Question

Find the square root of $$\sqrt{13-4\sqrt{3}}$$.

Answer 1

Hint: In the given question, we have been given to solve an expression containing the square root function. To solve this question, we are going to first factorize the expression inside the square root into a form so as to make it a square. Then we are going to apply the required operations and solve it.

Formula Used:
We are going to use the formula of whole square difference:
$${\left(a-b\right)}^2=a^2-2ab+b^2$$

Complete step by step solution:
We have to simplify the expression $$p=\sqrt{13-4\sqrt{3}}$$.
Let $$\sqrt{13-4\sqrt{3}}=\sqrt{x}-\sqrt{y}$$
Squaring both sides,
$$13-4\sqrt{3}=x+y-2\sqrt{xy}$$
Separating the two parts,
$$x+y=13$$ and $$2\sqrt{xy}=4\sqrt{3}\Rightarrow xy=12$$
Now, $$x+y=13$$
Squaring both sides,
$${\left(x+y\right)}^2=x^2+y^2+2xy={\left(13\right)}^2=169$$
Thus, $$x^2+y^2+2\times 12=169\Rightarrow x^2+y^2=145$$
Now, subtracting $$2xy=24$$ from both sides,
$$x^2+y^2-2xy=145-24=121$$
$${\left(x-y\right)}^2={\left(11\right)}^2$$
So, $$x-y=11$$
Now, $$x+y=13$$
Hence, $$x=12$$ and $$y=1$$
Now, we have to evaluate the value of $$\sqrt{x}-\sqrt{y}$$.
Hence, $$\sqrt{13-4\sqrt{3}}=\sqrt{\sqrt{12}-1}$$

Note: In the given question, we had to simplify an expression inside the square root. We did that by making the required amends so as to make the function a square. We did that by using the formula of whole square difference. So, it is very important that we know the formula, how to convert one expression and the procedure for doing the same.