Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a contest problem. It asks for all the real solutions of

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$,

for arbitrary real $p$.

From the eq. it's clear that $p$ cannot be negative. By squaring once, splitting the radicals from the other terms and squaring again, I found a formula for $x^2$:

$x^2 = 1+\frac{p^2}{16(1-p/2)}$

From here we have $p<2$ for the second radical to be defined, and by comparing this expression with $p$ we get that $p<2$ makes the first radical be defined as well. Yet solutions of the original equation can only be found when $p\leq 4/3$, and $4/3$ is the (double) root of $9p^2-24p+16=0$ (obtained from comparing $1+\frac{p^2}{16(1-p/2)}$ with $p$).

I don't understand why this happens. Perhaps I've done something wrong. If I haven't, I'd appreciate any help in proving that there are no solutions if $p$ is greater than $4/3$.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

When solving equations, by doing algebra like squaring etc, one could potentially include more solutions than the original equation.

For instance

$x = 1$ implies $x^2 = 1$, by squaring.

The second equation also has $x = -1$ as a solution, which is not a solution of the first one.

In your case, when you square etc, you might lose constraints like $x \ge 1$ and $x^2 \ge p$.

I suppose when you squared, you had the term $x - 2\sqrt{x^2 -1}$.

The constraint that $x \ge 2 \sqrt{x^2-1}$ (i.e. $x^2 \le 4/3$) gives us the constraint that $p \le 4/3$.

As you noticed,

$1 + \frac{p^2}{16(1-p/2)} -p \ge 0$ for all $p$.

Thus

$1 + \frac{p^2}{16(1-p/2)} \le 4/3$ implies that $0 \le 1 + \frac{p^2}{16(1-p/2)} -p \le 4/3 - p$ which implies $4/3 \ge p$.

share|improve this answer
    
thank you. I'm still confused, since 1+... >= p, 0<p<2 is satisfied for all p in [0,2). wolframalpha.com/input/?i=Plot[{1+%2B+p^2/(16*(1+-+(p/2))),+p},+{p,+0‌​,+3}] Anyway I'm probably doing something very wrong. Thank you in advance for any further clarification. –  Weltschmerz Oct 12 '10 at 0:35
    
@Welt: I have updated the answer. I had a mistake in algebra earlier. –  Aryabhata Oct 12 '10 at 0:51
    
very well, thank you. This problem struck me as somewhat odd. –  Weltschmerz Oct 12 '10 at 0:57
    
you stole my rep! And you have more than you know what to do with. I had my answer typed up and was 15 sec from posting when yours hit. Turned out you had covered everything I had to say, and more clearly. Good answer. –  Ross Millikan Oct 12 '10 at 13:33
    
@Ross: Thanks! Sorry about the wasted effort. –  Aryabhata Oct 12 '10 at 13:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.