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The question

This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:

$$frac{(x-a)sqrt{x-a}+(x-b)sqrt{x-b}}{sqrt{x-a}+sqrt{x-b}}=a-b;a>b.$$

My attempt

Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $sqrt{x-a}+sqrt{x-b}$:

$$begin{align*}
(x-a)sqrt{x-a}+(x-b)sqrt{x-b}&=(a-b)(sqrt{x-a}+sqrt{x-b})\
&=asqrt{x-a}+asqrt{x-b}-bsqrt{x-a}-bsqrt{x-b}
end{align*}$$

Now I have the following:

$$(x-a)sqrt{x-a}+(x-b)sqrt{x-b}=asqrt{x-a}+asqrt{x-b}-bsqrt{x-a}-bsqrt{x-b}$$

Simplifying the RHS as I was out of ideas at that point:

$$xsqrt{x-a}-asqrt{x-a}+xsqrt{x-b}-bsqrt{x-b}=asqrt{x-a}+asqrt{x-b}-bsqrt{x-a}-bsqrt{x-b}$$

I noticed that all one of the common factors $sqrt{x-a},sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $sqrt{x-b}$ terms on one side and $sqrt{x-a}$ terms on the other.

$$sqrt{x-b}(x-a)=sqrt{x-a}(2a-b-x)$$

I tried to then square both sides, but that led to quite a mess.

$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$

I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.

The $a>b$ hint is interesting, but I have no clue what implication it may have here.

I think the $(x-a)sqrt {x-a}$ patterns may mean something, perhaps I could do something with $asqrt a=sqrt{a^3}$, but at this point it is probably a dead end.

I appreciate any help.

Use the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.

We get:
$$frac{(x-a)sqrt{x-a}+(x-b)sqrt{x-b}}{sqrt{x-a}+sqrt{x-b}}=\
frac{(sqrt{x-a}+sqrt{x-b})((x-a)-sqrt{(x-a)(x-b)}+(x-b))}{sqrt{x-a}+sqrt{x-b}}=\
2x-a-b-sqrt{(x-a)(x-b)}=a-b Rightarrow \
(x-a)(x-b)=(2x-2a)^2 Rightarrow \
3x^2+(b-7a)x+4a^2-ab=0 Rightarrow \
x=frac{(7a-b)pm sqrt{(b-7a)^2-12(4a^2-ab)}}{6}=\
frac{7a-bpm (a-b)}{6}=\
frac{4a-b}{3}; a.$$

Hint: Define $$u=sqrt{x-a}\w=sqrt{x-b}$$therefore $${w^3+u^3over u+w}=w^2-u^2$$which yields to $$2u^3=uw^2-u^2w$$one answer is $u=0$ or $x=a$ which is valid. The others can be found by solving $$2u^2=w^2-uw$$or $$u^2+uw=a-b$$by substituting we obtain $$x-a+sqrt{(x-a)(x-b)}=a-b$$

Hint: Write your equation in the form
$$sqrt{x-a}(x+b-2a)=sqrt{x-b}(a-x)$$ and square it. We get in the case of $$a>b$$ $$x=a$$ or $$x=frac{1}{3}(4a-b)$$

Another way is as follows:

• Set $boxed{x = a + t(a-b)}$ for $t geq 0$
  $$begin{eqnarray*}
  frac{(x-a)sqrt{x-a}+(x-b)sqrt{x-b}}{sqrt{x-a}+sqrt{x-b}} & = & a-b \
  & Leftrightarrow & \
  frac{t(a-b)sqrt{t(a-b)}+(t+1)(a-b)sqrt{(t+1)(a-b)}}{sqrt{t(a-b)}+sqrt{(t+1)(a-b)}} & = & a-b \
  & Leftrightarrow & \
  frac{tsqrt{t}+(t+1)sqrt{t+1}}{sqrt{t}+sqrt{t+1}} & = & 1 \
  & Leftrightarrow & \
  (t(sqrt{t+1} + sqrt{t})+sqrt{t+1})(sqrt{t+1}-sqrt{t}) & = & 1 \
  & Leftrightarrow & \
  t+ t+1 - sqrt{t(t+1)} & = & 1 \
  & Leftrightarrow & \
  2t & = & sqrt{t(t+1)} \
  & Leftrightarrow & \
  t =frac{1}{3} & mbox{ or } & t= 0 \
  & stackrel{x = a + t(a-b)}{Leftrightarrow} & \
  boxed{x = a + frac{1}{3}(a-b)} &mbox{ or } & boxed{x= a}
  end{eqnarray*}$$