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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 ax2+bx+ c =0 are real and unequal, then b2−4ac <0 . Is it true?

Sol. No. For real and unequal roots, b2−4ac > 0

2. If ax2+bx+ c =0 has equal roots, then b2= 4ac. Is it true?

Sol. Yes, For equal roots discriminant = 0.

⇒ b2−4ac =0 ⇒ b2= 4ac3. Find the discriminant of

4. Find the discriminant of

=   8+ 24= 32

 5. If 4a2x2−4abx+k  =0 has equal roots, then find the value of k.

 Sol. We have, 4a2x2−4abx+k =0

For equal roots b2−4ac=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) x2 + x – 5 = 0

(c) x2 + 3x + 2 √2 = 0 (d) 5x2 – 3x + 1 = 0

 Sol. (b) Roots of quadratic equation will be real and distinct, if b2 – 4ac >0.

So, only in (b), x2 + x – 5

b2 – 4ac = (1)2 – 4.1.(–5) = 1 + 20 = 21 > 0

8. If the discriminant of 3x2 + 2x + a = 0 is double the discriminant of x2 – 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 b2ac = 0. Is it true?

Sol. We have, abx2 + (b2–ac) x – bc= 0 has equal roots then

For equal roots, discriminant = 0

12. If p2x2+ (p2+q2)x+ q2= 0 has equal roots, then find p2−q2 is equal to.

Sol. We have, p2x2+ (p2+q2)x+ q2= 0 = ( )

Since it has equal roots, therefore, b2 – 4ac = 0

13. If the quadratic equation 2x2− kx + 3=0 has equal roots, then find the value of k.

Sol. Since, have equal roots 2x2− kx + 3=0

⇒   b2 – 4ac = 0

(−k)2–4 x 2 x 3=0

K2= 24k

⇒   k= ± 2√6.

14. If 2x2 + k + 3/2 = 0 has equal roots, then find the value of k.

Sol. If 2x2 + k + 3/2 = 0 has equal roots, then b2 – 4ac = 0

15. Find the value of k, for which the quadratic equation 4x2 + 4 √3 x + k = 0 has equal roots.

Sol. For equal roots, b2 – 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, b2 – 4ac > 0

⇒ (–4)2 – 4.1.p > 0 ⇒ p < 4

17. Write the nature of roots of quadratic equation 4x2+4√3 x + 3 = 0

Sol. We have, 4x2+4√3 x + 3 = 0

D = b2 – 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 = b2 – 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

x2 + 1– 2x + 2x + 2 = 0

⇒ x2 + 3 = 0 ⇒ x2 + 0. x + 3 = 0

Now, D = b2 – 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 = 2x2 – 13x + 15 ⇒ 2x2 – 16x + 2 3 = 0

Now, D = b2 – 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) x2 + 1/ x2 = 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) x2 + 1/ x2 = 4

⇒   x4 – 4 x2 + 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)x2 + 2(k – 12)x + 2 = 0 have equal roots ?

Sol. D = b2 – 4 ac = [2(k – 12)]2 – 4(k – 12) (2)

= 4(k – 12)2 – 8(k – 12) = 4(k2 + 144 – 24k) – 8(k – 12)

= 4k2 + 576 – 96k – 8 k + 9 6 = 4 k2 – 104k + 672.

For equal roots,

D = 0

⇒ 4k2 – 104k + 672 = 0

⇒ 4(k2 – 26k + 168) = 0

⇒ k2 – 26k + 168 = 0

⇒ k2 – 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 x2 + 6x + 9 = 0.

Sol. LHS = x2 + 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) 3x2 + 5x + 6 = 0.

Sol. (i) We have,

a =√3, b = (−2√ 2)2 and c = −2√3

Now, D =b2 – 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) 4x2 + 7x + 2 = 0; (ii) x2 + 2x + 3 = 0.

Sol. (i) We have,

a = 4, b = 7 and c = 2

Now, D = b2 – 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 = b2 – 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) y2 – 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 = b2 – 4 ac

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

(ii) We have,

a = 2, b = – 4 and c = 3

Now, D = b2 – 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 Px2 + 8x – 2 = 0 has real roots.

Sol. We have,

a = P, b = 8 and c = – 2

Now, D = b2 – 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 Kx2 – 2x + 2 = 0 has real roots ?

Sol. We have,

a = K, b = –2 and c = 2

Now, D = b2 – 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 = b2 – 4 ac = (P)2 – 4 (2) (3) = P2 – 24

For real and equal roots,

D = 0 ⇒ P2 – 24 = 0

⇒ P2 = 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

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