nomadzkia0re

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

Beginner2022-04-02Added 16 answers

Step 1

The quadratic$a{x}^{2}+bx+c$ has two distinct real roots if the discriminant $\mathrm{\u25b3}={b}^{2}-4ac>0$ .

In this case we have the quadratic

$-2{x}^{2}+kx-2$ ,

so$b=k,\text{}a=-2$ and $c=-2$ .

Thus for the discriminant we have

${b}^{2}-4ac={k}^{2}-16$

$=(k-4)(k+4)>0$ ,

so$k>4$ or $k<-4$ .

The quadratic

In this case we have the quadratic

so

Thus for the discriminant we have

so

Kathleen Sanchez

Beginner2022-04-03Added 7 answers

Step 1

The equation in$a{x}^{2}+bx+c$ form for $-2{x}^{2}+kx-2$ is $a=-2,\text{}b=k,\text{}c=-2$ 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 $k\succ 4$ 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 $\left|k\right|>4$ so $k<-4$ or $k>4$ .

The equation in

(Not

So we need to have

The means

So

If they are both negative then

so

So either

Or we could do:

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