Solving surds without compairing

Question

Question: Let $a + \sqrt{2b} = 3 - 2\sqrt{2}$ .Find the value of $a - \sqrt{2b}$

What I did: I compared the whole numbers and the irrational numbers in both sides and calculated the answer $3 + 2\sqrt{2}$.

However, I am not very satisfied to do it this way. I could not manage to calculate it with the algebraic formulas.

How to do it?

you have to writ \sqrt{2} in between dollars to get $\sqrt{2}$ — , Jun 24 '14 at 08:21
i do not understand why you do not want to compare those to get value of $a-\sqrt{2b}$.. I do not see any better way of doing this... — , Jun 24 '14 at 08:26
It is not mentioned what kind of numbers they are. I just wanted to know whether it can be done in any other way. — Archisman Panigrahi, Jun 24 '14 at 08:33
If $a$ and $b$ are real, it can't be done anyway. For example, $a=\pi$, $b=(3-2\sqrt2-\pi)^2/2$ — ajotatxe, Jun 24 '14 at 08:37

ajotatxe · Accepted Answer · 2014-06-24T08:42:45.820

4

Assuming that $a$ and $b$ are integer (or rational), we have that $$a-3=2\sqrt2-\sqrt{2b}=\sqrt2(2-\sqrt b)$$ or $$\sqrt b=2-\frac{a-3}{\sqrt2}$$ Squaring, $$b=4-2\sqrt2(a-3)+\frac{(a-3)^2}2$$ That is, $$2\sqrt2(a-3)=4-b+\frac{(a-3)^2}2$$ R.H.S. is rational. Thus, $a-3=0$.

edited Jun 24 '14 at 08:42

answered Jun 24 '14 at 08:36

ajotatxe

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Solving surds without compairing

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