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# How do you simplify $\sqrt{99}?$

Last updated date: 25th Feb 2024
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Answer
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Hint: The given expression is a radical expression as it contains the square root.
For solving the given radical expression we have to use the multiplication property of radical expression.
That is the product of two numbers under square root is equal to the product of that under root number.
i.e. $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$
Using the above radical multipliable properly we have to simplify the given expression.

Complete step by step solution:
Given that, there is a number having square root as $\sqrt{99}$
This number is also called a radical expression.
We know the property of multiplication for the radical expression.
When $a,b\ge 0$
Then, $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$
Now using above radical property simplify the given equation (expression)
$\sqrt{99}=\sqrt{9\times 11}$
According to the above properly the number splitted.
$=\sqrt{9}\times \sqrt{11}$
$\Rightarrow \sqrt{{{3}^{2}}}\times \sqrt{11}$
$=3\sqrt{11}$

Therefore, the answer of given expression $\sqrt{99}$ is $3\sqrt{11}$

Additional Information:
The number of expressions having square root. Cube root or any other fractional component is known as radical expression.
There are some properties of radical expression i.e. product property,
It $a$ and $b$ are real numbers then
$\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}=\sqrt{ab}$
Analogous quotient properly.
If $a$and $b$ are real number and $b\ne 0$
Then, $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$
Also multiplication of two negative numbers will always result in positive numbers.
For example: $a\times a={{a}^{2}}$
And $\left( -a \right)\times \left( -a \right)={{a}^{2}}$
Therefore while taking square root we have to consider both positive and negative number i.e. $\pm \sqrt{ab}$

Note: In this numerical we used the multiplication property of radical expression.
As we have to simplify $\sqrt{99}$, according to properly it will be distinct into $\sqrt{9\times 11}$ because $\sqrt{9\times 11}=\sqrt{99}$. But the square root of $9$ is $3$ and do not here is no square root.
$\sqrt{9}$ is split into $\sqrt{{{3}^{2}}}$ as $\sqrt{{{3}^{2}}}=9$and therefore square root and square get cancelled out each other. And we get only $3$ but the square root remains the same.
Hence the answer we get is $3\sqrt{11}$