I have an equation like this:

$6y−\sqrt{2y^2-1} = 7y - 2$

The solution tells me to first move the $6y$ to the other side by subtracting it:

$-\sqrt{2y^2-1} = y - 2$

And then they suggest to divide (or multiply) by $-1$ to obtain:

$\sqrt{2y^2-1} = 2 - y$

​Now, at the second step, what always gets me is that I'm tempted to square both sides, instead of multiplying by $-1$ like so in order to get rid of the minus sign and the radical in one step:

$$-\sqrt{2y^2-1} = y - 2$$ $$(-\sqrt{2y^2-1})^2 = (y - 2)^2$$ $$2y^2-1 = (y - 2)^2$$

But that leads to a wrong solution. Why can't I get rid of the minus sign and the radical in one step and more generall, when does it make sense to multiply both sides of an equation with $-1$, what's the rule?

  • $\begingroup$ "But that leads to a wrong solution. " Why do you assume that? It does not lead to a wrong answer. $\endgroup$ – fleablood Jan 3 '18 at 8:46
  • $\begingroup$ To be flippant, it doesn't matter what you do to each side. As long as it is the same thing you will get a true statement. You can divide both sides by 37, add the square root of pi, and take the cosines of both sides for all anyone cares. (It won't help you solve for y, but you will get a true, albeit convoluted and useless, statement.) [ If $2x + 3 = 7$ then $\cos (\frac{2x+3}{37} + \sqrt{\pi}) = \cos (\frac{7}{37} + \sqrt{\pi})$. That IS true.... It just doesn't help you get the answer.] $\endgroup$ – fleablood Jan 3 '18 at 8:52

Why does it give a wrong solution? Continuing your books solution, let us square both sides to get: $$(\sqrt {2y^2-1})^2 =(2-y)^2 $$ $$\implies (\sqrt {2y^2-1})^2=y^2-4y+4=(y-2)^2$$

which is the same you got.

Note that an easy way to eliminate a $- $ sign when you encounter it in an equation is to square it on both sides. There is no hard-and-fast rule to deal with such situations.

  • $\begingroup$ Ah I see, yeah I think I got confused somewhere else then. But in general, it is ok then to square the radical to get rid of the minus (given I also square the other side of the equation)? $\endgroup$ – Max Jan 3 '18 at 8:24
  • $\begingroup$ @Max Yes, it's ok. But be aware that it will introduce wrong solutions in some situations. If $A=B$, it's true that $A^2=B^2$, but if $A^2=B^2$, then $(A+B)(A-B)=0$ and $A=B$ or $A=-B$ are both valid. $\endgroup$ – Jean-Claude Arbaut Jan 3 '18 at 8:26

No difference

$$(-\sqrt{2y^2-1})^2 = (y - 2)^2$$

Is the same as $$(\sqrt{2y^2-1})^2 = (y - 2)^2 = (-(y - 2))^2$$


You only have this kind of issues for inequalities.

Here it is important to check that the final solution is such that esists

$$\sqrt{2y^2-1}\implies y\geq\frac{\sqrt2}{2} \quad y\leq\frac{-\sqrt2}{2}$$


You don't have to multiply by $-1$. But you have to note that $-\sqrt{2y^2 - 1}$ is a negative or zero number, so $y-2 \le 0$.

If you multiply by $-1$ (which the book only does because it thinks it will be easier for you) you will get $\sqrt{2y^2 - 1} = 2-y$. In this case you have to note that $\sqrt{2y^2 - 1}$ is a positive of zero number so $2-y \ge 0$.

But $y-2 \le 0 \iff 2-y \ge 0 \iff y\le 2$. It doesn't matter which one you not as long as you note one of them.

Now to solve you must square both sides.

If you didn't multiple by $-1$ you get

$(-\sqrt{2y^2 - 1})^2 = (y-2)^2$

$2y^2 - 1 = y^2 -4y +4$.

$y^2 +4y - 5 = 0$

If you did multiply by $-1$ you get

$(\sqrt{2y^2 - 1})^2 = (2-y)^2$

$y^2 - 1 = 4 -4y + y^2$

$y^2 + 4y - 5 = 0$.

You see the difference? (That was a joke. There IS no difference. $(-k)^2 = k^2$ so $(-\sqrt{2y^2 - 1})^2 = \sqrt{2y^2 - 1}^2$ and $(y-2)^2 = (2-y)^2$.

But there are two things I must point out.

1) The book thinks it's easier to multiply by $-1$ to get rid of the minus sign to make things easier to explain. It's actually .... not important.

2) When you square both sides you add in an extra possible answer (maybe) that may not be correct. But this happens WHETHER OR NOT you multiply by $-1$.... and that is why you have to note $y \le 2$.


To finish

$y^2 +4y - 5 = 0$

$(y +5)(y-1) = 0$

So either $y+5 = 0$ OR $y-1 = 0$.

So either $y = -5$ OR $y=1$. As we know $y \le 2$ we know both of these answers are acceptable.

If $y = 1$ we can check either: Does $-\sqrt{2y^2 - 1} {? \over =} (y-2)$

$-\sqrt{2(1)^2 - 1} {? \over =} ((1)-2)$

$-\sqrt{2-1} {? \over =} -1$

$-\sqrt{1} {? \over =} -1$

$-1 = -1$. Check it does.


we could check:

Does $\sqrt{2y^2 - 1} {? \over =} (2-y)$

$\sqrt{2(1)^2 - 1} {? \over =} (2- (1))$

$\sqrt{2-1} {? \over =} 1$

$\sqrt{1} {? \over =} 1$

$1 = 1$. Check it does.

The only difference is in one you don't get distraction by the minus sign.

That may or may not make things easier.


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