2022-04-01

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

Lesly Fernandez

Step 1
The quadratic $a{x}^{2}+bx+c$ has two distinct real roots if the discriminant $\mathrm{△}={b}^{2}-4ac>0$.
In this case we have the quadratic
$-2{x}^{2}+kx-2$,
so and $c=-2$.
Thus for the discriminant we have
${b}^{2}-4ac={k}^{2}-16$
$=\left(k-4\right)\left(k+4\right)>0$,
so $k>4$ or $k<-4$.

Kathleen Sanchez

Step 1
The equation in $a{x}^{2}+bx+c$ form for $-2{x}^{2}+kx-2$ is so the discriminant is
${b}^{2}-4ac={k}^{2}-4\left(-2\right)\left(-2\right)={k}^{2}-16$.
(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 $\left(k+4\right)\left(k-4\right)>0$. So either both $k+4$ and $k-4$ are both positive or both negative. If they are both positive then $k+4>0$ and $k\succ 4$ and $k-4>0$ and $k>4$.
So $k>4\succ 4$ 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 $k>4$.

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