Dexter Odom

## Answered question

2022-04-04

Find all the real roots of
${x}^{2}+\frac{{x}^{2}}{{\left(x+1\right)}^{2}}=3$

### Answer & Explanation

pypberissootcu

Beginner2022-04-05Added 14 answers

Step 1
${x}^{2}+\frac{{x}^{2}}{{\left(x+1\right)}^{2}}=3$
$\left(x\ne 1\right)$
$⇒{x}^{2}+\frac{{x}^{2}}{{\left(x+1\right)}^{2}}-3=0$
Which simplifies as $\frac{{x}^{4}+2{x}^{3}-{x}^{2}-6x-3}{{\left(x+1\right)}^{2}}=0$
$⇒{x}^{4}+2{x}^{3}-{x}^{2}-6x-3=0$
Factorising, we get
$\left({x}^{2}-x-1\right)\left({x}^{2}+3x+3\right)=0$
For which we get two equations to solve
${x}^{2}-x-1=0$
${x}^{2}+3x+3=0$
Solving the first eqn. we get $x=\frac{1±\sqrt{5}}{2}$ The second equation has no (real) solution.
$\therefore x=\frac{1±\sqrt{5}}{2}$

Mikaela Winters

Beginner2022-04-06Added 14 answers

Step 1
Factorize the equation as follows
$0={x}^{2}+\frac{{x}^{2}}{{\left(x+1\right)}^{2}}-3$
$=\frac{1}{{\left(1+x\right)}^{2}}\left({x}^{2}{\left(x+1\right)}^{2}+{x}^{2}-3{\left(x+1\right)}^{2}\right)$
$=\frac{1}{{\left(1+x\right)}^{2}}\left({x}^{4}+2\left(x+1\right){x}^{2}-3{\left(x+1\right)}^{2}\right)$
$=\frac{1}{{\left(1+x\right)}^{2}}\left({x}^{2}-\left(x+1\right)\right)\left({x}^{2}+3\left(x+1\right)\right)$
which, from the factor ${x}^{2}-x-1$, yields the real solution
$x=\frac{1}{2}±\frac{\sqrt{5}}{2}$

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