How do you solve ${{x}^{2}}+5=167$ ?

Hint: To solve ${{x}^{2}}+5=167$ , we have to use algebraic rules. We have to first collect constant terms on one side. Then take the square root on both sides and solve for x.

Complete step-by-step answer:
We have to solve ${{x}^{2}}+5=167$ . Let us take the constant terms on one side. For this, we will move 5 from LHS to RHS. Therefore, its sign changes to negative.
$\begin{align}
  & {{x}^{2}}+5=167 \\
 & \Rightarrow {{x}^{2}}=167-5 \\
\end{align}$
On solving the RHS, we will get
$\Rightarrow {{x}^{2}}=162$
Let us take the square root on both the side.
$\begin{align}
  & \Rightarrow \sqrt{{{x}^{2}}}=\sqrt{162} \\
 & \Rightarrow x=\pm \sqrt{162} \\
\end{align}$
We have to take the square root of 162. We will do this by taking the LCM.
$\begin{align}
  & 2\left| \!{\underline {\,
  162 \,}} \right. \\
 & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }1 \\
\end{align}$
We obtained $162=2\times 3\times 3\times 3\times 3$ . Let us group two similar terms.
\[162=2\times \left( 3\times 3 \right)\times \left( 3\times 3 \right)\]
We will get the square root of 162 as $\sqrt{162}=3\times 3\times \sqrt{2}=9\sqrt{2}$ .
Therefore, we can write
$x=\pm 9\sqrt{2}$
We can also write the value in the decimals. We know that $\sqrt{2}=1.414$ . Therefore, the above value becomes
$x=\pm \left( 9\times 1.414 \right)=\pm 12.726$
Hence, the solution of the given equation is $\pm 9\sqrt{2}$ or $\pm 12.726$ .

Note: Students must be very careful when taking the square root. Do not miss the $\pm $ sign. We used $\pm $ since the square of a positive or negative number will always be positive. That is, ${{\left( +9\sqrt{2} \right)}^{2}}={{\left( -9\sqrt{2} \right)}^{2}}=162$ . Students must know algebraic rules to solve these types of questions. When a positive number is moved from one side to another, it will become negative. The converse is also true. Similarly, when we move a multiplier or multiplicand from one side to another, it will be the divisor. The converse is also true.