# Solve each equation. (x+5)^2−4=12, square roots x^2−4x+4=0, completing the square

Question
Solve each equation.
$$\displaystyle{\left({x}+{5}\right)}^{{2}}−{4}={12}$$, square roots
$$\displaystyle{x}^{{2}}−{4}{x}+{4}={0}$$, completing the square

2020-10-19
The square root method is to isolate the perfect square and then square root both sides to eliminate the square. Remember that when square rooting both sides of an equation, you MUST write a ±:
$$\displaystyle{\left({x}+{5}\right)}^{{2}}-{4}={12}$$
$$\displaystyle{\left({x}+{5}\right)}^{{2}}={16}$$
$$\displaystyle\sqrt{{{x}+{5}}}^{{2}}=\pm\sqrt{{16}}$$
x+5=+-4
x=-5+4
x=-5-4=-9
x=-5+4=-1
The completing the square method is done by writing one side of the equation as a perfect square and then square rooting both sides to eliminate the square. A perfect square is of the form $$\displaystyle{a}^{{2}}−{2}{a}{b}+{b}^{{2}}\ {\quad\text{or}\quad}\ {a}^{{2}}+{2}{a}{b}+{b}^{{2}}$$. Since $$\displaystyle{x}^{{2}}−{4}{x}+{4}={x}^{{2}}−{2}{\left({x}\right)}{\left({2}\right)}+{2}^{{2}}$$, then it is already a perfect square where a=x and b=2. Since $$\displaystyle{a}^{{2}}−{2}{a}{b}+{b}^{{2}}={\left({a}−{b}\right)}^{{2}}$$, then $$\displaystyle{x}^{{2}}−{4}{x}+{4}={x}^{{2}}−{2}{\left({x}\right)}{\left({2}\right)}+{2}^{{2}}={\left({x}−{2}\right)}^{{2}}$$. Therefore:
$$\displaystyle{x}^{{2}}−{4}{x}+{4}={0}$$ Given equation.
$$\displaystyle{\left({x}−{2}\right)}^{{2}}={0}$$ Factor.
$$\displaystyle\sqrt{{{x}−{2}}}^{{2}}=±{0}$$ Square root both sides.
x−2=0 Simplify.
x=2 Add 2 on both sides. ​ ​ ​ ​ ​

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