# Square-root equation with no solutions

There is an equation given:

$${\sqrt {3x-7}} + {\sqrt {2x-1}} = 0$$

Solving it algebraically:

$${(\sqrt {3x-7})^2} + {(\sqrt {2x-1})^2} = 0$$ $$3x-7 + 2x-1 = 0$$ $$5x = 8$$ $$x = \frac{8}{5}$$

But doing it $${\sqrt {3x-7}} = {-\sqrt {2x-1}}$$ $${(\sqrt {3x-7})^2} = {(-\sqrt {2x-1})^2}$$ $$3x-7 = 2x-1$$ $$x = 6$$

Both answer don't satisfy the given equation. I was told that you can't add two principle root's and expect to get a value zero, therefore there's no solution. But is that conclusion right?

• Your first step takes you from $\alpha+\beta=0$ to $\alpha^2+\beta^2=0$ without justification. Jan 27 '19 at 15:02

Talking about principle roots of reals, we have $$\sqrt x\ge 0$$ whenever it is defined (i.e., when $$x\ge 0$$). Hence both roots in the equation must be non-negative and so $$\underbrace{\sqrt{3x-7}}_{\ge 0}+\underbrace{\sqrt{2x-1}}_{\ge 0}=0$$ is only possible when both summands on the left are $$=0$$. This, however, leads to $$3x=7$$ and $$2x=1$$, which cannot hold at the same time.