Question

The square root of $\left( 19-8\sqrt{3} \right)$ is?

Answer 1

Hint: Assume the given expression as E. Now, write $8\sqrt{3}=2\times 4\times \sqrt{3}$ and compare it with $2\times a\times b$ to find the values of a and b. In the next step, split 19 into two terms such that it becomes of the form ${{a}^{2}}+{{b}^{2}}$. Finally, use the algebraic identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ and take the square root both the sides to get the answer.

Complete step by step solution:
Here we have been asked to find the square root of the expression $\left( 19-8\sqrt{3} \right)$. Let us assume this expression as E, so we have,
$\Rightarrow E=\left( 19-8\sqrt{3} \right)$
Now, we need to convert it in the form ${{a}^{2}}+{{b}^{2}}-2ab$ so that we can apply the algebraic identity and get the answer. In the expression ${{a}^{2}}+{{b}^{2}}-2ab$ we have three terms and in the given expression $\left( 19-8\sqrt{3} \right)$ we have only two terms. So we need to split one of the terms. When we compare the two expressions then we see that before $8\sqrt{3}$ we have minus sign so we need to compare it with 2ab. That means 19 should be split into \[{{a}^{2}}+{{b}^{2}}\]. So on comparing $8\sqrt{3}$ with 2ab we get,
\[\begin{align}
  & \Rightarrow 8\sqrt{3}=2\times a\times b \\
 & \Rightarrow 2\times 4\times \sqrt{3}=2\times a\times b \\
\end{align}\]
From the above relation we can assume $a=4$ and $b=\sqrt{3}$, so squaring and adding a and b we get,
\[\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}={{4}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=16+3 \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=19 \\
\end{align}\]
So we can write $\left( 19-8\sqrt{3} \right)$ in the form ${{a}^{2}}+{{b}^{2}}-2ab$ as:
\[\begin{align}
  & \Rightarrow 19-8\sqrt{3}={{4}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2\left( 4 \right)\left( \sqrt{3} \right) \\
 & \Rightarrow E={{4}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2\left( 4 \right)\left( \sqrt{3} \right) \\
\end{align}\]
Using the algebraic identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ we get,
\[\Rightarrow E={{\left( 4-\sqrt{3} \right)}^{2}}\]
Taking square root both the sides we get,
\[\Rightarrow \sqrt{E}=\left( 4-\sqrt{3} \right)\]
Hence the above expression is our answer.

Note: Remember that here we have considered $a=4$ and $b=\sqrt{3}$. You can also consider the opposite to it, that is $a=\sqrt{3}$ and $b=4$, then proceed using the same formula and identity. The only thing you have remember is that in the final step you have to take mod in the square root obtained as $\left| \sqrt{3}-4 \right|$ to get the answer positive. If there would have been expression like $\left( 19+8\sqrt{3} \right)$ then we would have used the algebraic identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$.