Question

The value of $x$, which satisfies $(x-3)(x+1)+3(x-3)\sqrt{\left(\frac{x+1}{x-3}\right)}-28=0$ is

• A. $1+2\sqrt{5}$
• B. $1-\sqrt{53}$
• C. $1-2\sqrt{5}$
• D. $1+\sqrt{53}$

Answer 1

Answer: (A), (B)

The correct options are
A $1+2\sqrt{5}$
B $1-\sqrt{53}$

The given equation is valid for $\frac{x+1}{x-3}>0$
$\therefore x\in (-\infty ,-1)\cup (3,\infty )$ ...(1)

And the given equation can be written as

$(x-3)(x+1)+3(x-3)\frac{\sqrt{\left|(x+1)\right|}}{\sqrt{|x-3|}}-28=0$ ...(2)

On that domain, we have

$3\frac{(x-3)\sqrt{|x+1|}}{\sqrt{|x-3|}}=3\sqrt{(x+1)(x-3)}$, if $x>3$ and

$3\frac{(x-3)\sqrt{|x+1|}}{\sqrt{|x-3|}}=-3\sqrt{(x+1)(x-3)}$, if $x<-1$

Therefore equation (2) is equivalent to the collection

$\{\begin{array}{c}(x-3)(x+1)+3\sqrt{(x+1)(x-3)}-28=0\text{if}x>3\\ (x-3)(x+1)-3\sqrt{(x+1)(x-3)}-28=0\text{if}x<-1\end{array}\iff \{\begin{array}{c}(\sqrt{(x-3)(x+1)}+7)(\sqrt{(x-3)(x+1)}-4)=0\text{if}x>3\\ (\sqrt{(x-3)(x+1)}-7)(\sqrt{(x-3)(x+1)}+4)=0\text{if}x,-1\end{array}$

First system from this Collection

$\sqrt{(x-3)(x+1)}+7>0\therefore \sqrt{(x-3)(x+1)}-4=0\text{If}x>3\Rightarrow (x-3)(x+1)=16\Rightarrow x^2-2x-19=0\therefore x=\frac{2\pm \sqrt{4+76}}{2}\therefore x=1\pm 2\sqrt{5}\text{But}x>3,\therefore x=1+2\sqrt{5}$

And second system from this collection

$\sqrt{(x-3)(x+1)}+4>0\therefore \sqrt{(x-3)(x+1)}-7=0\text{If}x>-1\Rightarrow (x-3)(x+1)=49\Rightarrow x^2-2x-52=0\therefore x=\frac{2\pm \sqrt{4+208}}{2}x=1\pm \sqrt{1+52}\therefore x=1\pm \sqrt{53}\text{But}x<-1,\therefore x=1-\sqrt{53}$

Combining above two results, we get the solution of the original equation as

$x_1=1+2\sqrt{5}\text{and}{x}_2=1-\sqrt{53}$.