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
 A: I would say just:
$kx^2+2x+2-k=0$
$4-4(2-k)(k)\ge0$
$4-8k+4k^2\ge0$
$4(k-1)^2\ge0$
which is obviously true.
Also you have to consider $k=0$ separately, when your equation is not quadratic, in which case it has the real root $x=-1$
A: Hint: write the equation as:
$$0= k(x^2-1) + 2(x+1)=k(x-1)(x+1)+2(x+1) = (x+1)(kx-k+2)$$
A: $$kx^2+2x+2-k=0$$
Consider the discriminant:
$$ b^2-4ac=4-4(k)(2-k)$$
$$b^2-4ac=4k^2-8k+4$$
$$ b^2-4ac=4(k^2-2k+1)$$
$$ b^2-4ac=4(k-1)^2$$
This mean $$ b^2-4ac \ge 0 $$
So when you plug in any value for $k$ $(k \neq 0)$the discriminat will always be $(\ge0)$ hence you will always have real roots
Another case is $k=0$ where $2x+2=0 \Leftrightarrow x=1$
A: 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, $1-2/k$ .
