Manipulation of square root function

[Albert1]

$\dfrac {\sqrt {3-3x}+\sqrt {x+6}}{\sqrt {1-4x}+\sqrt {2x+8}}=\dfrac {\sqrt {3-3x}-\sqrt {x+6}}{\sqrt {1-4x}-\sqrt {2x+8}}$
please find :$x$

 

[soroban]

Hello, Albert!

$\dfrac {\sqrt {3-3x}+\sqrt {x+6}}{\sqrt {1-4x}+\sqrt {2x+8}}\:=\:\dfrac {\sqrt {3-3x}-\sqrt {x+6}}{\sqrt {1-4x}-\sqrt {2x+8}}$

$\text{Solve for }x.$

Spoiler

Note that:

. . $\begin{Bmatrix}1-4x \:\ge\:0 & \Rightarrow & x\:\le\:\frac{1}{4}\\ 2x+8 \:\ge\:0 & \Rightarrow & x\:\ge \:\text{-}4 \end{Bmatrix}\;\;\;-4 \:\le x \:\le\:\frac{1}{4}$
We have: .\frac{a+b}{c+d} \:=\:\frac{a-b}{c+d}

. . which simplifies to: .$ad \,=\,bc$

That is: .$\sqrt{3-3x}\cdot\sqrt{2x+8} \:=\:\sqrt{x+6}\cdot\sqrt{1-4x}$

Then: .$(3-3x)(2x+8) \:=\: (x+6)(1-4x)$

. . $6x+24-6x^2-24x \;=\;x-4x^2+6-24x $

. . $2x^2-5x-18\:=\:0\quad\Rightarrow\quad (2x-9)(x+2) \:=\:0 $But $x = \tfrac{9}{2}$ is not in the domain.

Therefore, the only root is: .$x = \text{-}2.$

 

[kaliprasad]

Spoiler

if $\frac{a}{b}= \frac{c}{d}$ then both = $\frac{(a+c)}{(b + d)}$ also $\frac{(a-c)}{(b - d)}$

so if ($\frac{(a+c)}{(b + d)}$ =$\frac{(a-c)}{(b - d)}$
then$\frac{a}{b}= \frac{c}{d}$
taking $a = \sqrt{(3- 3x)}$ $b = \sqrt{(x+6)}$ $c = \sqrt{(1-4x)}$ $d=\sqrt{(2x+8)}$
we get $\sqrt{\frac{(3-3x)}{(x+6)}}$ = $\sqrt{\frac{(4x-1)}{(2x+8)}}$

square both sides and get
(3x-3)(2x+8) = (x+6)(4x-1)
expand to get $2x^2 – 5x – 18 = 0$
(x+2)(2x-9) = 0
x = - 2 or 9/2
x = 9/2 is erroneous so x = -2 is the solution

 

[Albert1]

Thanks Soroban and Kaliprasad , both of your answers are correct :)