Question:
Prove that there are two solutions of the equation $x^2 + kx = 3 - k $
The obvious solution would be finding the value of $b^2 - 4ac$
$x^2 + kx + k - 3=0$
$b^2 - 4 ac$
=$k^2 - 4(1)(k-3)$
=$k^2 - 4k + 12$
From here, $b^2 - 4ac$ have a algebraic value. If it is a constant, I would already be able to prove that the equation have $2$ real roots for every value of "$k$". So, how do I go about doing this question when I can't get $b^2 - 4ac$ to be a constant value?
Picking up from where you left off, $$k^2 - 4k + 12 \equiv (k-2)^2 + 8 \geq 0 +8 > 0$$
since squares are non-negative. Then it follows that your discriminant is positive for all values of $k \in \mathbb{R}$ and hence the quadratic equation always has two roots.