Question

How do you solve ${{x}^{2}}+20x+104=0$ by completing the square?

Answer 1

Hint: We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ to form a square for the left side of the equation ${{x}^{2}}+20x+104=0$. Then we take the square root on both sides of the equation. From that we subtract 10 to the both sides to find the value of $x$ for ${{x}^{2}}+20x+104=0$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}+20x+104=0$.
We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ to form the square.
$\begin{align}
  & {{x}^{2}}+20x+104=0 \\
 & \Rightarrow {{x}^{2}}+2\times 10\times x+{{10}^{2}}+4=0 \\
 & \Rightarrow {{\left( x+10 \right)}^{2}}=-4 \\
\end{align}$
We interchanged the numbers for $a=x,b=10$.
Now we have a quadratic equation ${{\left( x+10 \right)}^{2}}=-4$.
We need to find the solution of the given equation ${{\left( x+10 \right)}^{2}}=-4$.
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x+10 \right)}^{2}}}=\pm \sqrt{-4}=\pm 2\sqrt{-1} \\
 & \Rightarrow \left( x+10 \right)=\pm 2i \\
\end{align}$
Therefore, we subtract 10 to get
$\begin{align}
  & \left( x+10 \right)-10=\pm 2i-10 \\
 & \Rightarrow x=-10\pm 2i \\
\end{align}$

The given quadratic equation has two solutions and they are $x=-10\pm 2i$.

Note: We try to verify the value of the root of $x=-10\pm 2i$ for the equation ${{x}^{2}}+20x+104$.
Putting the value $x=-10+2i$ in the left side of the equation we get
$\begin{align}
  & {{x}^{2}}+20x+104 \\
 & \Rightarrow {{\left( -10+2i \right)}^{2}}+20\left( -10+2i \right)+104 \\
 & \Rightarrow 100+4{{i}^{2}}-40i-200+40i+104 \\
 & \Rightarrow -4+4 \\
 & \Rightarrow 0 \\
\end{align}$
Therefore, the value $x=-10+2i$ satisfies the equation ${{x}^{2}}+20x+104$ as $f\left( -10+2i \right)=0$.
Same thing can be said for $x=-10-2i$.