Question

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

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The possible values of k are -1 , 3.

The given equation quadratic equation is 4x² - 2(k+1)x+(k+1) = 0

For the equation to have equal roots the value of determinant should be equal to zero (D = 0).

For the quadratic equation - a = 4 , b= 2(k+1) , c = (k+1)

D = b² - 4ac = 0 , for the equation to have equal roots .

=> 4(k+1)²- 4 × 4 × (k+1) = 0

=> k² + 1 + 2k -4k -4 = 0

=> k² -2k -3 = 0

=> k² - 3k + k -3 = 0

=> k(k-3) +(k-3) = 0

=> k = 3 , -1

The possible values of k are 3 , -1 for which the quatratic equation has equal roots.