A problem with roots

**Coach** · September 26th 2007, 03:00 AM

This is from one of the Finnish national Math competitions MAOL http://www.maol.fi/fileadmin/users/K...vastaukset.pdf

prove that
$<br /> \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6$

I have a question

1) The sample answer splits 17 into 9 and 8. Yet it moves the eight to the "ends" of the equation $\sqrt{9-12{\sqrt2}+8}-\sqrt{9+12{\sqrt2}+8}=6$

Why is this?

Thank you!

**topsquark** · September 26th 2007, 04:18 AM

Originally Posted by Coach

This is from one of the Finnish national Math competitions MAOL http://www.maol.fi/fileadmin/users/K...vastaukset.pdf

prove that
$<br /> \sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}=6$

I have a question

1) The sample answer splits 17 into 9 and 8. Yet it moves the eight to the "ends" of the equation $\sqrt{9-12{\sqrt2}+8}-\sqrt{9+12{\sqrt2}+8}=6$

Why is this?

Thank you!

It's writing the problem in a suggestive form.

Note that
$(3 + \sqrt{8})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 + 12\sqrt{2} + 8$

and
$(3 - \sqrt{8})^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 - 12\sqrt{2} + 8$

-Dan

**Soroban** · October 12th 2007, 05:58 AM

Hello, Coach!

Dan has the best approach . . .

Prove that: . $\sqrt{17-12{\sqrt2}}-\sqrt{17+12{\sqrt2}}\:=\:6$

In case you can't "eyeball" those radicals, here's a tip.

Since the difference of those two radicals in an integer,
. . we might suspect that the radicands are perfect squares.

We conjecture: . $(a + b\sqrt{2})^2\:=\:17 + 12\sqrt{2}$ . . . . where $a$ is positive and rational

The left side is: . $a^2 + 2\sqrt{2}ab + 2b^2 \:=\:(a^2+b^2) + 2\sqrt{2}ab$

So we have: . $(a^2+2b^2) + 2\sqrt{2}ab \:=\:17 + 12\sqrt{2}$

Equate rational and irrational components:
. . $a^2+2b^2\:=\:17$ .[1]
. . $2\sqrt{2}ab \:=\:12\sqrt{2}\quad\Rightarrow\quad ab = 6$ .[2]
and solve the system of equations.

Solve [2] for $b\!:\;\;b \:=\:\frac{6}{a}$
Substitute into [1]: . $a^2 + 2\left(\frac{6}{a}\right)^2\:=\:17$

. . and we have: . $a^4 - 17a^2 + 72 \:=\:0$

. . which factors: . $(a^2-8)(a^2-9)\:=\:0$

. . and has roots: . $a\;=\:\pm2\sqrt{2},\:\pm3$

Using $a = 3$ , we have: . $b = 2$

Therefore: . $17 + 12\sqrt{2} \:=\:(3 + 2\sqrt{2})^2$ . . . . ta-DAA!

**red_dog** · October 12th 2007, 10:23 AM

We can use the following identities:
$\displaystyle\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+\sqrt {A^2-B}}{2}}+\sqrt{\frac{A-\sqrt{A^2-B}}{2}}$

$\displaystyle\sqrt{A-\sqrt{B}}=\sqrt{\frac{A+\sqrt{A^2-B}}{2}}-\sqrt{\frac{A-\sqrt{A^2-B}}{2}}$

Now, $\displaystyle\sqrt{17\pm 12\sqrt{2}}=\sqrt{17\pm \sqrt{288}}$ and plug $A=17, \ B=288$ in the formula above.

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