Write the following in the form $x + 2^{0.5}y$:

$\left(5.5 - 3(2)^{0.5}\right)^{0.5}$

I am not sure how to do it, I though about putting under the same denominator and then trying to put it in a quadratic completed square form (if that makes sense) - so that I get one bracket squared which should cancel out the square root.

Unfortunately I can't find how to do it, anyone have any ideas?

  • $\begingroup$ You can square it and then solve the equations . $\endgroup$
    – user252450
    Sep 15 '15 at 15:48
  • $\begingroup$ its not an equation though $\endgroup$ Sep 15 '15 at 15:58
  • $\begingroup$ the question is to write that in the form that I said, which requires some manipulation $\endgroup$ Sep 15 '15 at 15:58
  • 3
    $\begingroup$ I'll explain it better . Treat $x$ and $y$ like variables you want to find . Now square the equation to get $$\frac{11}{2}+5\sqrt{2}=x^2+2y^2+2xy\sqrt{2}$$ Now you want that : $$\frac{11}{2}=x^2+2y^2$$ and also $$2xy=5$$ so now just solve the system of equations . $\endgroup$
    – user252450
    Sep 15 '15 at 16:09
  • 2
    $\begingroup$ This is from squaring the following : $$\sqrt{\frac{11}{2}+5\sqrt{2}}=x+y\sqrt{2}$$ Isn't this the question ? $\endgroup$
    – user252450
    Sep 15 '15 at 16:18

It's a good way to start by assuming such a form for your radical and then use it to gain new information of $x$ and $y$ (this is a good strategy in general , instead of the guessing way ) :


Now square it : $$5.5-3\sqrt{2}=x^2+2y^2+2xy\sqrt{2}$$

Now with a little wishfull thinking (another good strategy ) you may require that the parts with $\sqrt{2}$ are equal and the parts without it are also equal (to make things simpler ) . So now solve the system of equations :

$$x^2+2y^2=5.5$$ and $$2xy=-3$$

I am sure that now you can solve this (for example by eliminating $y$ from the second and plugging in the first and then solving a quadratic )

Finally you should arrive at the only solution : $(-1,1.5)$ .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.