Don't make mistakes! #2

Algebra Level 5

Add up all integers $x$ that don't satisfy the below equation.

$\large \frac{\sqrt{x-2}}{\sqrt{x-9}}=\sqrt{\frac{x-2}{x-9}}$

Note:

We are taking the fundamental (complex) square root.
If you think all integers satisfy the equation, submit -999999 as your answer.
If you think there are infinitely many integers that don't satisfy the equation, submit 999999 as your answer.

This problem is a part of <Don't make mistakes!> series .

The answer is 42.

2 solutions

Boi (보이)
Jun 16, 2017

We know that $\dfrac{\sqrt a}{\sqrt b}=-\sqrt{\dfrac{a}{b}}$ only when $a\geq0$ and $b<0$ .

Then, the equation won't hold if $x-2>0$ and $x-9<0$ , since if $x-2=0$ , the equation holds.

$2<x<9$ .

So is the answer $3+4+5+6+7+8=\boxed{33}$ ?

No, we should remember that if $x-9=0$ , the equation itself doesn't make sense.

Therefore we should include the case $x=9$ .

The answer is $3+4+5+6+7+8+9=\boxed{42}$ .

FYI I edited the question to indicate that we're allowing for complex square roots. Otherwise, we consider just the (real) square root.

Calvin Lin Staff - 3 years, 11 months ago

Thanks! :D

Boi (보이) - 3 years, 11 months ago

Niranjan Khanderia
Jul 8, 2017

$If~\dfrac{x-2}{x-9}<0,~ or ~ x-9=0,~ we~ do~ not~ have~ a ~real~ solution.\\ \dfrac{x-2}{x-9}=1+\dfrac 7 {x-9}. \\ \therefore~obviously,~no~solution~for~x=9.\\ 1+\dfrac 7 {x-9}<0~~if~\dfrac 7 {x-9}< -1.~~possible~for~x=3, 4, 5, 6, 7, 8.\\ So~the ~required~sum=\displaystyle \sum_{n=3}^9n=\dfrac{(3+9)(8-3+1)} 2=\Large \color{#D61F06}{42.}$

Don't make mistakes! #2

The answer is 42.

2 solutions

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