# Solving $x\sqrt{2}=1-x$

Solve for $$x$$ of the equation,

$$x\sqrt{2}=1-x$$

According to the solution, $$x$$ is supposed to be $$\sqrt{2}-1$$. The solutions to this test have proven wrong from case to case, so I wonder ...

If you solve it like this: $$x\sqrt{2}+x = 1$$ and pull out $$x$$ to get $$x(\sqrt{2}+1) = 1$$. Then, dividing by $$\sqrt{2}+1$$, we get

$$x=\frac1{\sqrt{2}+1}$$

right?

• The source is incorrect; your answer is correct. Commented Aug 28, 2019 at 1:14
– lulu
Commented Aug 28, 2019 at 1:16
• $\frac 1{\sqrt 2 +1} = \sqrt 2 -1$ so both you and the book are correct. Commented Aug 28, 2019 at 1:16
• Your method is the correct way and I presume the way the book wanted you to do it. But if you remove the radical from the denominator you can express it as the book did. Commented Aug 28, 2019 at 1:20

The convention is, if the answer contains radicals in the denominator, they should be rationalized. Thus,

$$x=\frac1{\sqrt{2}+1} =\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}$$ $$=\frac{\sqrt{2}-1}{(\sqrt{2})^2-1}=\sqrt{2}-1$$

$$\frac 1{\sqrt 2 + 1}=\frac 1{\sqrt 2+1}\frac {\sqrt 2 -1}{\sqrt 2 -1} = \frac {\sqrt 2 -1}{\sqrt 2^2 + \sqrt 2 - \sqrt 2 - 1} = \frac {\sqrt 2 -1}{2-1} =\frac {\sqrt 2-1}1 = \sqrt 2 -1$$.

Yes, is correct $$x=\frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}=\frac{1-\sqrt{2}}{(1-2)}=\sqrt{2}-1$$

Your work seems right. But let's check the given solution:

$$(\sqrt2-1)\sqrt2 =2-\sqrt2=1-(\sqrt2-1)$$

It checks out!

But a linear equation of this form can have only one solution. We conclude that the two solutions must be equivalent. In fact,

$$\frac1{\sqrt2+1} = \frac1{\sqrt2+1}\cdot\frac{\sqrt2-1}{\sqrt2-1} = \frac{\sqrt2-1}{2-1}=\sqrt2-1$$