Question

Find value of k for which the quadratic equation 4x^2-2(k+1)+(k+4)=0 has equal roots.

Answer 1

For real and equal roots,
D = 0
b² - 4ac = 0

Here,
a = 4
b = -2(k + 1)
c = (k + 1)

=>  -2²(k + 1)² - 4 x 4 x (k + 1) = 0
=> 4(k² + 2k + 1) - 16(k + 1) = 0
=>4k² + 8k + 4 - 16k - 16 = 0
=> 4k² - 8k - 12 = 0
=> k² - 2k - 3 = 0
=> k² -3k + k - 3 = 0
=> k(k - 3) + 1(k - 3) = 0
=> (k + 1)(k - 3) = 0
k = -1 or k = 3

∴ The values of k are -1 and 3

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Answer 2

Here Is Your Answer

4x²-2(k+1)+(k+4) = 0
4x²-(2k+2)+(k+4)
D=b²-4ac
= (2K+2)²-4×4×(K+4)
= 4k²+4+8K-16×(K+4)
= 4k²+4+8k-16k-64
= 4k²-8k-60
4k²-8k-60=0
Dividing by 4
k²-2k-15=0
k²-(5-3)k-15=0
k²-5k+3k-15=0
k(k-5) +3(k-5)
(k-5) (k+3)
[k = 5] , [k = -3]

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