I have trouble solving this first degree equation:

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

I can't see a way to factorize this, so using a distributive aproach I get this:

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

Now my issue is that, how do I get rid of $ \sqrt{2} $ terms ? I tried to squared the whole equation this way: $ 2x² = (x \sqrt{2})² + 3$

But now I'm stuck, the expected result is (a non-detailed results is given, hence I'm looking for how to continue solving):

$ x = \frac{4 + 3 \sqrt{2}}{2} $

Am I not using the correct way to solve this, or I'm mistaking in the evaluation of: $(x \sqrt{2})²$ ?

Or should I even start by squaring both sides to this way:

$ 2(x-1)² = (x + 1) - 1$



2 Answers 2


You don't need to get rid of the $\sqrt2$ term. You're looking to solve this in $x$, so all you need to do is group all $x$ terms together. You have:


Which yields:

\begin{align} \iff & 2x-x\sqrt2=1+\sqrt2\\ \iff &x(2-\sqrt2)=1+\sqrt2 \end{align}

Now you can divide both sides by $(2-\sqrt2)$, because it is nonzero, to obtain

\begin{align} \iff &x=\frac{1+\sqrt2}{2-\sqrt2} \end{align}

At this point, we usually try and rationalize the denominator. In other words, we multiply numerator and denominator both by the same number $c$, so that the value of $x$ is not changed. We choose $c$ such that the denominator (after multiplication) simplifies to a rational number.

This case is particularly easy, and we take advantage of the identity $(a-b)(a+b)=a^2-b^2$. Our choice of constant $c$ is hence simply $2+\sqrt2$. We get

\begin{align} \iff &x=\frac{1+\sqrt2}{2-\sqrt2}\cdot\frac{2+\sqrt2}{2+\sqrt2}\\ \iff &x=\frac{(1+\sqrt2)(2+\sqrt2)}{2^2-{(\sqrt2)}^2}\\ \iff &x=\frac{2+\sqrt2+2\sqrt2+2}{4-2}\\ \iff &x=\frac{4+3\sqrt2}{2} \end{align}

which is the answer sought.


Collect $x$ terms on the LHS and constants on the RHS: $$\big(2-\sqrt{2}\big)x=\sqrt 2+1\Longrightarrow x=\frac{\sqrt 2+1}{2-\sqrt 2}=\frac{3\sqrt 2+4}{2}.$$


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