The Equation $-2x^2 +kx -2 = 0$ has $2$ different real solutions. Find the set of possible values of $k$

Question

I found what I think the discriminant is $k^2-16k = 0$ but I am unsure where I should go from here to answer the question itself, please help.

Answer 1

The quadratic $ax^2+bx+c$ has two distinct real roots if the discriminant $\Delta=b^2-4ac>0.$

In this case we have the quadratic $-2x^2+kx-2$, so $b=k$, $a=-2$ and $c=-2$.

Thus for the discriminant we have $b^2-4ac=k^2-16=(k-4)(k+4)>0$, so $k>4$ or $k<-4$.

Answer 2

I believe you made a mistake in thinking that $"4(a)(c)"$ under the radical included $k.$

We know $\quad -2x^2 +kx -2\quad =\quad 2x^2 -kx +2 = 0\quad $ solved for x is

\begin{equation} x= \frac{k\pm\sqrt{k^2-4(2)(2)}}{2(2)}=\frac{k\pm\sqrt{k^2-16}}{4} =\frac{k\pm\sqrt{(k-4)(k+4)}}{4} \implies k\ge 4 \lor k\le -4 \end{equation}

because if $\quad-4 < k < 4\quad$ then the radical yields an imaginary number.

Answer 3

The equation in $ax^2 + bx + c$ form for $-2x^2 +kx -2$ is $a=-2,b=k,c=-2$ so the discriminant is $b^2 - 4ac = k^2-4(-2)(-2) = k^2 -16$. (Not $k^2-16k$.... and not $blah = 0$; just the $blah$ itself; the discriminant is a value-- not a statement about the value-- the value itself)

So we need to have $k^2 -16 > 0$.

The means $(k+4)(k-4) > 0$. So either both $k+4$ and $k-4$ are both positive or both negative. If they are both positive then $k + 4 > 0$ and $k > -4$ and $k -4 >0$ and $k >4$. So $k > 4 > -4$ or in other words $k > 4$. If they are both negative then $k+4 < 0$ and $k < -4$ and $k-4 < 0$ so $k < 4$.. so $k < -4 < 4$ or in other words $k < -4$.

So either $k > 4$ or $k < -4$.

Or we could do: $k^2 -16 > 0$ so $k^2 > 16$ so $|k| > 4$ so $k < -4$ or $k > 4$.