Find all the real roots of $x^2 + \frac{x^2}{(x+1)^2}=3$.
I got a 4th power equation after simplifying but didn't understand what to do afterwards
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I think it is a good idea to add the 4th degree equation you get to the question. – Etemon Dec 29 '20 at 04:55
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1The real roots have exceptionally complicated forms (see Wolfram), so I doubt there's must you really can do except using the quartic formula. I do notice you can factor $x^2$ on the left hand side, too, but I don't think you'll get anywhere useful that way. I imagine the best you can realistically do is approximate – PrincessEev Dec 29 '20 at 04:57
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2Is it $x^2+x^2(x+1)^2=3$ or is it $x^2+\frac{x^2}{(x+1)^2}=3$? Be clear. – Ottavio Dec 29 '20 at 05:06
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THE SECOND ONE. – will the wise Dec 29 '20 at 05:07
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5Does this answer your question? Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$ – lab bhattacharjee Dec 29 '20 at 05:07
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1The substitution $t=1+x$ bring your equation to a more symmetric form $(1-t)^2+\left(1-\dfrac1t\right)^2=3.$ In particular, this shows that if $t$ is a solution, then $\dfrac1t$ is also a solution. – Bumblebee Dec 29 '20 at 05:32
3 Answers
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Factorize the equation as follows
\begin{align} 0= & x^2 + \frac{x^2}{(x+1)^2}-3\\=& \frac1{(1+x)^2}\left(x^2(x+1)^2+x^2-3(x+1)^2 \right) \\ = &\frac1{(1+x)^2}\left(x^4 +2(x+1)x^2-3(x+1)^2 \right) \\ = &\frac1{(1+x)^2}(x^2-(x+1))(x^2+3(x+1)) \\ \end{align} which, from the factor $x^2-x-1$, yields the real solution $$x=\frac12\pm \frac{\sqrt5}2$$
Quanto
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$x^2+\frac{x^2}{(x+1)^2}=3$ (x≠1)
$\Rightarrow x^2+\frac{x^2}{(x+1)^2}-3=0$
Which simplifies as $\frac{x^4+2x^3-x^2-6x-3}{(x+1)^2}=0$
$\Rightarrow x^4+2x^3-x^2-6x-3=0$
Factorising, we get
$(x^2-x-1)(x^2+3x+3)=0$
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.
$\therefore x=\frac{1±\sqrt{5}}{2}$
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$$x^2 + \frac{x^2}{(x+1)^2}=3\implies \frac{x^2}{x^2+2x+1}+x^2-3= 0\\ \implies x^2+(x^4+2x^3+x^2)-(3x^2+6x+3)\\ = x^4 + 2 x^3 - x^2 - 6 x - 3=0\\ \text{ which factors as follows}\\ =(x^2 - x - 1) (x^2 + 3 x + 3) = 0$$
Both of these can be solve by the quadratic equation.
$$(x^2 - x - 1)=0\quad x =\frac{(1 \pm \sqrt{5})}{2}\approx 0.618\lor1.618=\phi$$
The other two roots are complex.
poetasis
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In which exam are you supposed to solve a fourth degree eqn by hand? – Ishraaq Parvez Dec 29 '20 at 05:04
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You can approximate the intervals in which the roots lie, by checking the signs of f(x) for certain integer values. – Ishraaq Parvez Dec 29 '20 at 05:05
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