Question

Use the quadratic formula to solve over complex numbers Show all the steps Make sure final answer has a simplified radical

Answer 1

The quadratic equation is,

`4x^2-2x+5=0`

The quadratic equation ax^2 + bx + c = 0 can be solved by quadratic formula as,

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

The value of a is 4, value of b is -2 and value of c is 5.

Solve the quadratic equation by quadratic formula.

`\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\cdot4\cdot5}}{2\cdot4} \\ =\frac{2\pm\sqrt[]{4-80}}{8} \\ =\frac{2\pm\sqrt[]{-76}}{8} \\ =\frac{2\pm2\sqrt[]{19}i}{8} \\ =(2)/(8)\pm\frac{2\sqrt[]{19}}{8}i \\ =(1)/(4)+i\frac{\sqrt[]{19}}{4},(1)/(4)-i\frac{\sqrt[]{19}}{4} \end{gathered}`

So answer is,

`x=(1)/(4)+i\frac{\sqrt[]{19}}{4},(1)/(4)-i\frac{\sqrt[]{19}}{4}`