Question

Find the square root of the given equation ${{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$?
(a) ${{x}^{2}}\left( a+b \right)$,
(b) ${{x}^{\left( \dfrac{a+b}{2} \right)}}$,
(c) ${{x}^{\dfrac{{{\left( a+b \right)}^{2}}}{2}}}$,
(d) ${{x}^{a+b}}$.

Answer 1

Hint: We start solving the problem by assigning variable for the square of ${{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$. We then use law of exponents ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$ to proceed through the problem. After using the law, we make necessary arrangements in the exponents and use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to proceed through the problem. We then apply square for the obtained result and use the law ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to proceed through the problem. We then make necessary arrangements and calculations to get the required result.

Complete step by step answer:
According to the problem, we need to find the square root of ${{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$. Let us consider the square root to be ‘y’.
So, we have $y=\sqrt{{{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}}$.
$\Rightarrow {{y}^{2}}={{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$ ---(1).
From the law of exponents, we know that ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$. We use this result in equation (1).
$\Rightarrow {{y}^{2}}={{x}^{{{b}^{2}}+{{b}^{2}}+2ab+{{a}^{2}}-{{b}^{2}}}}$.
$\Rightarrow {{y}^{2}}={{x}^{{{b}^{2}}+2ab+{{a}^{2}}}}$.
$\Rightarrow {{y}^{2}}={{x}^{{{\left( b \right)}^{2}}+2\left( a \right)\left( b \right)+{{\left( a \right)}^{2}}}}$ ---(2).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we use this result in equation (2).
$\Rightarrow {{y}^{2}}={{x}^{{{\left( a+b \right)}^{2}}}}$.
$\Rightarrow y={{\left( {{x}^{{{\left( a+b \right)}^{2}}}} \right)}^{\dfrac{1}{2}}}$ ---(3).
From the law of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. We use this result in equation (3).
\[\Rightarrow y={{x}^{\left( {{\left( a+b \right)}^{2}} \right)}}^{\times \left( \dfrac{1}{2} \right)}\].
\[\Rightarrow y={{x}^{\dfrac{{{\left( a+b \right)}^{2}}}{2}}}\].
So, we have found the square root of ${{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$ as \[{{x}^{\dfrac{{{\left( a+b \right)}^{2}}}{2}}}\].
∴ The square root of ${{x}^{{{b}^{2}}}}{{x}^{{{b}^{2}}+2ab}}{{x}^{{{a}^{2}}-{{b}^{2}}}}$ is \[{{x}^{\dfrac{{{\left( a+b \right)}^{2}}}{2}}}\].
The correct option for the given problem is (c).
Note:
We should not make confuse $\sqrt{{{x}^{{{\left( a+b \right)}^{2}}}}}$ with ${{x}^{\left( a+b \right)}}$ as that is wrong and doesn’t follow the laws of exponents. We can check this by assuming numerical values for ‘x’, ‘a’ and ‘b’ and find the square root for the required form. We should not confuse while applying formulas and make mistakes while doing calculations and arrangements. Similarly, we can expect problems that contain cubic equations in exponents.