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18 votes
Find the zeros by using the quadratic formula and tell whether the solutions are real or imaginary. F(x)=8x^2–4x+10

User AfikDeri
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1 Answer

24 votes
24 votes

We have a quadratic equation and we have to find the zeros.

We can start by simplyfing:


\begin{gathered} f(x)=8x^2-4x+10=0 \\ 2(4x^2-2x+5)=0 \\ 4x^2-2x+5=0 \end{gathered}

Then, we apply the quadratic formula:


\begin{gathered} x=(-(-2))/(2\cdot4)\pm\frac{\sqrt[]{(-2)^2-4\cdot4\cdot5}}{2\cdot4}=(2)/(8)\pm\frac{\sqrt[]{4-80}}{8}=(1)/(4)\pm\frac{\sqrt[]{-76}}{8} \\ x=(1)/(4)\pm\frac{\sqrt[]{76(-1)}}{8}=(1)/(4)\pm\frac{\sqrt[]{76}}{8}\cdot i=(1)/(4)\pm\frac{\sqrt[]{4\cdot19}}{8}\cdot i=(1)/(4)\pm(1)/(4)\sqrt[]{19}i \end{gathered}
\begin{gathered} x_1=(1)/(4)+(1)/(4)\sqrt[]{19}i\approx0.25+1.09i \\ x_2=(1)/(4)-(1)/(4)\sqrt[]{19}i\approx0.25-1.09i \end{gathered}

The roots are imaginary.

Their values are approximately x1=0.25+1.09i and x2=0.25-1.09i.

User Kabb
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