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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?

YuiTo Cheng
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Arthur
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  • @Hans Lundmark, I think that is too much for a beginner... – UmbQbify Jul 12 '20 at 17:51
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    @UmbQbify-Key20-: Well, they can just pick a simpler one from the long list of linked questions then.. https://math.stackexchange.com/questions/linked/196155?lq=1 – Hans Lundmark Jul 12 '20 at 17:54
  • the answer isn't especially obvious here. It's $\frac {3-\sqrt 7}{\sqrt 2}$. The usual guess, $a+b\sqrt 7$ for $a,b\in \mathbb Q$ does not work. Maybe it would help if you gave us some idea of the level of the material you are studying. – lulu Jul 12 '20 at 17:57
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    $ \sqrt{8-\sqrt{63}} \times\sqrt{8+\sqrt{63}} =\pm1 $ – Narasimham Jul 12 '20 at 17:58
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    @Narasimham, If I may point something out, the expression cannot be $-1$ – UmbQbify Jul 12 '20 at 18:07
  • $ \sqrt{...} $ can be either positive or negative. – Narasimham Jul 12 '20 at 18:09
  • @UmbQbify-Key20- : Narasimham does not claim OP's expression is $-1$. – Eric Towers Jul 12 '20 at 18:10
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    @Narasimham : No. The Square root function is unambiguously nonnegative. The complete set of solutions to $x^2 = a$ requires that we know that both the square root and its negative are solutions. – Eric Towers Jul 12 '20 at 18:11
  • Instead, try representing the problem as $\frac{1}{\sqrt{2}}\sqrt{16-2\sqrt{9\cdot 7}}$. Note the coefficient of $2$ in front of the inner square root [hint, hint]. – John Joy Jul 13 '20 at 21:27

5 Answers5

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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

$$ \sqrt{8-\sqrt{63}} = \frac{3-\sqrt{7}}{\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{14}}{2}. \quad \blacksquare $$

AT1089
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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

$$\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}2 } -\sqrt{\frac{a-\sqrt{a^2-c}}2 } $$

Quanto
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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

$$ \color{red}{ x = \frac{- \sqrt{14} + \sqrt{18}}{2} } $$

Will Jagy
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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

JustTolerateMe
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$$ 8-3\sqrt{7} = a^2 + b^2 - 2ab $$ 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

Tim Crosby
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