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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?

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

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