Q1. If the roots of the quadratic equation − ax^{2} + bx + c = 0 are equal then show that b^{2} = 4ac.
Sol. ∵ For equal roots, we have
b^{2} − 4ac =0
∴ b^{2} = 4ac
Q2. Find the value of ‘k’ for which the quadratic equation kx^{2} − 5x + k = 0 have real roots.
Sol. Comparing kx^{2} − 5x + k = 0 with ax^{2} + bx + c = 0, we have:
a = k
b = − 5
c = k
∴ b^{2} − 4ac =(− 5)^{2} − 4 (k) (k)
= 25 − 4k^{2}
For equal roots, b^{2} − 4ac = 0
∴ 25 − 4k^{2} = 0
⇒ 4k^{2} = 25
⇒ k^{2 }= 25/4
⇒
∴ b^{2} = 4ac
Q3. If 2 is a root of the equation x^{2} + kx + 12 = 0 and the equation x^{2} + kx + q = 0 has equal roots, find the value of q.
Sol. Since, 2 is a root of x^{2} + kx + 12 = 0
∴ (2)^{2} + k(2) + 12 = 0
or 4 + 2k + 12 = 0
⇒ 2k = −16 or k = − 8
Roots of x^{2} + kx + q = 0 are equal
∴ k^{2} − 4(1) (q) = 0 ^{or }k^{2} − 4q = 0
But k = −8, so (−8)^{2} = 4q or q = 16
Q4. If − 4 is a root of the quadratic equation x^{2} + px − 4 = 0 and x^{2} + px + k = 0 has equal roots, find the value of k.
Sol. ∵ (–4) is a root of x^{2 }+ px − 4 = 0
∴ (− 4)^{2} + p (− 4) = 0
⇒ 16 − 4p − 4= 0
⇒ 4p = 12 or p = 3
Now, x^{2} + px + k = 0
⇒ x^{2} + 3x + k = 0 [∵ p = 3]
Now, a = 1, b = 3 and c = + k
∴ b^{2} − 4ac = (3)^{2} − 4 (1) (k)
= 9 − 4k
For equal roots, b^{2} − 4ac = 0
⇒ 9 − 4k =0 ⇒ 4k = 9
⇒ k = 9/4
Q5. If one root of the quadratic equation 2x^{2} − 3x + p = 0 is 3, find the other root of the quadratic equation. Also, find the value of p.
Sol. We have: 2x^{2} − 3x + p = 0 ...(1)
∴ a = 2, b = − 3 and c = p
Since, the sum of the roots = b/a
∵ One of the roots = 3
∴ The other root
Now, substituting x = 3 in (1), we get
2 (3)^{2} − 3 (3) + p =0
⇒ 18 − 9 + p = 0
⇒ 9 + p = 0
⇒ p = − 9
Q6. If one of the roots of x^{2} + px − 4 = 0 is − 4 then find the product of its roots and the value of p.
Sol. If − 4 is a root of the quadratic equation,
x^{2} + px − 4=0
∴ (− 4)^{2} + (− 4) (p) − 4 = 0
⇒ 16 − 4p − 4 = 0
⇒ 12 − 4p = 0
⇒ p = 3
Now, in ax^{2} + bx + c = 0, the product of the roots = c/a
∴ Product of the roots in x^{2} − px − 4= 0
= 4/1 = 4
Q7. For what value of k, does the given equation have real and equal roots? (k + 1) x^{2} − 2 (k − 1) x + 1 = 0.
Sol. Comparing the given equation with ax^{2} + bx + c = 0, we have:
a = k + 1
b = − 2 (k − 1)
c = 1
For equal roots, b^{2} − 4ac = 0
∴ [− 2 (k − 1)]^{2 }− 4 (k + 1) (1) = 0
⇒ 4 (k − 1)^{2 } − 4 (k + 1) = 0
⇒ 4 (k^{2 } + 1 − 2k) − 4k − 4 = 0
⇒ 4k^{2 } + 4 − 8k − 4k − 4 = 0
⇒ 4k^{2 } − 12k = 0
⇒ 4k (k − 3) = 0
⇒ k = 0 or k = 3
Q8. Using quadratic formula, solve the following quadratic equation for x:
x^{2} − 2ax + (a^{2} − b^{2}) = 0
Sol. Comparing x^{2} − 2ax + (a^{2} − b^{2}) = 0, with ax^{2} + bx + c = 0, we have:
a = 1, b = − 2a, c = a^{2} − b^{2}
∴ x =( a + b) or x = (a − b)
Q9. If one of the roots of the quadratic equation 2x^{2} + kx − 6 = 0 is 2, find the value of k. Also, find the other root.
