Question

Find all real solutions of the equation.

${\left(x+2\right)}^2={\left(x-4\right)}^2$

Short answer

Real solution of the given equation: $x=1$

Step-by-step solution

Step 1. Simplify both sides.

Here we shall use the identity ${\left(x+y\right)}^2=x^2+2xy+y^2$

$\begin{array}{c}{\left(x+2\right)}^2={\left(x-4\right)}^2\\ x^2+4x+4=x^2-8x+16\end{array}$

Step 2. Now we shall move all variables on left hand side.

$\begin{array}{l}{x}^2+4x+4=x^2-8x+16\\ x^2+4x-x^2+8x=16-4\end{array}$

Step 3. Simplify the equation.

We shall simplify both sides to find the real solution.

$\begin{array}{c}\cancel{x^2}+4x-\cancel{x^2}+8x=16-4\\ 12x=12\\ x=\frac{12}{12}\\ x=1\end{array}$