<s>[INST] Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
 [/INST] $$ 8-3\sqrt{7}  = a^2 + b^2 - 2ab $$
Let $3\sqrt{7}= 2ab$
$$ab = 1.5\sqrt{7}$$
$$b = \frac{1.5\sqrt{7}}{a}$$
$$a^2 + b^2 = 8$$
$$a^2 +\frac{15.75}{a^2} = 8$$
$y = a^2$
$$y + \frac{15.75}{y} = 8$$
$$y^2 + 15.75 = 8y$$
$$y^2 + 15.75-8y = 0$$
Solve and get
$$y = \frac{7}{2}$$
$$y = \frac{9}{2}$$
Case 1
$$y = \frac{7}{2}$$
$$a = \pm \sqrt \frac{7}{2}$$
$$b = \pm 1.5\sqrt{2}$$
Just remember the sign of a and b are opposite
$$a -b = \sqrt \frac{7}{2}-1.5\sqrt{2}= \frac{\sqrt{7} -3}{\sqrt{2}}   $$or
$$\frac{ 3-\sqrt{7}}{\sqrt{2}}$$
Case 2 :
$$y = \frac{9}{2}$$
$$a =  \pm\frac{3}{\sqrt2}$$
$$b = \pm \frac{1.5\sqrt{14}}{3}=\pm\frac{ \sqrt{14}}{2}=\pm\frac{ \sqrt{7}}{\sqrt{2}} $$
which is the identical solution as above
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