Question

How do you complete the square to solve ${{x}^{2}}+4x=12?$

Answer 1

Hint: We will make both the left-hand side and the right-hand side of the given algebraic equation the perfect squares. We will add or subtract a quantity so that both sides become the perfect squares. Then we will find the square roots of the terms in both sides.

Complete step by step solution:
Consider the given algebraic equation ${{x}^{2}}+4x=12.$
We are asked to complete the square to solve the above equation for the variable $x.$
Our first task is to make both the left-hand side and the right-hand side of the equations the perfect squares.
Here, to make them the perfect squares, we will add $4$ to the terms on the left-hand side as well as the right-hand side. This procedure makes no difference in the given equation because the same quantity is added to the terms on both the sides.
Now the equation will become ${{x}^{2}}+4x+4=12+4.$
We know that an equation of the form ${{x}^{2}}+ax+b=c$ can be converted into $\left( x+p \right)\left( x+q \right)=c,$ if there are numbers $p$ and $q$ such that $p+q=a$ and $pq=b.$
Here, $a=4$ and $b=4$ implies $p=q=2.$ Because, $2+2=4$ and $2\times 2=4.$
So, our equation will become $\left( x+2 \right)\left( x+2 \right)=16.$
Furthermore, this will become ${{\left( x+2 \right)}^{2}}={{4}^{2}}.$
Thus, both the sides of the equation have become the perfect squares of some numbers, that is, $x+2$ and $4.$
Let us find the square root of the terms on both the sides of the equation.
Then we will get, $x+2=\pm \sqrt{{{4}^{2}}}=\pm 4.$
From this we can write $x+2=-4$ or $x+2=4.$
So, we will get $x=-4-2=-6$ or $x=4-2=2.$
Hence the solution of the equation is $x=-6$ or $x=2.$

Note: There may be situations when we have to subtract a quantity from both sides of the equation. Remember that we need to add (subtract) the values to (from) the terms on both sides. If we forget to add (subtract) to (from) either of the sides, then we cannot find the solution of the given equation.