If the expression is $\sqrt{12+\sqrt{140}}-\sqrt{8+2\sqrt{15}}-2\sqrt{10-2\sqrt{21}}=a\sqrt{3}+b\
\sqrt{7}$, then find the value of $\sqrt{8a-b}$.
A. 7
B. 6
C. 3
D. 1

Hint: We express the given root forms into squares to simplify the result. We individually find the root values and find the L.H.S part. We equate the solution with R.H.S to find the value of a and b. We put values in the equation to find the answer of the problem.

Complete step-by-step answer:
The given root form can be expressed in simpler form. The given expressions like $\
\sqrt{12+\sqrt{140}},\sqrt{8+2\sqrt{15}},\sqrt{10-2\sqrt{21}}$ can be expressed into square
form to eliminate the square root.
All the three individual terms are of the form $\sqrt{x+y\pm 2\sqrt{xy}}$ or $\sqrt{x+y\pm \
\sqrt{4xy}}$.
The root values will be $\sqrt{x+y\pm 2\sqrt{xy}}=\sqrt{x+y\pm \sqrt{4xy}}$
$\begin{align}
  & \sqrt{x+y\pm \sqrt{4xy}}=\sqrt{x+y\pm 2\sqrt{xy}} \\
 & =\sqrt{{{\left( \sqrt{x} \right)}^{2}}+{{\left( \sqrt{y} \right)}^{2}}\pm 2\sqrt{x}\sqrt{y}} \\
 & =\sqrt{{{\left( \sqrt{x}\pm \sqrt{y} \right)}^{2}}} \\
 & =\sqrt{x}\pm \sqrt{y} \\
\end{align}$
Now we expand them individually.
For the first term $\sqrt{12+\sqrt{140}}$, we try to put the general form to find the value of x
and y.
$\sqrt{12+\sqrt{140}}$ is of the form $\sqrt{x+y+\sqrt{4xy}}$.
So, $\sqrt{12+\sqrt{140}}=\sqrt{5+7+\sqrt{4\times 5\times 7}}$. Here $x=5,y=7$.
We get $\sqrt{12+\sqrt{140}}=\sqrt{{{\left( \sqrt{5}+\sqrt{7} \right)}^{2}}}=\sqrt{5}+\sqrt{7}$.
For the second term $\sqrt{8+2\sqrt{115}}$, we try to put the general form to find the value of x and y.
$\sqrt{8+2\sqrt{115}}$ is of the form $\sqrt{x+y+2\sqrt{xy}}$.
So, $\sqrt{8+2\sqrt{15}}=\sqrt{5+3+2\sqrt{5\times 3}}$. Here $x=5,y=3$.
We get $\sqrt{8+2\sqrt{15}}=\sqrt{{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}=\sqrt{5}+\sqrt{3}$.
For the third term $\sqrt{10-2\sqrt{21}}$, we try to put the general form to find the value of x and y.
$\sqrt{10-2\sqrt{21}}$ is of the form $\sqrt{x+y-2\sqrt{xy}}$.
So, $\sqrt{10-2\sqrt{21}}=\sqrt{7+3-2\sqrt{7\times 3}}$. Here $x=7,y=3$.
We get $\sqrt{10-2\sqrt{21}}=\sqrt{{{\left( \sqrt{7}-\sqrt{3} \right)}^{2}}}=\sqrt{7}-\sqrt{3}$.
Now we put the values in the equation.
$\begin{align}
  & \sqrt{12+\sqrt{140}}-\sqrt{8+2\sqrt{15}}-2\sqrt{10-2\sqrt{21}}=a\sqrt{3}+b\sqrt{7} \\
 & \Rightarrow \left( \sqrt{5}+\sqrt{7} \right)-\left( \sqrt{5}+\sqrt{3} \right)-2\left( \sqrt{7}-\
\sqrt{3} \right)=a\sqrt{3}+b\sqrt{7} \\
\end{align}$
We solve the equation and interchange the sides.
$\begin{align}
  & a\sqrt{3}+b\sqrt{7}=\left( \sqrt{5}+\sqrt{7} \right)-\left( \sqrt{5}+\sqrt{3} \right)-2\left( \
\sqrt{7}-\sqrt{3} \right) \\
 & \Rightarrow a\sqrt{3}+b\sqrt{7}=\sqrt{5}+\sqrt{7}-\sqrt{5}-\sqrt{3}-2\sqrt{7}+2\sqrt{3}=\
\sqrt{3}-\sqrt{7} \\
\end{align}$
Now we equate the both coefficient parts of the equation for similar form.
So, $a=1,b=-1$.
We need to find $\sqrt{8a-b}$.
Putting value, we get $\sqrt{8\times 1-\left( -1 \right)}=\sqrt{8+1}=\sqrt{9}=3$.
So, the correct option is (C)

So, the correct answer is “Option C”.

Note: We need to remember the expansion of squares under roots. The form of the given is crucial. We don’t need to use both roots, positive roots are sufficient enough to find the solution. Equating the coefficients are only possible when the surds’ values are equal.