What are the values of $k$ when $kx^2+x+k$, has equal zeroes? Show that the quadratic equation $kx^2 + 2(x+1)=k$ has real roots for all the values of $k\in \mathbb{R}$
what i did
$kx^2+2x+2-k=0$
$4-4(2-k)(k)>0$
$4-8k+4k^2>0$
$(64±64)÷ 4(2)>0$
$128÷ 8>0$
$16>0$
please help me check

We can just compute the roots using factorization:
$$
kx^2+2(x+1) = k \\
kx^2 +2x + 2 - k = 0 \\
kx^2 -x^2 + x^2 +2x + 1 + 1-k = 0 \\
(k-1)x^2 + (x+1)^2 + 1-k = 0 \\
(k-1)(x^2 - 1) + (x+1)^2  = 0 \\
(k-1)(x+1)(x - 1) + (x+1)^2  = 0 \\
(x+1)((k-1)(x - 1) + (x+1))  = 0 \\
(x+1)(kx - k - x + 1 + x+1)  = 0 \\
(x+1)(kx - k + 2)  = 0 \\
$$
Hence the roots are $-1$ and, if $k$ is non-zero, $1-2/k$ .
