What are the values of $k$ when $kx^2+x+k$, has equal zeroes?

Question

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

Answer 1

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 $x=-1$

Answer 2

Hint: write the equation as:

$$0= k(x^2-1) + 2(x+1)=k(x-1)(x+1)+2(x+1) = (x+1)(kx-k+2)$$

Answer 3

$$kx^2+2x+2-k=0$$

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 $2x+2=0 \Leftrightarrow x=1$

Answer 4

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, $1-2/k$ .