Question

How do you solve the equation ${{x}^{2}}+2x=7$ by completing the square?

Answer 1

Hint: Solving a quadratic equation by completing the square is easy after going through some rearrangements of the terms. To have a complete square in the form of ${{\left( x+a \right)}^{2}}$ we will add some numbers on both the side and convert the left-hand side into a complete square. Now, the value of $x$ can be found by further simplifying the equation.

Complete step by step answer:
The given equation is
${{x}^{2}}+2x=7....\text{expression}1$
To have a complete square in the left-hand side of the equation let’s start by taking the square ${{\left( x+a \right)}^{2}}$ for comparison.
We know, ${{\left( x+a \right)}^{2}}={{x}^{2}}+2\cdot a\cdot x+{{a}^{2}}....\text{expression2}$
As, the first two terms of the expression $\left( {{x}^{2}}+2\cdot a\cdot x+{{a}^{2}} \right)$ are ${{x}^{2}}$ and $2x$ , we compare $\text{expression}1$ with the first two terms in the right-hand side of $\text{expression2}$ and we get $a=1$ .
Hence, to get the square term ${{\left( x+1 \right)}^{2}}$ we add $1$ to the both sides of $\text{expression}1$
$\Rightarrow {{x}^{2}}+2x+1=7+1$
After further simplification of the above expression, we get
 $\Rightarrow {{x}^{2}}+2x+1=8$
The above equation can be also written as
$\Rightarrow {{x}^{2}}+2\cdot 1\cdot x+1=8$
$\Rightarrow {{\left( x+1 \right)}^{2}}=8$
Now taking square root on both the sides of the equation and also keeping both the values after doing the square root we get
$\Rightarrow x+1=\pm \sqrt{8}$
Further simplifying we get
$\Rightarrow x=-1\pm 2\sqrt{2}$
Therefore, the solution of the equation ${{x}^{2}}+2x=7$ are $x=-1+2\sqrt{2}$ and $x=-1-2\sqrt{2}$

Note:
We must be careful that if any number is added to the left-hand side of the equation, we must add the same number on the right-hand side too. Otherwise, both the sides of the equation will not be balanced. Also, while doing square root on both sides of the equation, we must keep in mind that we have to take both positive and negative signs. As, both are the real solutions of the given equation.