Question

Solve $x^2+\frac{x^2}{(x+1{)}^2}=3$

• A. $x=\frac{1\pm \sqrt{5}}{2},\frac{-3\pm i\sqrt{3}}{2}$
• B. $x=\frac{7\pm \sqrt{5}}{2},\frac{-5\pm i\sqrt{3}}{2}$
• C. $x=\frac{3\pm \sqrt{5}}{2},\frac{-3\pm \sqrt{3}}{2}$
• D. $x=\frac{1\pm \sqrt{3}}{2},\frac{-3\pm i\sqrt{2}}{2}$

Answer 1

The correct option is A $x=\frac{1\pm \sqrt{5}}{2},\frac{-3\pm i\sqrt{3}}{2}$

We have,

$x^2+\frac{x^2}{(x+1{)}^2}=3$

$x^2+{\left(\frac{x}{x+1}\right)}^2=3$

$x^2+{(1-\frac{1}{x+1})}^2=3$

$x^2+1-\frac{2}{x+1}+{\left(\frac{1}{x+1}\right)}^2=3$

$x^2+2-\frac{2}{x+1}+{\left(\frac{1}{x+1}\right)}^2=4$

$x^2+\frac{2x+2-2}{x+1}+{\left(\frac{1}{x+1}\right)}^2=4$

$x^2+\frac{2x}{x+1}+{\left(\frac{1}{x+1}\right)}^2=4$

${(x+\frac{1}{x+1})}^2=4$

$(x+\frac{1}{x+1})=\pm 2$

So,

$x=\frac{1\pm \sqrt{5}}{2},\frac{-3\pm i\sqrt{3}}{2}$

Hence, this is the answer.