Find all real roots of the irrational equation $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3} = \sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ .

Question

Find all real roots of the irrational equation $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3} = \sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ .

My try :

Given, $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3} = \sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ .

Let $u = x^2-2x-3, v = 3x^2+3x+5, w = x^2 + 3$ . Then, $3x^2+x-1 = u + v - w$ .

So, the above equation becomes $\sqrt{u+v-w} + \sqrt{u} = \sqrt{v} + \sqrt{w} => \sqrt{u + v - w} = \sqrt{v} + \sqrt{w} - \sqrt{u} => u + v - w = u + v + w - 2\sqrt{uv} + 2\sqrt{vw} - 2\sqrt{uw} => 2w - 2\sqrt{w}(\sqrt{u} - \sqrt{v}) - 2\sqrt{uv} = 0 => w - \sqrt{w}(\sqrt{u} - \sqrt{v}) - \sqrt{uv} = 0 => \sqrt{w} = \sqrt{v}, -\sqrt{u}$ .

If $\sqrt{w} = \sqrt{v} => w = v => x^2 + 3 = 3x^2+3x+5 $ which have no real roots .

Now if $\sqrt{w} = -\sqrt{u} => w = u => x^2 + 3 = x^2-2x-3 => x = -3$ .

Hence, the only real root so far satisfying the equation is $x = -3$ .

But I'm not sure if it's the only possible real root .

Answer 1

One of the real roots is $x=-3$ since both $3x^2+x-1=3x^2+3x+5$ and $x^2-2x-3=x^2+3$ are satisfied by $\color{red}{x=-3}$.

Now, we have to check whether other real roots also exist.

For $x>-3$, it is clear that $3x^2+3x+5>3x^2+x-1$ and $x^2+3>x^2-2x-3$. So, $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3} < \sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ for $x>-3$, and therefore no real root in $(-3,\infty)$.

For $x<-3$, it is clear that $3x^2+3x+5<3x^2+x-1$ and $x^2+3<x^2-2x-3$. So, $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3} > \sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ for $x<-3$, and therefore no real root in $(-\infty,-3)$.

Thus, $\color{red}{\boxed{x=-3}}$ is the only real root of the given equation. $\blacksquare$

PS: The domain of $\sqrt{3x^2+x-1} + \sqrt{x^2-2x-3}$ is $(-\infty,-1]\cup[3,\infty)$, whereas that of $\sqrt{3x^2+3x+5} + \sqrt{x^2+3}$ is $(-\infty,\infty)$.

Answer 2

Observe that $$(3x^2+3x+5)-(3x^2+x-1)=(x^2+3)-(x^2-2x-3)\ \ \ \ (0)$$

By the given condition, $$\sqrt{3x^2+3x+5}-\sqrt{3x^2+x-1}=-(\sqrt{x^2+3}-\sqrt{x^2-2x-3})\ \ \ \ (1)$$

What if $\sqrt{3x^2+3x+5}-\sqrt{3x^2+x-1}=0$

else by $(0)/(1),$

$$\sqrt{3x^2+3x+5}+\sqrt{3x^2+x-1}=-(\sqrt{x^2+3}+\sqrt{x^2-2x-3})\ \ \ \ (2)$$

But for real $y\ge0,$ the principal value of $\sqrt y\ge0 $