Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$

Question

The equation is $$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$

I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?

I got to:

$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$

Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :) — mrtaurho, Oct 21 '18 at 09:14
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from ${-\sqrt{12},-1,0,\sqrt{12},4}$ may be of interest... — Raymond Manzoni, Oct 21 '18 at 09:29
... and that's why "squaring both sides" is the $\boxed{??}\text{-th}$ thing one must do when he's solving an equation. What number goes in the box? — , Oct 21 '18 at 09:30
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them. — Oscar Lanzi, Oct 21 '18 at 09:33
@OscarLanzi Most radical equations require the same considerations, which may become rather complicated when you actually get to compute them. — , Oct 21 '18 at 09:36
If you start squaring, I suppose that you end with a polynomial of degree $14$. It would be funny. — Claude Leibovici, Oct 21 '18 at 13:49

nonuser · Accepted Answer · 2018-10-21T10:00:45.763

7

Left side exist only iff $${\frac{4-x}x}\geq 0\iff x\in (0,4]$$ and $${\frac{x-4}{x+1}}\geq 0\iff x\in (-\infty,-1)\cup [4,\infty)$$

So the only legitimate value for left side is $4$ which works.

edited Oct 21 '18 at 10:00

answered Oct 21 '18 at 09:53

nonuser

91,557

score 2 · Answer 2 · answered Oct 21 '18 at 09:30

2

Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right

$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$

Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get

$$0=2-\sqrt{4^2-12}\Rightarrow 0=2-\sqrt{16-12}\Rightarrow 0=0$$

And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_{2/3}= -0.889727 \pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.

answered Oct 21 '18 at 09:30

mrtaurho

16,663

1

Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations. – Oct 21 '18 at 09:33
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus. – mrtaurho Oct 21 '18 at 09:37
1

@SaucyO'Path what do you think it should be done before squaring? – Aleex_ Oct 21 '18 at 09:49
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs. – Oct 21 '18 at 10:01

Oscar Lanzi · Answer 3 · 2018-10-21T10:05:50.750

Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.

There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-\sqrt{x^2-12}$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.

So the only place to look for a real solution is $x=4$.

Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$

3 Answers3