**

If ax^{2} + bx + c = 0 has equal roots, then c =

**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 = \(\frac { { b }^{ 2 } }{ 4a }\)

**Question 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 x a x 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

**Question 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

**Question 7.**

**Solution:**

**(b)**

Which is not possible

x = 3 is correct

**Question 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)**

**Question 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)**

**Question 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 = \(\frac { 2ac }{ a + c }\)

(d) b = ac

**Solution:**

**(b)**

**Question 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)**

**Question 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

**Question 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)**

**Question 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

**Question 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

**Question 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)**

**Question 17.**

If the sum of the roots of the equation x^{2} – x = λ (2x – 1) is zero, then λ =

(a) -2

(b) 2

(c) – \(\frac { 1 }{ 2 }\)

(d) \(\frac { 1 }{ 2 }\)

**Solution:**

**(c)**

**Question 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)**

**Question 19.**

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

**Solution:**

**(a)**

**Question 20.**

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

**Solution:**

**(b)**

**Question 21.**

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

**Solution:**

**(b)**

**Question 22.**

If sin α and cos α 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)**

**Question 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)**

**Question 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)

**Question 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 x 1 + b = 0

=> b = -2

Now a + b = 1 – 2 = -1

**Question 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 x (-2) = – 4

Equation will be

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

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

=> x^{2} – 4 = 0

**Question 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)**

**Question 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 = \(\frac { c }{ a }\) = \(\frac { 4 }{ 2 }\) = 2

Second root = \(\frac { 2 }{ 2 }\) = 1

**Question 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 = \(\frac { c }{ a }\) = \(\frac { 3 }{ 1 }\) = 3

Second root = \(\frac { 3 }{ 1 }\) = 3

**Question 30.**

If one root of the equation 4x^{2} – 2x + (λ – 4) = 0 be the reciprocal of the other, then λ =

(a) 8

(b) -8

(c) 4

(d) -4

**Solution:**

**(a)**

**Question 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) – \(\frac { 1 }{ 2 }\)

(c) 6

(d) -3 **[CBSE 2012]**

**Solution:**

**(a)**

**Question 32.**

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

(a) 6, – \(\frac { 1 }{ 6 }\)

(b) 36, -36

(c) 6, -6

(d) \(\frac { 3 }{ 4 }\) , – \(\frac { 3 }{ 4 }\) **[CBSE 2014]**

**Solution:**

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

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

Now D = b^{2} – 4ac = (4k)^{2} – 4 x 16 x 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

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