Question

The roots of x² + kx+k=0 are real and equal find k?​

Answer 1

Answer:

For equal roots, condition is

b^2-4ac=0

Here b=k, c=k and a=1. Putting these values in the above condition, we get

k^2 -4k=0

k(k-4)=0

So either k=0 or k-4=0

k=0 not possible, then k-4=0. So k=4 Ans

Answer 2

Answer:

k =  $\frac{-x^{2} }{x + 1}$

Step-by-step explanation:

Change $x^{2}$ to the other side which will become $-x^{2}$.

The equation will become kx + k = $-x^{2}$.

Factor out variable k, it will become: k(x + 1) = $-x^{2}$

Transfer (x + 1) to $-x^{2}$, it will be k = $\frac{-x^{2} }{x + 1}$.