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 :)

 

[CellCoree]

I would first analyze the given equation and determine its properties. The equation involves the manipulation of square root functions, specifically in the numerator and denominator.

Upon closer examination, it appears that the equation is a rational function, with two square root functions in the numerator and two in the denominator. The numerator and denominator also have similar terms, with one term being the opposite of the other (i.e. $\sqrt {3-3x}$ and $-\sqrt {3-3x}$).

Using algebraic techniques, I can manipulate the given equation to solve for $x$. I would first simplify the equation by rationalizing the denominator, which means multiplying both the numerator and denominator by the conjugate of the denominator. This would eliminate the square root functions in the denominator and result in a simpler equation.

Next, I would square both sides of the equation to eliminate the remaining square root functions. This would result in a quadratic equation, which can be solved using traditional methods such as factoring, completing the square, or using the quadratic formula.

Once I have obtained the solutions for $x$, I would then check for any extraneous solutions by plugging them back into the original equation. If the solutions satisfy the original equation, then they are the valid solutions for $x$. However, if they do not, then they are extraneous solutions and should be discarded.

In conclusion, as a scientist, I would approach the manipulation of the square root function in a systematic and logical manner, using algebraic techniques to solve for $x$ and verifying the solutions to ensure their validity.