Find $a$ such that $\dfrac{25}{\sqrt{x-1}} + \dfrac{4}{\sqrt{a-2}} = 14 - \sqrt{x-1}-\sqrt{a-2}$ has at least 1 solution.

Question

I tried letting by letting $u = \sqrt{x-1}$ and $v = \sqrt{a-2}$. Simplified to: $$uv^2 - u^2v +14uv -25v -4u =0 $$

$0< u + v < 14$

From here I can't complete the squares or proceed further. Any hints on how to approach this?

Answer 1

Notice $$ \begin{align} &\frac{25}{u} + \frac{4}{v} = 14 - u-v\\ \iff & \frac{25}{u} - 10 + u + \frac{4}{v} - 4 + v = 0\\ \iff & \frac{25 -10u + u^2}{u} + \frac{4 -4v + v^2}{v} = 0\\ \iff & \frac{(5-u)^2}{u} + \frac{(2-v)^2}{v} = 0\\ \iff & u = 5\land v = 2 \end{align} $$ In order for the equation to have at least a solution in $x$, $a$ need to satisfy $\sqrt{a-2} = v = 2 \iff a = 6$.

If $a$ do equal to $6$, the equation have an unique solution in $x$ given by $\sqrt{x-1} = u = 5 \implies x = 26$.

Answer 2

With $u=\sqrt {x-1}$ and $v=\sqrt {a-2}$, we have $25/u +u=14-4/v-v.$

Let $b=14-4/v-v.$

So $25/u +u=b,$ so $u^2-bu+25=0.$ By the Quadratic Formula, if $b\in\Bbb R$ then this has a solution $u\in\Bbb R$

iff $b^2\ge 100$ iff $|b|\ge 10$ iff

iff $|14-4/v-v|\ge 10$ iff $(4/v+v\ge 24 \lor 4/v+v\le 4)$ iff

$(\bullet )$ iff $(v>0\land [4/v+v\ge 24 \lor [ 4/v+v\le 4])$ iff

iff $(v>0\land [v^2-24v+4\ge 0 \lor v^2-4v+4\le 0).$

I leave the rest to you. When you find the range of $v$, you can get the range of $a$. The insertion of $\quad$ "$v>0$" in $(\bullet )$ comes from $v=\sqrt {a-2}\ge 0$ and from the occurrence of $4/v$ in the first line..

Answer 3

Since square roots are strictly positive (else denominators would vanish and equation would be ill defined),

then let set $\begin{cases}\sqrt{x-1}=5e^u\\\sqrt{a-2}=2e^v\end{cases}$

Reporting in the equation gives: $$5e^{-u}+2e^{-v}=14-5e^u-2e^v\iff 10\cosh(u)+4\cosh(v)=14$$

But since $\cosh \ge 1$ with equality only in $0$ this has an unique solution which is $u=v=0$

Substituting back gives $x=26$ and $a=6$

Answer 4

Since $u,v>0,$

$u+\frac{25}u\ge10$ and $v+\frac4v\ge4$

hence

$$u+\frac{25}u+v+\frac4v\ge14,$$

with equality iff $u=5$ and $v=2,$ i.e. $x=26$ and $a=6.$