Is there any way to simplify expressions with roots in it?

Hi Zhidong.
There's no general way of simplifying things like this, but in this particular case we can proceed as follows:

Then $x^2=7+\sqrt{13}+7-\sqrt{13}-2\sqrt{(7+\sqrt{13})(7-\sqrt{13})}$

$x^2=14-2\sqrt{49-13}=14-2\sqrt{36}$

$x^2=2$

so $x=\pm \sqrt{2}$

can you see that $x$ has to be positive?

Therefore $x=\sqrt{2}$.

This sort of thing would work in general for an expression like $\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}$ but for more general sorts of expressions we'd still be left with lots of nasty roots all over the place.

We could try repeatedly squaring to get rid of them.

eg. try and show that if $x=\sqrt{5}-\sqrt{2}$ then

$x^4-14x^2+9=0$

which isn't nearly as nice as $x^2=2$!

By Paul Smith on Wednesday, July 03, 2002 - 05:52 pm:

Yes, there is a method. Just set x = (7 + 13^1/2 )^1/2 - (7 - 13^1/2 )^1/2 , then:

x² = 7 + 13^1/2 - 2(7 + 13^1/2 )^1/2 (7 - 13^1/2 )^1/2 + 7 - 13^1/2
x² = 14 - 2(49 - 13)^1/2
x² = 14 - 12 = 2

However, by squaring the original equation we introduced a second (incorrect) value of x. Clearly, since 7 + 13^1/2 > 7 - 13^1/2 , x > 0, and thus, x = 2^1/2