0
$\begingroup$

So I realised that you can factor $x^2-(a+b)x+ab$ and $x^2-(b+c)x+bc$ to $(x-a)(x-b)$ and $(x-b)(x-c)$ and then you get $$\sqrt{(x-a)(x-b)}+\sqrt{(x-b)(x-c)}=0$$ This is where I got stuck. I tried squaring both sides but I couldn't find the solution.

$\endgroup$
3
  • $\begingroup$ Isn't $x=b$ an obvious solution? $\endgroup$ Feb 11 '18 at 11:47
  • $\begingroup$ Yes, but how do I know that that is the only solution? $\endgroup$
    – Ayy Lmao
    Feb 11 '18 at 11:48
  • $\begingroup$ @AyyLmao There could be two solutions but only if $a=c$, check out my answer. $\endgroup$
    – Yanko
    Feb 11 '18 at 11:49
1
$\begingroup$

So you've got this: $$\sqrt{(x-a)(x-b)}+\sqrt{(x-b)(x-c)}=0$$ And the $\sqrt{x}$ is always non-negative. So you are adding $2$ non-negative numbers together. When can that sum be $0$? When both of the terms are $0$. So you should have both $\sqrt{(x-a)(x-b)}=0$ and $\sqrt{(x-b)(x-c)}=0$. But $\sqrt{x}=0$ only if $x=0$, so you should have $(x-a)(x-b)=0$ and $(x-b)(x-c)=0$. From this, $x=b$ is a solution, and if $a=c$, then $x=a=c$ is also a solution.

$\endgroup$
2
$\begingroup$

You almost done, $x=b$ is a solution, if $x\not = b$ then you can divide by $\sqrt{(x-b)}$ and you have $\sqrt{(x-a)}+\sqrt{(x-c)} = 0$ now a square is always positive so a sum of two squares is zero if both are zero, hence $x=a$ and $x=c$.

So conclusion: $x=b$ is always a zero, and if $a=c$ then $x=a$ is another solution.

$\endgroup$
-1
$\begingroup$

Multiplying both sides with $\sqrt{(x-a)(x-b)}-\sqrt{(x-c)(x-b)}$, we see that necessarily $$(x-a)(x-b)-(x-c)(x-b)=0,$$ i.e., $$ (x-b)(c-a)=0.$$ So $x=b$ (which we verify to actually be a solution), or we need $a=c$. But with $a=c$, the original equation becomes $$ 2\sqrt{(x-a)(x-b)}=0,$$ hence $x=a$ (or again $x=b$)

$\endgroup$
1
  • 1
    $\begingroup$ How do you know that $\sqrt{(x-a)(x-b)}-\sqrt{(x-c)(x-b)}$ isn't zero? If I recall correctly, you can't multiply both sides with zero $\endgroup$
    – Ayy Lmao
    Feb 11 '18 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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