Question

simplify this:
$1/(\sqrt{7-4\sqrt{3} } )$

Answer 1

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$(a-b)^2=7–4\sqrt{3}$ where $a,b\in\R$

$\implies a^2–2ab+b^2=7–4\sqrt{3}$

Comparing the positive number:

$a^2+b^2=7\tag{1}$

Comparing the negative number with the root

$ab=2\sqrt{3}\tag{2}$

From equation $(2)$

$b=\dfrac{2\sqrt{3}}{a}\tag{3}$

Putting this into equation $(1)$

$a^2+\dfrac{12}{a^2}=7$

$\implies a^4–7a^2+12=0$

$\implies (a^2–4)(a^2–3)=0$

$\implies a^2=4$

$\implies a=\pm 2$

Putting these values into equation $(3)$ gives

$b=\dfrac{2\sqrt{3}}{a}=\pm \sqrt{3}$

So, we have $(a,b)=(2,\sqrt{3}),(-2,-\sqrt{3})$

Now, using the first pair I have

$\sqrt{7–4\sqrt{3}}=\sqrt{(2-\sqrt{3})^2}=\boxed{2-\sqrt{3}}$

Using the second pair I have

$\sqrt{7–4\sqrt{3}}=\sqrt{(-2-\sqrt{3})^2}$

This gives $7+4\sqrt{3}$ inside the square root, which is not what’s given here.

The only solution is $\boxed{2-\sqrt{3}}$