Question

find k so that 4x sq - 2 (k+1)x + (k+4) =0 has real and equal roots​

Answer 1

Step-by-step explanation:

$4 {x}^{2} - 2(k + 1)x + k + 4 = 0 \\ 4 {x}^{2} - 2kx - 2x + k = - 4 \\ 4x - 2kx - 2x + k = \sqrt{ - 4 } \\ 2x - 2kx + k = - 2 \\ x - 2kx + k = \frac{ - 2}{2} = - 1 \\ x - kx + k = \frac{ - 1}{2} \\ x + k = \frac{ - 1}{2} + kx \\ x + k = \frac{ - 1 + 2xk}{2}$

$k = \frac{ - 1 + 2xk}{2} - x \\ k = \frac{ - 1 + 2kx - 2x}{2}$

Solve using value