The equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ has how many solutions?

Question

What would be the correct approach to this problem?Squaring seems to make it even more complicated!

P.S:Sorry,at first I wanted to avoid invalid roots.But thanks for your help.Its a good idea to check at the end.

Really? Have you actually tried squaring it? Square once and rearrange. Then square again. It is not at all hard. — almagest, Jul 05 '16 at 05:11
After squaring, do a little manipulation and square again. You will get a pleasant equation. Solve. Then check whether the solution(s) is/are solutions of the original equation (the squarings could introduce "extraneous" roots that are not roots of the original equation). — André Nicolas, Jul 05 '16 at 05:12
@almagest I thought squaring may cause "extraneous" roots.So...But i think its a good idea to just check at the end! — , Jul 05 '16 at 05:17
@ZOZ That is a whole different issue. Please make your questions clearer. Yes, of course, squaring may introduce bogus roots. You carry out a series of implications: if $x$ is a root of this, then it is a root of this, and so on. Then at the end, it is essential to check whether the roots you have found are roots of the original equation. For example, if you found $x=-2$, that would not work, because the second square root is not defined for $x=-2$ (assuming you are not working in the complex domain). — almagest, Jul 05 '16 at 05:21

score 6 · Accepted Answer · answered Jul 05 '16 at 05:14

Like someone said, just do it. The original equation is equivalent to: $$\frac{2}{\sqrt{x-1}+\sqrt{x+1}}=\sqrt{4x-1} $$ where every term makes sense iff $x\geq 1$. With such assumption, however, the LHS is $\leq\sqrt{2}$ while the RHS is $\geq\sqrt{3}$. So, no solutions.

score 4 · Answer 2 · answered Jul 05 '16 at 05:13

4

First, we should have $x \ge 1$. Now, the equation can be re-written as:

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

The RHS is $\ge \sqrt3 \times \sqrt2 > 2$. So, the equation has no solutions.

answered Jul 05 '16 at 05:13

score 2 · Answer 3 · answered Jul 05 '16 at 07:08

You don't really need to calculate with square roots at all: you need only know that $\sqrt x<\sqrt y$ whenever $1\le x<y$. The expression as a whole makes sense if and only if $x\ge 1$, in which case

$$\sqrt{x+1}-\sqrt{x-1}\le\sqrt{x+1}<\sqrt{4x-1}\;,$$

where the last inequality follows from the fact that $1\le x+1<4x-1$. Thus, there is no solution. (In fact $x+1<4x-1$ for $x>\frac{2}3$.)

score 1 · Answer 4 · answered Jul 05 '16 at 05:15

1

Hint: first note the domain of the equation is the interval $[1,\infty)$. Define $$f(x):=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1}$$

Now note that $f(1)<0$ and $f$ is decreasing.

answered Jul 05 '16 at 05:15

symplectomorphic

18,604

Ross Millikan · Answer 5 · 2016-07-05T05:26:04.827

1

Squaring helps. You cut three square roots to one, then isolate it and square again. $$\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}\\x+1-2\sqrt{x^2-1}+x-1=4x-1\\ -2\sqrt{x^2-1}=2x-1\\4x^2-4=4x^2-4x+1\\x=\frac 54$$ As we have squared, we have to check the solution we have found, which fails as $\sqrt{\frac 54+1}-\sqrt{\frac 54-1}=\frac 32-\frac 12=1 \neq \sqrt{4 \cdot \frac 54-1}=2$ . There is no solution. This is not surprising. The two terms on the left are not very different, so their difference will be too small to equal the term on the right.

edited Jul 05 '16 at 05:26

answered Jul 05 '16 at 05:15

Ross Millikan

374,822

It should be $x=5/4$. – Takahiro Waki Jul 05 '16 at 05:16
Because the answer needs to satisfy $1-2x>0$. – Takahiro Waki Jul 05 '16 at 05:30
@TakahiroWaki: I dropped a sign which led to the $3/4$ instead of $5/4$. I don't see why we need $1-2x \gt 0$. We need all the radicals to be positive, which means we need $x \ge 1$ – Ross Millikan Jul 05 '16 at 13:26
On third equation, LHD is negative, so RHD must be negative. Then 5/4 is not smaller than 1/2, so 5/4 is not answer. – Takahiro Waki Jul 06 '16 at 06:47
@TakahiroWaki: Yes, I get that it is not the answer by another route. You are right I could have stopped there. – Ross Millikan Jul 06 '16 at 15:23

Tushar Rakheja · Answer 6 · 2016-07-05T05:26:53.303

0

No solution.

Squaring actually does not make it much worse! If you square both sides you get.

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

Rearrange this to get.

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

Square both sides now. You'll be left with a linear equation in x.

When you solve it, you'll get 5/4, which won't satisfy the original equation. So nothing more to be done.

edited Jul 05 '16 at 05:26

answered Jul 05 '16 at 05:20

Tushar Rakheja

166
6

The equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ has how many solutions?

6 Answers6