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

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 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

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 = 4b²
⇒ b² = 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

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 9x² + 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 ax² + bx + c = 0 has equal roots, then c =
(a) -b/2a
(b) b/2a

(c) -b²/2a

(d) b²/2a

Solution

(d) In the equation ax² + bx + c = 0
D = b²– 4ac
Roots are equal
D = 0
⇒ b²– 4ac = 0
⇒ 4ac = b²
⇒ c = b²/4a

5. If the equation ax² + 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 ax² + 2x + a = 0
D = b²– 4ac = (2)²– 4×a× a = 4 – 4a²
Roots are real and equal
D = 0
⇒ 4 – 4a² = 0
⇒ 4 = 4a²
⇒ 1 = a²
⇒ a² = 1
⇒ a² = (±1)²
⇒ a = ±1

6. The positive value of k for which the equation x² + kx + 64 = 0 and x²– 8x + k = 0 will both have real roots, is
(a) 4
(b) 8
(c) 12
(d) 16
Solution
(d) In the equation x² + kx + 64 = 0
a = 1, b = k, c = 64
D = b² - 4ac = k² - 4×1×64
⇒ k² - 256
∵ The roots are real.
∴ D ≥ 0
⇒ k² - 256 ≥ 0
⇒ k² ≥ 256
⇒ k² ≥ (±16)²
⇒ k ≥ 16 ...(1)
Only positive value is taken.
Now in second equation,
x² - 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)

which is not possible
∴ x = 3 is correct.

8. If 2 is a root of the equation x² + bx + 12 = 0 and the equation x² + bx + q = 0 has equal roots, then q =
(a) 8
(b) – 8
(c) 16
(d) -16
Solution
(c) 16

9. If the equation (a² + b²) x²– 2 (ac + bd) x + c² + d² = 0 has equal roots, then

(a) ab = cd
(b) ad = bc
(c) ad = √bc
(d) ab = √cd

Solution
(b)
In the equation,

(a² + b²)x² - 2(ac + bd)x + (c² + d²) = 0
D = B² - 4AC
= [-2(ac bd)]2 - 4 (a² + b) (c² + d²)
= 4[a²c² + b²d² + 2abcd] - 4[a²c² + a²d² + b²c² + b²d²]
= 4a²c² + 4b²d² + 8abcd - 4a²c² - 4a²d² - 4b²c² - 4b²d²
= 8abcd - 4a²d² - 4b²c²
= -4[a²d² + b²c² - 2abcd]
= -4 (ad - bc)²
∵ Roots are equal
∴ D = 0
-4 (ad - bc)² = 0
⇒ ad - bc = 0
⇒ ad = bc