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

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 .

• I m just asking if some more real roots are possible to this equation or not ? Commented Jun 13 at 12:35
• I suppose you meant 'implies' when you used a character pair $=>$. If so, you may want to replace it with the LaTeX/MathJax command \implies, which renders as $\implies$. :) Commented Jun 13 at 14:31

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 $$x^2+3. So, $$\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 $$(-\infty,\infty)$$.

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 $$\sqrt y\ge0$$