**Class 10 Quadratic Equations exercise 4 extra questions-answers**

**Topic– 4: Nature of roots **

** Very Short Answer Type Questions: [1 Mark] **

**1. **If the roots of ax^{2}+bx+ c =0 are real and unequal, then b^{2}−4ac <0 .
Is it true?

**Sol. **No. For real and unequal roots, b^{2}−4ac > 0

**2. **If ax2+bx+ c =0 has equal roots, then b^{2}= 4ac. Is it true?

**Sol.
**Yes, For equal roots discriminant = 0.

⇒ b^{2}−4ac =0 ⇒ b^{2}= 4ac**3. **Find the discriminant of

**4. **Find the discriminant of

= 8+ 24= 32

**5. **If 4a^{2}x^{2}−4abx+k =0 has
equal roots, then find the value of k.

**Sol. **We
have, 4a^{2}x^{2}−4abx+k =0

For equal roots b^{2}−4ac=0

**6. **Determine the nature of the roots of the quadratic equation

Therefore, the given equation has no real roots.

**7. **Which of the following equations has two distinct real roots?

(b) x^{2} + x –
5 = 0

(c) x^{2} + 3x +
2 √2 = 0 (d) 5x^{2} – 3x +
1 = 0

**Sol. **(b) Roots of quadratic equation will
be real and distinct, if b^{2}
– 4ac >0.

So, only in (b), x^{2} + x –
5

b^{2} – 4ac =
(1)^{2} – 4.1.(–5) = 1 + 20 = 21 > 0

**8. **If the discriminant of 3x^{2} + 2x + a = 0 is double the discriminant of x^{2}
– 4x + 2 = 0, then find the value of a.

**Sol.
**(2)^{2} – 4.3.a =
2[(–4)^{2} – 4.1.2]

⇒ 4 – 12a = 2(16 – 8)

⇒ 4 – 12a = 16 ⇒ –12a = 12 ⇒ a = –1

**10. Determine the nature of the roots of **

So, roots are real and equal.

**11.
****If
the quadratic equation abx2 + (b2 – ac)x – bc = 0 has equal roots then b ^{2}ac**

**=**

**0**

**. Is it true?**

**Sol.
**We have, abx^{2}
+ (b^{2}–ac) x – bc= 0 has
equal roots then

For equal roots, discriminant = 0

**12. **If p^{2}x^{2}+
(p^{2}+q^{2})x+ q^{2}= 0 has equal roots, then find p^{2}−q^{2} is equal to.

**Sol.
**We have, p^{2}x^{2}+
(p^{2}+q^{2})x+ q^{2}= 0 = ( )

Since it has equal roots, therefore, b^{2} – 4ac = 0

**13.
**If the quadratic equation 2x^{2}− kx + 3=0 has equal roots, then find the value of k.

**Sol.
**Since, have equal roots 2x^{2}− kx +
3=0

⇒ b^{2} –
4ac = 0

(−k)^{2}–4
x 2 x 3=0

K^{2}= 24k

⇒ k= ± 2√6.

**14.
**If 2x^{2 }+ k
+ 3/2 = 0 has equal roots, then
find the value of k.

**Sol. **If 2x^{2 }+ k + 3/2 = 0 has equal roots, then b^{2} – 4ac = 0

**15.
**Find the value of k, for which the
quadratic equation 4x^{2} + 4 √3 x + k = 0 has equal roots.

**Sol.
**For equal roots, b^{2}
– 4ac = 0.

⇒ (4 √3 )^{2} – 4.4.k = 0

⇒ 48 – 16 k = 0 ⇒ k = 3

**16.
**Find the value of p for which the
quadratic equation x(x – 4) + p = 0 has real roots.

**Sol.
**For real roots, b^{2}
– 4ac > 0

⇒ (–4)2 – 4.1.p > 0 ⇒ p < 4

**17.
**Write the nature of roots of quadratic
equation 4x^{2}+4√3 x + 3 = 0

**Sol.
**We have, 4x^{2}+4√3 x + 3 = 0

D = b^{2} – 4ac = (4√3)2 – 4.4.3 =48 – 48 =0

As D = 0, the equation has real and equal roots

**Short Answer Type Questions: [2 Marks] **

**1. ****Find the nature of
the roots of the quadratic equations ****2×2 – **√**5x – 2 = 0 using the quadratic formula**

**Sol.
**We have,a = 2, b = –√**5**
and c = –2

Now,D = b^{2}
– 4ac =(–√**5**)^{2}– 4 (2) (–2) =
5 + 16 = 21

Since,D > 0

So, roots are real and distincts.

**2. Does (x – 1)**^{2}** + 2(x + 1) = 0 have real roots? Justify your answer. **

**Sol.
**After simplifying the given equation, we get

(x – 1)2 + 2(x + 1 ) = 0

x^{2} + 1–
2x + 2x + 2 = 0

⇒ x^{2} + 3 = 0 ⇒ x^{2} + 0. x + 3 =
0

Now, D = b^{2} – 4 ac = (0)^{2}
– 4(1) (3) [a = 1, b = 0, c = 3 ]

= 0 – 12 = –12

Since, D < 0. So, it has no real roots.

have a real roots ? Justify your answer.

**Sol. **We have,

3x – 8 =
2x^{2} – 13x + 15 ⇒ 2x^{2} – 16x + 2 3
= 0

Now, D = b^{2} – 4 ac = (–16)^{2}
– 4 × 2 × 23 = 256 – 184 = 72

Since, D > 0. So, roots are real.

