Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$ We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})^2}=34-2\sqrt{289-120}=34-2\sqrt{169}=34-2.13=8=(2\sqrt2)^2, $$ so $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$ In the hints the authors have written that I should use the fact that the LHS is positive and square it. What would be the problem if it wasn't positive? The identity obviously won't hold because LHS<0, RHS>0...
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2You equated the squares of the LHS and RHS. To posit that LHS = RHS, you need to say they have the same sign. I think that's what the hint was about. – Stefan Lafon Apr 06 '21 at 21:30
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1Oh, so I did show that $\text{LHS}^2=\text{RHS}^2$ but this can mean that $\text{LHS}=-\text{RHS}$ or vice versa? – kormoran Apr 06 '21 at 21:32
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3It is possible to show that $\sqrt{17 + 2\sqrt{30}} = \sqrt{15} + \sqrt{2}$ and that $\sqrt{17 - 2\sqrt{30}} = \sqrt{15} - \sqrt{2}$, from which the result follows. – N. F. Taussig Apr 06 '21 at 21:33
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@N.F.Taussig, a nice alternative solution. Thank you. My question is still valid, right? – kormoran Apr 06 '21 at 21:34
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1@Medi Yes, that is true. You need the fact that both sides are positive to draw the conclusion that $\sqrt{17 + 2\sqrt{30}} - \sqrt{17 - 2\sqrt{30}} = 2\sqrt{2}$ from your calculations. – N. F. Taussig Apr 06 '21 at 21:34
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4Does this answer your question? Show that $A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}$ is a whole number – lone student Apr 06 '21 at 21:44
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@lonestudent, this is in fact a question of mine?! So no it doesn't. – kormoran Apr 06 '21 at 21:45
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@lonestudent, which question?! The question I ask in this post is far more different... Here we are trying to prove an identity. I really don't see your point. As a user said, you are giving me a hard time. – kormoran Apr 06 '21 at 21:48
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@lonestudent, If you had read my question, you would have known that I am asking something much more specific here! All the best! I am not asking for methods for solving the given problem...? – kormoran Apr 06 '21 at 21:56
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both of your original real numbers are roots of $$ x^4 - 34 x^2 + 169 $$
Standard bit for quartic with no cubic term and no linear, $$ (x^2 - 13)^2 - 8 x^2 = x^4 - 34 x^2 + 169 $$ $$ (x^2 - 13)^2 - (x \sqrt 8)^2 = x^4 - 34 x^2 + 169 $$
This becomes ( because a difference of squares) $$ (x^2 - x \sqrt 8 - 13)(x^2 + x \sqrt 8 - 13) $$ so that your numbersare two out of four values $$ \frac{\pm \sqrt 8 \pm \sqrt {60}}{2} $$ or $$ \pm \sqrt 2 \pm \sqrt {15} $$
This leads to checking $$ (\sqrt 2 + \sqrt {15} )^2 = 17 + 2 \sqrt{30} $$ along with $$ (-\sqrt 2 + \sqrt {15} )^2 = 17 - 2 \sqrt{30} $$
Your original expression is equal to $$ (\sqrt 2 + \sqrt {15} ) - (-\sqrt 2 + \sqrt {15} ) $$
Will Jagy
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Since the roots on the LHS get "annihilated", it makes very much sense to look for natural $m,n$ such that
\begin{eqnarray*} 17+2\sqrt{30} & = & \left(\sqrt n + \sqrt m\right)^2 & = & n + m + 2\sqrt{nm}\\ 17-2\sqrt{30} & = & \left(\sqrt n - \sqrt m\right)^2 & = & n + m - 2\sqrt{nm} \end{eqnarray*}
From this approach you get \begin{eqnarray*} n+m & = & 17\\ \sqrt{nm} & = & \sqrt{30} \end{eqnarray*}
Obviously $n = 15, m=2$ fit, hence
$$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}= \sqrt{15}+\sqrt 2 - (\sqrt{15}-\sqrt 2)=2\sqrt{2}$$
trancelocation
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Thank you for the response! May I ask you why if $17+2\sqrt{30}=\left(\sqrt{n}+\sqrt{m}\right)^2$ then $17-2\sqrt{30}=\left(\sqrt{n}\color{red}{-}\sqrt{m}\right)^2$ where $n$ and $m$ are the same natural numbers? – kormoran Apr 07 '21 at 10:44
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@Medi At the beginning I do not know this. But I set up two equations to test, whether I can find such $m$ and $n$. The test is quick and turns out to be successful. So, I can find the solution without any further more complicated calculations. The motivation behind this quick test lies just in the binomial formula $(a \color{blue}{\pm} b)^2 = a^2+b^2\color{blue}{\pm} 2ab$. – trancelocation Apr 07 '21 at 11:15