Question

Use quadratic formula to find the roots of x^2+2x-7

Answer 1

Okay, here we have this:

`x^2+2x-7=0`

We will solve using the general formula, then we obtain:

`\begin{gathered} x_(1,2)=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot(-7)}}{2\cdot1} \\ =\frac{-2\pm\sqrt[]{4+28}}{2} \\ =\frac{-2\pm\sqrt[]{32}}{2} \\ =\frac{-2\pm4\sqrt[]{2}}{2} \end{gathered}`

Let's separate the solutions:

`\begin{gathered} x_1=\frac{-2+4\sqrt[]{2}}{2} \\ =\frac{2(-1+2\sqrt[]{2})}{2} \\ =-1+2\sqrt[]{2} \end{gathered}`
`\begin{gathered} x_2_{}=\frac{-2-4\sqrt[]{2}}{2} \\ =\frac{2(-1-2\sqrt[]{2})}{2} \\ =-1-2\sqrt[]{2} \end{gathered}`

Finally we obtain that the roots are: -1+2√2 and -1-2√2.