**4. **Which of the following are quadratic equations:

(i) (x + 8) (x + 4) = x(x + 3) + 7; (ii) x^{2} + 1/ x^{2} = 4?

**Sol.
**(i) (x + 8) (x + 4) = x (x + 3) +
7

⇒ x(x + 4) + 8 (x + 4) = x2 + 3x + 7

⇒ x2 + 4x + 8x + 32 = x2 + 3x + 7

⇒ 12x – 3 x + 32 – 7 = 0

⇒ 9x + 25 = 0, which is a linear equation.

Hence, the given equation is not a quadratic equation.

(ii) x^{2} + 1/ x^{2} = 4

⇒ x^{4} – 4 x^{2} + 1 = 0 It is a
polynomial of degree 4.

Hence, the given equation is not a quadratic equation.

**5. **For what value of k does (k – 12)x^{2}
+ 2(k – 12)x + 2 = 0 have equal roots ?

**Sol. **D = b^{2} – 4 ac = [2(k – 12)]^{2} –
4(k – 12) (2)

= 4(k – 12)^{2} – 8(k – 12) = 4(k^{2} + 144 – 24k) – 8(k – 12)

= 4k^{2} + 576 – 96k – 8 k + 9 6 = 4 k^{2} – 104k + 672.

For equal roots,

D = 0

⇒ 4k^{2} – 104k + 672
= 0

⇒ 4(k^{2} – 26k + 168)
= 0

⇒ k^{2} – 26k + 168
= 0

⇒ k^{2} – 14k – 12k + 168
= 0

⇒ k(k – 14) – 12 (k – 14) = 0

⇒ (k – 12) (k – 14) = 0

⇒ k = 14 ( a ≠ 0 ⇒ k ≠ 12)

**6. **Show that x = –3 is a solution of equation x^{2} + 6x + 9 = 0.

**Sol.
**LHS = x^{2} + 6x + 9 = (–3)^{2} + 6(–3) + 9

= 9 – 18 + 9 = 18 – 18 = 0 = RHS

∴ x = –3 is a solution of given equation.

**7. **Write the discriminant of each of the following quadratic equations.

(i) ; (ii) 3x^{2} + 5x +
6 = 0.

Sol. (i) We have,

a =√3, b = (−2√ 2)^{2} and c = −2√3

Now, D =b^{2} – 4ac = (−2√ 2)^{2} −4 (−√3)(–√3) = 8+24=32

(ii) We have,

a = 3, b = 5 and c = 6

Now, D = b2 – 4 ac = (5)^{2}
– 4 (3) (6) = 25 – 72 = – 47.

**8. **Comment upon the nature of roots of the following
equations :

(i) 4x^{2} + 7x + 2 = 0; (ii) x^{2} + 2x + 3 = 0.

**Sol.
**(i) We have,

a = 4, b = 7 and c = 2

Now, D = b^{2} – 4ac = (7)^{2}
– 4 (4) (2) = 49 – 32 = 17

Since, D > 0. So, roots are real and distincts.

(ii) We have,

a = 1, b = 2 and c = 3

Now, D = b^{2} – 4ac = (2)^{2}
– 4 (1) (3) = 4 – 12 = –8

Since D < 0. So, roots are not real (imaginary).

**9. **Without solving determine whether the following equations have real roots or not.

(i) y^{2} – 2/3 +1/9 y = 0
(ii) 2×2 – 4x + 3 = 0.

**Sol.
**(i) We have,

a = 1, b =– 2/3 and c =1/9

Now, D = b^{2} – 4 ac

Since, D = 0. So, roots are real and equal.

(ii) We have,

a = 2, b = – 4 and c = 3

Now, D = b^{2} – 4 ac = (–4)^{2}
– 4 (2) (3) = 16 – 24 = –8

Since, D < 0. So, roots are not real (Imaginary).

**10.
**For what value of P, the quadratic
equation Px^{2} + 8x – 2 = 0 has real roots.

**Sol.
**We have,

a = P, b = 8 and c = – 2

Now, D = b^{2} – 4 ac = (8)^{2}
– 4 (P) (–2) = 64 + 8P

For real roots,

D ≥ 0 ⇒ 64 + 8P ≥ 0 ⇒ 8 P ≥ –64 ⇒ P ≥ –8

**11.
**For what value of K, the given equation
Kx^{2} – 2x + 2 = 0 has real roots ?

**Sol.
**We have,

a = K, b = –2 and c = 2

Now, D = b^{2} – 4 ac = (–2)^{2}
– 4 (K) (2) = 4 – 8K

For real roots,

D ≥ 0 ⇒ 4 – 8K ≥ 0

**12. **Find the value of P,
so that equation 2×2 + Px + 3 = 0 has two equal real roots.

**Sol.
**We have,

a = 2, b = P and c = 3

Now, D = b^{2} – 4 ac = (P)^{2}
– 4 (2) (3) = P^{2} – 24

For real and equal roots,

D = 0 ⇒ P^{2} – 24
= 0

⇒ P^{2} = 24 ⇒ P = ± 2** √6**

**13. **For what value of k, are the roots of the quadratic equation

real and equal ?

a = 3, b = 2k, c = 27

Since, roots are real and equal.

∴ D = 0