Question

Complete the square to write the quadratic equation in vertex form: x^2+4x-5=0

Answer 1

Let's solve the equation:

`\begin{gathered} x^2+4x-5=0 \\ x^2+4x=5 \\ x^2+4x+((4)/(2))^2=5+((4)/(2))^2 \\ (x+2)^2=5+(16)/(4) \\ (x+2)^2=(36)/(4) \\ x+2=\pm\sqrt[]{(36)/(4)} \\ x+2=\pm3 \\ \text{ then} \\ x=-2+3 \\ x=1 \\ or\text{ } \\ x=-2-3 \\ x=-5 \end{gathered}`

Therefore, the solutions of the equation are x=1 and x=-5