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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 ) :

$$\sqrt{5.5-3\sqrt{2}}=x+y\sqrt{2}$$

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 : $(-1,1.5)$ .

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