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Solve the quadratic equation. Write complex solutions in standard form. X2+35=5z

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The quadratic is


x^2+35=5x

Subtract 5x from both sides


\begin{gathered} x^2-5x+35=5x-5x \\ x^2-5x+35=0 \end{gathered}

To solve it we will use the formula


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

a is the coefficient of x^2

b is the coefficient of x

c is the numerical term

Since a = 1, b = -5, c = 35

Then substitute them in the formula above


x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(1)(35)}}{2(1)}

Simplify it


\begin{gathered} x=\frac{5\pm\sqrt[]{25-140}}{2} \\ x=\frac{5\pm\sqrt[]{-115}}{2} \end{gathered}

since the square root of a -ve number is imaginary, then we will use i


\begin{gathered} x=\frac{5\pm\sqrt[]{115}.\sqrt[]{-1}}{2} \\ x=\frac{5\pm\sqrt[]{115}i}{2} \\ \sqrt[]{-1}=i \end{gathered}

The solutions are


\mleft\lbrace\frac{5+\sqrt[]{115}i}{2},\frac{5-\sqrt[]{115}i}{2}\mright\rbrace

User Adil Naseem
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