Question

for what value of k the eq. 4x^2-2(k+1)x+(k+1)=0 has real and 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

Hope This Helps you!

Answer 2

Answer:

The roots are 3 and -1

Step-by-step explanation:

As it has real and equal roots, therefore the discriminant is zero.

b 2 −4ac=0

⇒(2(k+1)) 2 −4.4(k+1)=0

⇒(2k+2) 2 −16(k+1)=0

⇒4k 2 +8k+4−16k−16=0

⇒4k 2 −8k−12=0

⇒4(k2 −2k−3)=0

⇒k 2 −2k−3=0

⇒k 2 −3k−k−3=0

⇒k(k−3)+1(k−3)=0

⇒(k−3)(k+1)=0

k−3=0 k+1=0

k=3 k=−1

∴ The roots are 3 and −1.