Sol. Given equation:
2x^{2} + kx − 6= 0
one root = 2
Substituting x = 2 in 2x^{2 }+ kx − 6 = 0
We have:
2 (2)^{2} + k (2) − 6= 0
⇒ 8 + 2k − 6= 0
⇒ 2k + 2 = 0 ⇒ k = − 1
∴ 2x^{2} + kx − 6 = 0 ⇒ 2x^{2} − x − 6 = 0
Sum of the roots = b/a = 1/2
∴ other root
= 3/2
Q10. Determine the value of k for which the quadratic equation 4x^{2} − 4kx + 1 = 0 has equal roots.
Sol. We have:
4x^{2} − 4kx + 1 = 0
Comparing with ax^{2} + bx + c = 0,
we have
a = 4, b = − 4k and c = 1
∴ b^{2} − 4ac =(− 4k)^{2} − 4 (4k) (1)
= 16k^{2} − 16
For equal roots
b^{2} − 4ac = 0
∴ 16k^{2} − 16 = 0
⇒ 16k^{2} = 16 ⇒ k^{2} = 1
⇒ k = ± 1
Q11. For what value of k, does the quadratic equation x^{2} − kx + 4 = 0 have equal roots?
Sol. Comparing x^{2} − kx + 4 = 0 with ax^{2} + bx + c = 0, we get
a = 1
b = − k
c = 4
∴ b^{2} − 4ac =(− k)^{2} − 4 (1) (4) = k^{2} − 16
For equal roots,
b^{2} − 4ac =0
⇒ k^{2} − 16 = 0
⇒ k^{2} = 16
⇒ k = ± 4
Q12. What is the nature of roots of the quadratic equation 4x^{2} − 12x + 9 = 0?
Sol. Comparing 4x^{2} − 12x + 9 = 0 with ax^{2} + bx + c = 0 we get
a = 4
b = − 12
c = 9
∴ b^{2} − 4ac =(− 12)^{2} − 4 (4) (9)
= 144 − 144 = 0
Since b^{2} − 4ac = 0
∴ The roots are real and equal.
Q13. Write the value of k for which the quadratic equation x^{2} − kx + 9 = 0 has equal roots.
Sol. Comparing x^{2 }− kx + 9 = 0 with ax^{2} + bx + c = 0, we get
a = 1
b = − k
c = 9
∴ b^{2} − 4ac = (− k)^{2} − 4 (1) (9)
= k^{2} − 36
For equal roots, b^{2} − 4ac = 0
⇒ k^{2} − 36 = 0 ⇒ k^{2} = 36
⇒ k = ± 6
Q14. For what value of k are the roots of the quadratic equation 3x^{2} + 2kx + 27 = 0 real and equal?
Sol. Comparing 3x^{2} + 2 kx + 27 = 0 with ax^{2} + bx + c = 0, we have:
a = 3
b = 2k
c = 27
∴ b^{2} − 4ac = (2k)^{2} − 4 (3) (27)
= 4k^{2} − (12 × 27)
For the roots to be real and equal
b^{2} − 4ac = 0
⇒ 4k^{2} − (12 × 17) = 0
⇒ 4k^{2} = 12 × 27
⇒
⇒ k = ± 9
Q15. For what value of k are the roots of the quadratic equation kx^{2} + 4x + 1 = 0 equal and real?
Sol. Comparing kx^{2} + 4x + 1 = 0, with ax^{2} + bx + c = 0, we get
a = k
b = 4
c = 1
∴ b^{2} − 4ac = (4)^{2 }− 4 (k) (1)
= 16 − 4k
For equal and real roots, we have
b^{2} − 4ac =0
⇒ 16 − 4k = 0
⇒ 4k = 16
⇒ k = 16/4 = 4
Q16. For what value of k does (k − 12) x^{2} + 2 (k − 12) x + 2 = 0 have equal roots?
Sol. Comparing (k − 12) x^{2} + 2 (k − 12) x + 2 = 0 with ax^{2} + bx + c = 0, we have:
a = (k − 12)
b = 2 (k − 12)
c = 2
∴ b^{2} − 4ac = [2 (k − 12)]^{2} − 4 (k − 12) (2)
= 4 (k − 12)^{2} − 8 (k − 12)
= 4 (k − 12) [k − 12 − 2]
= 4 (k − 12) (k − 14)
For equal roots,
b^{2} − 4ac =0
⇒ 4 (k − 12) [k − 14] = 0
⇒ Either 4 (k − 12) = 0 ⇒ k = 12
or k − 14 = 0 ⇒ k = 14
But k = 12 makes k − 12 = 0 which is not required
∴ k ≠ 12
⇒ k = 14
Q17. For what value of k does the equation 9x^{2} + 3kx + 4 = 0 has equal roots?
Sol. Comparing 9x^{2} + 3kx + 4 = 0 with ax^{2} + bx + c = 0, we get
a = 9
b = 3k
c = 4
∴ b^{2} − 4ac =(3k)^{2} − 4 (9) (4)
= 9k^{2} − 144
For equal roots,
b^{2} − 4ac = 0
⇒ 9k^{2} − 144 = 0
⇒9 k^{2} = 144
⇒
⇒ k = ± 4
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