Reduce the radical

Question

Calculate $\sqrt{34 - 24\sqrt{2}}$

I thought of turning $\sqrt{34 - 24\sqrt{2}}$ into the form $ x + y \sqrt {2} $. So by squaring, we get $ x ^ 2 + 2xy \sqrt {2} + 2y ^ 2 $. Turning it into a system we have: $ xy = -12 $ and $ x ^ 2 + 2y ^ 2 = 34 $. I will get 4 solutions for $ x $ and $ y $,

how can I end the problem using this method?

Answer 1

As $xy=-12<0$ they will have opposite signs

$$x^2(2y^2)=2(-12)^2=288$$

So, $x^2,2y^2$ are roots of $$0=t^2-34t+288=(t^2-16)(t^2-18)$$

If $x^2=t^2=16, x=\pm4, y=-\dfrac{12}x=\mp3$

What if $x^2=t^2=18?$

Answer 2

there might be integer solutions, not just rational.... Since $$17 \equiv 1 \pmod 8, $$ there is a solution to $u^2 + 2 v^2 = 17,$ namely $u=3, v = 2,$ giving $9+2\cdot 4 = 17.$ Since $2$ is also represented by $x^2 + 2 y^2,$ we can compose to get $4^2 + 2 \cdot 3^2 = 16 + 18 = 34.$ Finally, switching sign to $x=4, y=-3$ leads to $2xy = -24$

To check: what is $$ \left( 4 - 3 \sqrt 2 \right)^2 \; \; ? \; $$

Answer 3

You have $xy=-12$ and $x^2+2y^2=34$.

To end it from there, solve $x^2+2\left(-\dfrac {12}x\right)^2=34$.

(This becomes a quadratic equation in $u=x^2$.)