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

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Lesly Fernandez · Accepted Answer

Step 1The quadratic ax2+bx+c has two distinct real roots if the discriminant △=b2−4ac&amp;gt;0.In this case we have the quadratic−2x2+kx−2,so b=k, a=−2 and c=−2.Thus for the discriminant we haveb2−4ac=k2−16=(k−4)(k+4)&amp;gt;0,so k&amp;gt;4 or k&amp;lt;−4.

Kathleen Sanchez · Accepted Answer

Step 1The equation in ax2+bx+c form for −2x2+kx−2 is a=−2, b=k, c=−2 so the discriminant isb2−4ac=k2−4(−2)(−2)=k2−16.(Not k2−16k 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 k2−16&amp;gt;0.The means (k+4)(k−4)&amp;gt;0. So either both k+4 and k−4 are both positive or both negative. If they are both positive then k+4&amp;gt;0 and k≻4 and k−4&amp;gt;0 and k&amp;gt;4.So k&amp;gt;4≻4 or in other words k&amp;gt;4.If they are both negative then k+4&amp;lt;0 and k&amp;lt;−4 and k−4&amp;lt;0 so k&amp;lt;4so k&amp;lt;−4&amp;lt;4 or in other words k&amp;lt;−4.So either k&amp;gt;4 or k&amp;lt;−4.Or we could do: k2−16&amp;gt;0 so k2&amp;gt;16 so |k|&amp;gt;4 so k&amp;lt;−4 or k&amp;gt;4.

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

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