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I'm familiar with equations like:

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

Has no solutions, it's just an example off the top of my head

Just move the negative square root to the other side, square both sides and solve.

$\sqrt{x+1} = \sqrt{x+2}$

$x+1 = x+2$

0 = 1


My question is, if there are two square roots on one side, then can I still square both sides in this way:

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

$x+1 - (x+2) = x+3$

$x+1 - (x-2) = x+3$

$x = -4$

Or does squaring both sides cause something strange to happen on the left hand side?

altec
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    Note that actually $\left(\sqrt{x+1}- \sqrt{x+2}\right)^2 \color{red}{\neq} (x+1)-(x+2)$. To square $\sqrt{x+1}- \sqrt{x+2}$ correctly, try using the rule $(a-b)^2 = a^2 -2ab + b^2$. – Minus One-Twelfth Feb 15 '19 at 23:07
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    Notice, by the way, that your second example can't possibly have any solutions because the left-hand side is always negative and the square root function is generally defined to have the positive square root. This is a good reminder to confirm any potential solutions you may find by substituting them back into your original equation. – Robert Shore Feb 15 '19 at 23:10
  • It is called Freshman's dream – farruhota Feb 16 '19 at 06:06

4 Answers4

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No, what you did was wrong. this is because :

$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ does not imply $(x+1)-(x+2)=x+3$.

You need to square both sides, which means:

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

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

As a sidenote, you can plug x=-4 into the original equation to check whether x=-4 is not a solution.

Locally unskillful
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Hint: $(\sqrt{x+1} - \sqrt{x+2})^2 = (x+1) - 2\sqrt{x+1}\sqrt{x+2} + (x+2)$

Max
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Yes, you can square both sides. But you do actually have to square both sides.

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

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

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

$-2\sqrt{x+1}\sqrt{x+2} = -x$

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

$4(x+1)(x+2) = x^2$

$4x^2 +12x + 8 = x^2$

$3x^2 + 12x + 8 = 0$

$x = \frac {-12\pm{144-4*3*8}}{6} =4

$-2 \pm \frac{\sqrt{48}}6 = -2 \pm \frac 2{\sqrt 3}$.

BUt note: Squaring both sides gives extraneous solutions because it ignores whether terms given are positive or negative. These answers all don't actually work if we plug them back in.

.....

Obviously you can't just square parts of each side and expect the result to make any sense.

$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ obviously does not mean

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

Because $(\sqrt{x+1} - \sqrt{x+2})^2 \ne (\sqrt{x+1})^2 - (\sqrt{x+2})^2$.

fleablood
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  • Ah yeah this makes much more sense. Thanks! As a side note, can extraneous solutions come up after squaring something, even without radicals being in the equation? – altec Feb 16 '19 at 00:11
  • Yes, they can. For example, try squaring both sides of the equation $x=-1$, and you should see an extraneous solution pop up. – Minus One-Twelfth Feb 16 '19 at 00:58
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How I explain it to myself is negative if you want to multiply both sides of the equation by the same number, say 3, you can multiply-across through the pluses and minuses by 3, or take it in stages by putting a bracket around both sides, but the 3 at the start of the brackets on both sides and then multiply across in through the brackets(effectively inflating both sides by the same amount)... but you can’t square across, you need to square both sides, and I find using Brackets here are always a big help. And if you’ve got say, two square roots (radicals) added together (or subtracted) on one side then it’s an easier calculation to slide one over to the other side, change the plus to a minus , and vice versa, and then square both sides eg $\sqrt{2x-1} + \sqrt{x-1} = 5$ next step $\sqrt{2x-1} = 5 – \sqrt{x-1}$

Jose M Serra
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keepitassimpleasyoucan
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