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$$ \sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1} $$

How many solutions does it have for $x \in \mathbb{R}$?

I squared it once, then rearranges terms to isolate the radical, then squared again. I got a linear equation, which yielded $x = \frac54$, but when I put that back in the equation, it did not satisfy.

So I think there is no solution, but my book says there is 1.

Can anyone confirm if there is a solution or not?

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  • $\begingroup$ Mhhh, are you sure it's the real equation ?? No copy-paste mistake ? Because this $$\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}$$ does not have any real nor complex solutions ! But for example, this : $$\sqrt{x+1} - \sqrt{x-1} = \sqrt{x-4}$$ has a real solution... ($x = 2/3*(2+\sqrt{19})$) $\endgroup$
    – Lery
    Commented Sep 7, 2013 at 13:23
  • $\begingroup$ Yes, question is copied correctly. $\endgroup$
    – xylon97
    Commented Sep 7, 2013 at 13:30

4 Answers 4

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$$x+1+x-1-2\sqrt{x+1}\sqrt{x-1}=4x-1\implies(2x-1)^2=4(x^2-1)\implies$$

$$4x^2-4x+1=4x^2-4\implies 4x=5\implies x=\frac54$$

But, indeed

$$\sqrt{\frac54+1}-\sqrt{\frac54-1}\stackrel ?=\sqrt{5-1}\iff\frac32-\frac12=2$$

which is false, thus no solution.

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Square both sides to obtain $2x-2\sqrt{x+1}\sqrt{x-1}=4x-1$ or $-2\sqrt{x+1}\sqrt{x-1}=2x-1$.

Since the left side is not positive, $2x-1\leq0$ or $ x\leq \dfrac12$. But $x\geq 1$, so there is no real solution.

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    $\begingroup$ @DonAntonio Yes, thank you. I've edited my answer. $\endgroup$
    – Alraxite
    Commented Sep 7, 2013 at 15:35
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Simply square both sides twice to remove radicals $$ \begin{align} 2x-2\sqrt{x^2-1}&=4x-1\\ 1-2x&=2\sqrt{x^2-1}\\ 4x^2-4x+1&=4x^2-4\\ 5&=4x\end{align} $$ Plugging $x=\frac54$ into the original equation yields $1=2$. Thus, there is no solution

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As $(x+1)-(x-1)=2$ and given that $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}\ \ \ \ (1)$

$$\sqrt{x+1}+\sqrt{x-1}=\frac2{\sqrt{4x-1}}\ \ \ \ (2)$$

On addition, $$2\sqrt{x+1}=\sqrt{4x-1}+\frac2{\sqrt{4x-1}}=\frac{4x+1}{\sqrt{4x-1}}$$

$$\implies 2\sqrt{x+1} \sqrt{4x-1}=4x+1 \ \ \ \ (3)$$

Squaring we get $4x=5\iff x=\frac54$ which does not satisfy $(1)$ (the given equation) and $(2)$ but satisfies $(3)$

In fact, $x=\frac54$ is a root of $\sqrt{x+1}+\sqrt{x-1}=\sqrt{4x-1}$ and is an Extraneous root of $(1),(2)$

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    $\begingroup$ I am trying to figure out how an Extraneous root is introduced here even before squaring $\endgroup$ Commented Sep 7, 2013 at 13:18

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