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How do you simplify square root of $x$ to the ${{6}^{th}}$ power over $4900?$

Answer
VerifiedVerified
543.3k+ views
Hint: In any expression is with square root, when root or other fractional exponent we say that they are radical expressions. Here in the question we have to simplify it properly and while simplifying it also we have to power given in question. Generally we do not have the radicals at the denominators.

Complete step-by-step answer:
Here,
We have to simplify square root of $x$ to the ${{6}^{th}}$ power over $4900.$
So, from the given statement,
The expression becomes.
$\sqrt{\dfrac{{{x}^{6}}}{4900}}$
Now, we have to give the square root for both numerator and denominator we get,
$\sqrt{\dfrac{{{x}^{6}}}{4900}}=\dfrac{\sqrt{{{x}^{6}}}}{\sqrt{4900}}$
After simplifies it we get,
$=\dfrac{{{x}^{\dfrac{6}{2}}}}{\sqrt{49}.\sqrt{100}}$
$=\dfrac{{{x}^{3}}}{70}$
$\sqrt{\dfrac{{{x}^{6}}}{4900}}=\dfrac{{{x}^{3}}}{70}$

By simplifying the square root of $x$ to the ${{6}^{th}}$ power over $4900$ we get $\sqrt{\dfrac{{{x}^{6}}}{4900}}=\dfrac{{{x}^{3}}}{70}$

Additional Information:
Finding the square roots and converting them to exponents is a common operation in algebra. The square root, which is uses here the radical symbol, which are also non-binary operations. (Which involves only just one another?) To convert the square root to an exponent, you can use a fraction in the power to indicate which are roots or a radical. In case of add or subtract variables you must follow the same rule, even though there are various operations when adding or subtracting terms that have exactly the same variable, you either add or subtract the coefficient. Let the result stand with the variables.

Note:
When you find out the square roots, the symbol for that operation is a radical which is represented as $\left( \sqrt{{}} \right)$ use fraction in the power to indicate that the expression stands for a root. You have to multiply the exponents, when their rising power to power but, the bases are there is the same. Because rising from a power to power is that you have to multiply the exponents as long as. The bases are the same so you do not get any problems. So, this is some key point in which you have to remember it for solving the problems.