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?
 A: Note that $63=9 \times 7$ and $8=\frac{1}{2}(9+7)$. Therefore,
$$ 9+7-2\sqrt{9 \times 7} = (\sqrt{9}-\sqrt{7})^2, $$
so that
$$ 8-\sqrt{63} = \frac{1}{2}(16-2\sqrt{63}) = \frac{1}{2}(3-\sqrt{7})^2 $$
and
$$ \sqrt{8-\sqrt{63}} = \frac{3-\sqrt{7}}{\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{14}}{2}. \quad \blacksquare $$
A: Note
$$\sqrt{8-\sqrt{63}}
= \sqrt{\frac{16-2\sqrt{63}}2}
= \sqrt{\frac{(\sqrt9-\sqrt7)^2}2}
= \frac{3-\sqrt7}{\sqrt2}$$
Alternatively, apply the denest formula
$$\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}2 }
-\sqrt{\frac{a-\sqrt{a^2-c}}2 }
$$
A: if $x = \sqrt{8-\sqrt{63}},$  then $0<x<1$ and $x^2 - 8 = - \sqrt{63},$ then $x^4 - 16 x^2 + 64= 63,$  then
$$ x^4 - 16 x^2 + 1 = 0.  $$ Also
$$  x^2 - 16 + \frac{1}{x^2} = 0 $$
Taking $$ u = x + \frac{1}{x} $$
we get $u^2 - 18 = 0 $ and $$  u = \sqrt {18} $$
and
$$  x = \frac{3 \sqrt 2 \pm \sqrt{14}}{2} $$
and $x<1$ gives
$$ \color{blue}{ x = \frac{3 \sqrt 2 - \sqrt{14}}{2} } $$
Let's see, I could have chosen $$  v = \frac{1}{x} - x > 0 $$
with $v^2-14 = 0,$ then $v = \sqrt{14}$ or
$$  x^2 + \sqrt{14} x - 1 = 0, $$
$$  x = \frac{- \sqrt{14} \pm \sqrt{18}}{2} $$
and $x>0$ gives
$$ \color{red}{ x = \frac{- \sqrt{14} + \sqrt{18}}{2} } $$
A: Take
$\sqrt{8-\sqrt{63}}=\sqrt{a}-\sqrt{b}$
squaring both sides you get
$8-\sqrt{63} = a+b-2\sqrt{ab}$
which gives you two equations
$a+b=8..(i) and \sqrt{63}=2\sqrt{ab}...(ii)$
you can solve the rest
A: $$ 8-3\sqrt{7}  = a^2 + b^2 - 2ab $$
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
