How do you simplify \[\sqrt{7x}\left( \sqrt{x}-7\sqrt{7} \right)\]?

Hint: The expression \[\sqrt{7x}\left( \sqrt{x}-7\sqrt{7} \right)\] given in the above question can be simplified by using the distributive property of the algebraic multiplication which is given by $a\left( b+c \right)=ab+ac$. Then using the property of exponent, given by ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$, we can separate the bases raised to the same power. Lastly, on using the exponent property given by ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$, we will get the final simplified form of the given expression.

Complete step by step solution:
Let us write the expression given in the above question as
$\Rightarrow E=\sqrt{7x}\left( \sqrt{x}-7\sqrt{7} \right)$
We know that the square root of a number is equal to the number raised to the power of half. So the above expression can also be written as
$\Rightarrow E={{\left( 7x \right)}^{\dfrac{1}{2}}}\left( {{\left( x \right)}^{\dfrac{1}{2}}}-7{{\left( 7 \right)}^{\dfrac{1}{2}}} \right)$
Now, from the distributive property of the algebraic multiplication we know that $a\left( b+c \right)=ab+ac$. So the above expression now becomes
$\Rightarrow E={{\left( 7x \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}-{{\left( 7x \right)}^{\dfrac{1}{2}}}7{{\left( 7 \right)}^{\dfrac{1}{2}}}$
Now, we know from the properties of the exponents that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$. Substituting $a=7$, $b=x$ and $m=1$ we get \[{{\left( 7x \right)}^{\dfrac{1}{2}}}={{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}\]. Putting this in the above expression we get
\[\begin{align}
  & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}-{{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}7{{\left( 7 \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}-7{{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}} \\
\end{align}\]
From the properties of exponents we also know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$. So we can write the above expression as
\[\begin{align}
  & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}+\dfrac{1}{2}}}-7{{\left( 7 \right)}^{\dfrac{1}{2}+\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{1}}-7{{\left( 7 \right)}^{1}}{{\left( x \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}x-7\left( 7 \right){{\left( x \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow E={{\left( 7 \right)}^{\dfrac{1}{2}}}x-49{{\left( x \right)}^{\dfrac{1}{2}}} \\
\end{align}\]
Writing the above expression in the form of square roots, we finally obtain the given expression as
$\Rightarrow E=\sqrt{7}x-49\sqrt{x}$
Hence, the given expression is simplified.

Note: We may think of taking the square of the given expression, so as to remove the square roots, simplify the resulting expression, and then finally take the square root to get the final simplified expression. But this approach is wrong as we can never get rid of all the square roots even by squaring the given expression. The term \[\left( \sqrt{x}-7\sqrt{7} \right)\] will be squared according to the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. But since $a$ and $b$ both are square roots, the term $-2ab$ will contain the square root making the expression even more complicated.