The Equation \(\displaystyle-{2}{x}^{{{2}}}+{k}{x}-{2}={0}\) has 2 different real



Answered question


The Equation 2x2+kx2=0 has 2 different real solutions. Find the set of possible values of k.

Answer & Explanation

Lesly Fernandez

Lesly Fernandez

Beginner2022-04-02Added 16 answers

Step 1
The quadratic ax2+bx+c has two distinct real roots if the discriminant =b24ac>0.
In this case we have the quadratic
so b=k, a=2 and c=2.
Thus for the discriminant we have
so k>4 or k<4.
Kathleen Sanchez

Kathleen Sanchez

Beginner2022-04-03Added 7 answers

Step 1
The equation in ax2+bx+c form for 2x2+kx2 is a=2, b=k, c=2 so the discriminant is
(Not k216k 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 k216>0.
The means (k+4)(k4)>0. So either both k+4 and k4 are both positive or both negative. If they are both positive then k+4>0 and k4 and k4>0 and k>4.
So k>44 or in other words k>4.
If they are both negative then k+4<0 and k<4 and k4<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: k216>0 so k2>16 so |k|>4 so k<4 or k>4.

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