How to correctly find domain of this irrational equality?

Question

(Not sure if "domain" is a right term, but it seems to be closest to Russian term "Область допустимых значений".)

It seems like my previous post lacked details and caused confusion, so there is more detailed question.

I have equation $$\sqrt{4X+2\sqrt{3X^2+4}}=X+2$$

It has three solutions, namely $$ X_1=2;X_2=-2;X_3=0 $$

I was able to find all three solutions. I threw away $$X_3=0$$ because it was outside the domain (in other words, the original equation would just stop make sense or would be false with X=0). But it turned out that X=0 is a valid solution.

Here is story about how I got my domain wrong, if you are interested. Feel free to skip it. In order to find the domain I tried to solve $$4X+2\sqrt{3X^2+4}\geq0$$

After some manipulations I got $$X\geq-0.5\sqrt{ 3X^2+4 }$$

Then things gone wrong. Here I give my actions step by step:

$$X^2\geq0.25(3X^2+4) $$ $$X^2\geq0.75X^2+1 $$ $$0.25X^2\geq1 $$ $$X^2\geq4 $$ $$X^2-4\geq0 $$ $$(X+2)(X-2)\geq0 $$ $$X\in(-∞, -2]\cup[2, ∞)$$

As you can see, X can't be equal to zero. But in fact, it can because the true domain should be $$X\in[-2, ∞)$$.

Here is story of my wrong path to find the domain. How to correctly find domain of this irrational equality? What is the right way?

Answer 1

As you already noted, the step $X^2\geq0.25(3X^2+4)$ is incorrect, since $x\geq y\implies x^2\geq y^2$ holds only for $x,y\geq 0$.

What you need to do is case analysis: first look at the inequality $X\geq-0.5\sqrt{ 3X^2+4 }$ for $X\geq 0$. This always holds, since the right hand side is negative, so it's correct if $X\in [0,\infty)$

Now we look at $X<0$. Since both sides are negative and $x\geq y\implies x^2\leq y^2$ for $x,y<0$, we obtain $X^2\leq 0.25(3X^2+4)$ if $X<0$. This implies $(X+2)(X-2)\leq0$, or $X\in[-2,2]$ if $X<0$, so $X\in[-2,0)$.

So the ОДЗ is $[-2,\infty)$.

Answer 2

Since the equation has taken the positive sign for the radical you must have first $$x+2\ge 0\iff x=-2+\epsilon\quad \text {where } \epsilon\ge 0$$ Take now $x=-2+\epsilon$ and verify if this value satisfies $LHS$. You have

$$\sqrt{-8+8\epsilon+8\sqrt{1+\frac{3\epsilon^2-12\epsilon}{16}}}$$ Now make $\epsilon\to 0^+$ and you get $LHS\ge 0$ what agrees with $RHS$.

Thus the domain is $[-2,\infty)$

Answer 3

First of all I assume your teachers DO NOT require you to solve the equation over the complex numbers (is it right?). Than let's observe possible limitations, which could make the domain of this equality smaller than $(-\infty;+\infty$). There are three such limitations in your equality:

1. $3x^2+4$ should be $\geq0$ (which is true for ($-\infty;+\infty$))
2. LHS is a square root which is $\geq0$, so RHS should also be $\geq0$ (which is true for $[-2;+\infty$))
3. $4x+2\sqrt{3x^2+4}$ should be $\geq0$, which leads to $2\sqrt{3x^2+4} \geq-4x$. It expands in two ways ($-4x\lt0$ (a) and $-4x\geq0$ (b)):

3a) for $x\gt0$ the solution is any valid LHS root (and we already know it is so for $x$ in ($-\infty;+\infty$)) and the common part of the two is $(0;+\infty)$
3b) for $x\leq0$ we square both sides and solve it via $x^2\leq4$ to $[-2;2]$ and the answer for 3b is $[-2;0]$

, so for 3a & 3b in conjunction the domain is $[-2;+\infty)$.

Composing the three big points above we have the domain equal to $[-2;+\infty)$.