Question

What is the simplest radical form of the square root of $80$ ?

Answer 1

Hint: To solve the question we need to know the concept of square root. In this question we will prime factorise the number to find the square root of the number. We will use the Prime Factorization method to get the factors of the number, this helps in finding the factors of the number easily. We can recheck the solution by squaring the answer which we get after solving.

Complete step by step solution:
The question asks us to write the value of the square root of $80$ in terms of radical format. $''\sqrt{{}}''$ is the radical sign. So if we write $\sqrt{a}$ , mathematically it means $''\sqrt{{}}''$ is the radical sign, and $''a''$ id the radicand. To solve the question the first step would be to find the square root of the value:
$\sqrt{80}$
To find the square root of the number we write the number in product of all the prime factors associated with it.
We can find the value of factors by prime factorisation of the given number. So it would be written as,
$80=2\times 2\times 2\times 2\times 5$
Now, on substituting $80$ with the product of its prime factors, we get:
$\Rightarrow \sqrt{2\times 2\times 2\times 2\times 5}$
If $b=a\times a$, then it is the square root of $b$, $\sqrt{b}=\sqrt{a\times a}$ , which is equal to $a$ . So applying the same, on the given value:
$\Rightarrow \sqrt{2\times 2\times 2\times 2\times 5}$
$\Rightarrow 2\times 2\sqrt{5}$
$\Rightarrow 4\sqrt{5}$

$\therefore $ The radial form of $\sqrt{80}$ is $4\sqrt{5}$.

Note: We can check whether the answer is correct or not. To check this we will square the number and check that the number matches the question or not. Let us square the value $4\sqrt{5}$, the result we get after squaring is
\[\Rightarrow {{\left( 4\sqrt{5} \right)}^{2}}\]
On expanding we get,
\[\Rightarrow 4\times 4\times \sqrt{5}\times \sqrt{5}\]
$\Rightarrow 16\times 5$
$\Rightarrow 80$
So the value which we get by squaring the answer is the same as the question. So our solving is correct for the question.