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When solving quadratic equations like $\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x + 1}$ we are told to solve naively, for example we would get $x \in \{\frac{-\sqrt{5}}{2},\frac{\sqrt{5}}{2}\}$, even though the first solution doesn't work, and then try all the solutions and eliminate the extraneous ones. This is not a very elegant algorithm! How would one use the fact that $\sqrt{x}^2= |x|$ to avoid having to check answers?

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  • $\begingroup$ Do you really consider methods that solve absolute value equations to be more elegant than testing for extraneous solutions? $\endgroup$ – Rory Daulton Dec 13 '14 at 22:11
  • $\begingroup$ I thought maybe there was an elegant way to do it without checking solutions. $\endgroup$ – Elliot Gorokhovsky Dec 13 '14 at 22:11
  • $\begingroup$ There are many questions here at MathSE on solving absolute value equations. These can be fairly tricky to solve and usually involve checking multiple possibilities. I personally consider checking for extraneous solutions to be easier to understand and to do. $\endgroup$ – Rory Daulton Dec 13 '14 at 22:14
  • $\begingroup$ why is not checking is elegant? in fact you just have the faith that whatever the method you applied is correct.there is no reason not to error check the answers. it should be part of problem solving. $\endgroup$ – abel Dec 13 '14 at 23:39
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If you ensure that $$ \begin{cases} x+1\ge0\\ x-1\ge0\\ 2x+1\ge0 \end{cases} $$ then you can square both sides, because they are guaranteed to exist and, when $a,b\ge0$, $a=b$ if and only if $a^2=b^2$.

The conditions above are equivalent to $x\ge1$.

Squaring we get $$ x+1+2\sqrt{x^2-1}+x-1=2x+1 $$ that simplifies to $$ 2\sqrt{x^2-1}=1 $$ and you can square again, because both sides are non negative. This gives $$ 4x^2=5. $$ Since you know that $x\ge1$, the only solution is $$ x=\frac{\sqrt{5}}{2}. $$

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  • $\begingroup$ So in general I would just calculate the domain of the original equation and apply that to my solution set? $\endgroup$ – Elliot Gorokhovsky Dec 14 '14 at 0:28
  • $\begingroup$ @RenéG You must be careful also in the intermediate passages before squaring, because one side might be negative. $\endgroup$ – egreg Dec 14 '14 at 0:36
  • $\begingroup$ @RenéG Your proposed correction is wrong: $\sqrt{x^2-1}$ is defined (because $x\ge1$) and nonnegative by definition. $\endgroup$ – egreg Dec 14 '14 at 0:48
  • $\begingroup$ Then what did you mean by "you can square again, because both sides are non-negative"? $\endgroup$ – Elliot Gorokhovsky Dec 14 '14 at 0:51
  • $\begingroup$ @RenéG The same as before: when $a,b\ge0$, the equality $a=b$ is the same as $a^2=b^2$ (you don't add extraneous solutions, in other words). $\endgroup$ – egreg Dec 14 '14 at 0:52
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Try to square both sides.

$$ \sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1} \Leftrightarrow $$ $$ \Leftrightarrow \left ( \sqrt{x+1} + \sqrt{x-1} \right )^{2} = \left ( \sqrt{2x+1} \right )^{2} \Leftrightarrow $$ $$ \Leftrightarrow \left ( \sqrt{x+1} \right )^2 + 2\sqrt{x+1}\sqrt{x-1} + \left ( \sqrt{x-1} \right )^2 = 2x+1 $$ $$ \Leftrightarrow (x+1) + 2\sqrt{(x+1)(x-1)} + (x-1) = 2x+1 \Leftrightarrow $$ $$ \Leftrightarrow \sqrt{(x+1)(x-1)} = \frac{1}{2} \Leftrightarrow $$ $$ \Leftrightarrow (x+1)(x-1) = \frac{1}{4} \Leftrightarrow $$ $$ \Leftrightarrow x^{2} - 1 - \frac{1}{4} = 0 \Leftrightarrow $$ $$ \Leftrightarrow x^{2} - \frac{5}{4} = 0 $$

And now you can just use the quadratic formula.

$$ x = \frac{-(0)\pm \sqrt{(0)^2-4(1)(-\frac{5}{4}})}{2(1)} \Leftrightarrow $$ $$\Leftrightarrow x = \frac{\pm \sqrt{4\times \frac{5}{4}}}{2} \Leftrightarrow $$ $$\Leftrightarrow x = \frac{\pm \sqrt{5}}{2} $$

Also, $ -\frac{\sqrt{5}}{2} $ is not included in the final solution, try to see why...

I hope I have helped. Saclyr.

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