Question

Simplify the following: $\left| \sqrt{7-4\sqrt{3}} \right|$

Answer 1

Hint: Simplification of the above expression will be done if we can make the expression written inside the square root as a perfect square then that square in power and square root will be cancelled out and it will be easier to solve this expression. To do that, we can write 7 in $\left| \sqrt{7-4\sqrt{3}} \right|$ as $4+3$. After that the expression written inside the square root will like: $\left| \sqrt{4+3-4\sqrt{3}} \right|$. Now, we can write 4 as ${{2}^{2}}$ and 3 as ${{\left( \sqrt{3} \right)}^{2}}$ by doing that, the expression written inside the square root will become in the form of: ${{a}^{2}}+{{b}^{2}}-2ab$ which is the algebraic identity for ${{\left( a-b \right)}^{2}}$.

Complete step by step answer:
The expression given in the above problem which we have to simplify is as follows:
$\left| \sqrt{7-4\sqrt{3}} \right|$
Now, to simplify the above expression, we are trying to make the expression written inside the square root as the perfect square. For that, we are going to write 7 as $4+3$ in the above expression and we get,
$\left| \sqrt{4+3-4\sqrt{3}} \right|$
Now, we can write 4 as ${{2}^{2}}$ and 3 as ${{\left( \sqrt{3} \right)}^{2}}$ in the above expression and we get,
$\left| \sqrt{{{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-4\sqrt{3}} \right|$
The expression written inside the square root can become a perfect square if we write $4\sqrt{3}$ as $2.2.\sqrt{3}$. Writing this change in the above expression and we get,
$\left| \sqrt{{{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2.2.\sqrt{3}} \right|$
If you look carefully then the expression written in the above is in the form of:
${{a}^{2}}+{{b}^{2}}-2ab$
And the values of $a\And b$ are:
$\begin{align}
  & a=2; \\
 & b=\sqrt{3} \\
\end{align}$
And there is an algebraic identity we know that:
${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$
Now, substituting the values of $a\And b$ in the above equation we get,
${{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2.2.\sqrt{3}={{\left( 2-\sqrt{3} \right)}^{2}}$
Using this relation in the given expression we get,
$\left| \sqrt{{{\left( 2-\sqrt{3} \right)}^{2}}} \right|$
We can write the square root as the power of $\dfrac{1}{2}$ in the above expression and we get,
$\left| {{\left( {{\left( 2-\sqrt{3} \right)}^{2}} \right)}^{\dfrac{1}{2}}} \right|$
In the above expression, 2 written in the exponent will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & \left| {{\left( {{\left( 2-\sqrt{3} \right)}^{1}} \right)}^{\dfrac{1}{1}}} \right| \\
 & =\left| 2-\sqrt{3} \right| \\
\end{align}$

From the above solution, we have got the simplification of the given expression as $2-\sqrt{3}$.

Note: You might be thinking why we have stopped till $2-\sqrt{3}$ because sometimes in the exam, there are options in which you can see the answer to $2-\sqrt{3}$. There is no wrong if you further solve the problem by putting the value of $\sqrt{3}=1.732$.