# Radical equation $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$

The Question: $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$

Only thing I can take from this is that $x^2 -1=(x+1)(x-1)$, but I don't think that would help in any way.

I know the answer, but I don't know how to use that to work backwards.

$x= \frac54$

Enforce the substitution $u=\sqrt{x+1}$ and $v=\sqrt{x-1}$. Then, we see that

$$(u+v)(u+v-2)=0 \tag 1$$

Since $u\ge 0$ and $v\ge 0$, the only solution to $(1)$ is solution to $u=2-v$ or

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

Squaring both sides of $(2)$ yields

$$1+x=4+(x-1)-4\sqrt{x-1}\implies x=\frac54$$

• and here is was squaring both sides +1! – zz20s May 18 '16 at 3:21
• @zz20s That would be the first inclination by most. And thank you for the up vote!! -Mark – Mark Viola May 18 '16 at 3:23
• @Dr.MV, $$\sqrt{x+1}+\sqrt{x-1}=2$$ $$\iff\sqrt{x+1}-\sqrt{x-1}=1$$ Adding we get $$\sqrt{x+1}=\dfrac32$$ Now square – lab bhattacharjee May 18 '16 at 5:10
• @labbhattacharjee Yes, Lab. There is always more than one way to "skin a solution." ;-)) – Mark Viola May 18 '16 at 15:09
• @labbhattacharjee How did you get that second equation? – suomynonA May 19 '16 at 0:08

You have already seen a solution which is much more clever. But let us try whether this can be done with repeated squaring - which is usually one of the first thing which comes to mind in problems like this.

Do not forget that when you square an equation, you can get an extraneous solutions. (See here, here or here.)

We first move $\sqrt{x^2+1}$ to the RHS. We hope to get nicer expressions if we do not square something containing three square roots. \begin{align*} \sqrt{x+1}+\sqrt{x-1}&=x+\sqrt{x^2-1}\\ (\sqrt{x+1}+\sqrt{x-1})^2&=(x+\sqrt{x^2-1})^2\\ (x+1)+2\sqrt{(x+1)(x-1)}+(x-1)&=x^2+2x\sqrt{x^2-1}+(x^2-1)\\ 2x+2\sqrt{x^2-1}&=2x^2-1+2x\sqrt{x^2-1}\\ 2(1-x)\sqrt{x^2-1}&=2x^2-2x-1 \end{align*} Let me also mention that if we look at the LHS, we can see that $(\sqrt{x+1}+\sqrt{x-1})^2$ is precisely twice the expression $x+\sqrt{x^2-1}$. So we got some kind of relation between LHS and the RHS of the equation we started with. Perhaps this is one of possible ways to discover the clever substitution from Dr. MV's answer. But if we do not notice this, we can simply try to continue with squaring.

Now we only have one square root in our equation. We rearrange it in such way that all terms containing the square root are on one side of the equation and the remaining terms are on the other sited. \begin{align*} 2(1-x)\sqrt{x^2-1}&=2x^2-2x-1\\ 4(1-x)^2(x^2-1)&=(2x^2-2x-1)^2\\ 4(x-1)^2(x-1)(x+1)&=(2x(x-1)-1)^2 \end{align*} It looks that at this moment the substitution $x-1=t$ might simplify things a bit. (Of course, we could also continue without this substitution. You can try it that way, if you prefer.) \begin{align*} 4t^3(t+2)&=(2t(t+1)-1)^2\\ 4t^3(t+2)&=(2t^2+2t-1)^2\\ 4t^4+8t^3&=4t^4+8t^3-4t+1\\ 0&=4t+1\\ t&=\frac14\\ x&=t+1=\frac54 \end{align*} So we get $$x=\frac54.$$ Since we used squaring in the process, we also have to check whether $5/4$ fulfills the original equation. And we find out that it does.
The LHS is $\sqrt{\frac94}+\sqrt{\frac14}-\sqrt{\frac{25}{16}-1}=\sqrt{\frac94}+\sqrt{\frac14}-\sqrt{\frac{9}{16}}=\frac32+\frac12-\frac34=2-\frac34=\frac54$. And we get $\frac54$ on the RHS, too.