Question

How do you complete the square to solve $x^2+6x=7$?

Answer 1

Step 1: Complete the squares.

Given, $x^2+6x=7$

To complete the squares, add $9$ on both sides of the equation.

$$\begin{gathered} x^2+6x+9=7+9 \\ \Rightarrow x^2+2·3·x+3^2=16 \\ \Rightarrow {\left(x+3\right)}^2=16\left[\because x^2+2xy+y^2={\left(x+y\right)}^2\right] \\ \Rightarrow \left(x+3\right)=\pm \sqrt{16}\left[\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{sides}\right] \\ \therefore \left(\mathrm{x}+3\right)=\pm 4 \end{gathered}$$

Step 2: Solve for $x$.

If $x+3=4$,

$$\begin{gathered} x+3=4 \\ \Rightarrow x=4-3\left(\mathrm{Subtract}3\mathrm{on}\mathrm{from}\mathrm{sides}\right) \\ \therefore x=1 \end{gathered}$$

And if $x+3=-4$,

$$\begin{gathered} x+3=-4 \\ \Rightarrow x=-4-3\left(\mathrm{Subtract}3\mathrm{on}\mathrm{from}\mathrm{sides}\right) \\ \therefore x=-7 \end{gathered}$$

Hence, the solutions are $x=-7$, $x=1$.