The square of the solution of the equation...

Question

The square of the solution of the equation

$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$

is equal to: ...

$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$

$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$

$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$

$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$

$\implies x^2 = 2$

I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks!

Answer 1

The formula $(a-b)^2=a^2+b^2-2ab$ is used.

Here, $a=\sqrt{8+3\sqrt7}$ and $b=\sqrt{8-3\sqrt7}$.

Answer 2

Note that $$(a-b)^2 = a^2 + b^2 - 2ab $$

Your confusion is the $$-2ab = - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}} $$

Answer 3

Alt. hint:   show first that $\;\sqrt{8 \pm 3\sqrt{7}} = \dfrac{1}{\sqrt{2}}\left(3 \pm \sqrt{7}\right)\,$, then no squaring is needed.