Question

How do you simplify $\dfrac{\sqrt{63{{x}^{15}}{{y}^{9}}}}{\sqrt{7x{{y}^{11}}}}$ ?

Answer 1

Hint: We know that $\dfrac{\sqrt{a}}{\sqrt{b}}$ is equal to $\sqrt{\dfrac{a}{b}}$ and where b is not equal to 0 and $\sqrt{ab}$ is equal to product of $\sqrt{a}$ and $\sqrt{b}$. Using this we can write $\dfrac{\sqrt{63{{x}^{15}}{{y}^{9}}}}{\sqrt{7x{{y}^{11}}}}$ equal to $\sqrt{\dfrac{63{{x}^{15}}{{y}^{9}}}{7x{{y}^{11}}}}$ . We know that $\dfrac{{{x}^{n}}}{{{x}^{m}}}$ is equal to ${{x}^{n-m}}$ where x is a nonzero number, using this we can simplify the $\sqrt{\dfrac{63{{x}^{15}}{{y}^{9}}}{7x{{y}^{11}}}}$ by writing $\dfrac{{{x}^{15}}}{x}$ as ${{x}^{14}}$ and $\dfrac{{{y}^{9}}}{{{y}^{11}}}$ as $\dfrac{1}{{{y}^{2}}}$.

Complete step by step answer:
The given equation is $\dfrac{\sqrt{63{{x}^{15}}{{y}^{9}}}}{\sqrt{7x{{y}^{11}}}}$
We know that $\dfrac{\sqrt{a}}{\sqrt{b}}$ is equal to $\sqrt{\dfrac{a}{b}}$ and where b is not equal to 0, so $\dfrac{\sqrt{63{{x}^{15}}{{y}^{9}}}}{\sqrt{7x{{y}^{11}}}}$ = $\sqrt{\dfrac{63{{x}^{15}}{{y}^{9}}}{7x{{y}^{11}}}}$
We know that $\dfrac{{{x}^{n}}}{{{x}^{m}}}$ is equal to ${{x}^{n-m}}$ so if we put n equal to 15 and m equal to 1 for x , n equal to 9 and m equal to 11 for y we get $\dfrac{63{{x}^{15}}{{y}^{9}}}{7x{{y}^{11}}}$ is equal to$9{{x}^{14}}{{y}^{-2}}$ . we know that ${{x}^{-n}}$ is equal to $\dfrac{1}{{{x}^{n}}}$, so ${{y}^{-2}}$ is equal to $\dfrac{1}{{{y}^{2}}}$ and $9{{x}^{14}}{{y}^{-2}}$ is equal to $\dfrac{9{{x}^{14}}}{{{y}^{2}}}$ where x and y both are not equal to 0.
Now square root of $\dfrac{9{{x}^{14}}}{{{y}^{2}}}$ is equal to $\dfrac{\sqrt{9{{x}^{14}}}}{\sqrt{{{y}^{2}}}}$ . square root of $9{{x}^{14}}$ is equal to $3{{x}^{7}}$ and square root of ${{y}^{2}}$ is y. so square root of $\dfrac{9{{x}^{14}}}{{{y}^{2}}}$ is equal to $\dfrac{3x}{y}$ where y is not equal to 0.
$\dfrac{3x}{y}$ is the simplified form of $\dfrac{\sqrt{63{{x}^{15}}{{y}^{9}}}}{\sqrt{7x{{y}^{11}}}}$ where y is not equal to 0.

Note:
Always remember we can not make denominators of a fraction 0. So $\dfrac{\sqrt{a}}{\sqrt{b}}$ is equal to $\sqrt{\dfrac{a}{b}}$ will not valid when b is equal to 0 , b can be a complex number. ${{x}^{-n}}$ is equal to $\dfrac{1}{{{x}^{n}}}$ , x is non zero . To solve these type questions, remember all the exponential formulas.