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How do you simplify square root of $10$ times square root of $6?$

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Last updated date: 13th Jun 2024
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Answer
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Hint: A square root of a number $'x'$ is number $'y'$ that ${{y}^{2}}=x$ in other words, a number $'y'$ whose square is $X-$ For example $4$ and $-4$ are square roots of $16$ because ${{4}^{2}}={{\left( -4 \right)}^{2}}=16$
Factorisation is defined as breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity or factors, which when multiplied together given the original number or matrix, etc.
Example: $24=4\times 6$
$4$ and $6$ are the factors of $24.$

Complete step by step solution:
As per the problem square root of $10$ times square root of $6.$
Here, square root of $10$ is represented by $\sqrt{10}$ and square root of $6$ is represented by $\sqrt{6}$ and times is represented by $'X'$ multiplication.
Therefore the representation of statement is,
$\Rightarrow \sqrt{10}\times \sqrt{6}$
Here, you can multiply $10$ by $6.$
$=\sqrt{60}$
To factorize the $60$. This method can be adopted.
Factors of $60$ are $2\times 2\times \,3\times 5.$
Therefore you can write in this form.
$=\sqrt{2\times 2\times 3\times 5}$
$2\times 2=4$ or ${{2}^{2}}=4$ as you can write ${{2}^{2}}$ on the place of $2\times 2$
$=\sqrt{{{2}^{2}}\times 3\times 5}$
Here, apply rule of $\sqrt{{{a}^{2}}}=a$ means you can write $\sqrt{{{2}^{2}}}=2$
Therefore the modified equation will be,
$=2\sqrt{3\times 5}$
$=2\sqrt{15}$

Hence the simplification of square root of $10$ times square root of $6$ is $2\sqrt{15}$

Additional Information:
When you have a root square root for example in the denominator of a friction you can ‘remove’ it multiplying and dividing the fraction for the same quantity. The idea is to avoid the rational number in the denominator.
Consider $\dfrac{3}{\sqrt{2}}$
You can remove the square root by multiplying and dividing by $\sqrt{2}$
$\dfrac{3}{\sqrt{2}}\dfrac{\sqrt{2}}{\sqrt{2}}$
The operation does not change the value of you fraction because $\dfrac{\sqrt{2}}{\sqrt{2}}=1$ anyway and you fraction does not change by multiplying $1$ to it.
Now you can multiply in the numerator and denominator.
$\dfrac{3}{\sqrt{2}}.\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{3\sqrt{2}}{\left( \sqrt{2} \right).\left( \sqrt{2} \right)}=\dfrac{3\sqrt{2}}{2}$

Note: Multiplication and division radicals
$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}$
Or $\sqrt{a}\sqrt{b}=\sqrt{a\times b}$
Remember that you can multiply the two separate square root terms. Factorization can be done by using LCM method also.
Write $2\times 2={{2}^{2}}$
Apply rule $\sqrt{{{a}^{2}}}=a$