Question

Find the value of k for which the quadratic equation 4xsquare-2 (k+1)x+(k+1)was equal roots

Answer 1

Answer:

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:

3

Step-by-step explanation:

for any equation in quadratic ,, if it has real and equal roots then the discriminant of the given equation is zero

=> b^2-4ac=0,, for the given equation b= coefficient of 'x' term= -2(k+1),, a= coefficient of 'x^2' = 4 ,, c= constant= k+1

now ,the value of b^2-4ac=(2(k+1))^2-4(4)(k+1)

=>4(k+1)^2-4(k+1)(4)=0 (according to condition )

=>4(k+1)[(k+1)-4]=0 =>k+1-4=0 => k-3= 0

=>k =3