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Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$

I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.

By multiplying both sides by $(x+1)^2$ we get,

$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$ $$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$ But I can't factor the third degree polynomial.

Prime Mover
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Amirali
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  • See https://math.stackexchange.com/questions/2020139/solve-the-equation-x2-frac9x2x32-27 Observe that $$\left(\dfrac{x+1}x\right)-\dfrac1x$$ is a constant – lab bhattacharjee Jul 15 '22 at 14:05
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    By rational root theorem, try values $\pm\dfrac 12, \pm\dfrac 52, \pm\dfrac 54, \pm\dfrac14, \pm 5, \pm 1$. You’ll get $-\dfrac 12$ as a solution. – insipidintegrator Jul 15 '22 at 14:06
  • Exactly, use the hint from @labbhattacharjee, this is a beatiful move. – OnTheWay Jul 15 '22 at 14:37
  • Same logic works for negatives, LHS is obviously decreasing and $x=-\frac12$ is a root (the terms merely switch). – Macavity Jul 15 '22 at 14:38
  • @labbhattacharjee By your hint I got $$a^2+b^2=\frac54 \quad\text{and}\quad\frac1a-\frac1b=1\rightarrow b-a=ab\rightarrow \frac54-2ab=(ab)^2$$$ab=\frac12$ or $ab=-\frac52$. And here only $ab=\frac12$ is acceptable. Hence $a+b=\pm\frac32$ then from here we solve $x+\frac{x}{x+1}=\pm\frac32$ is this the calculation you meant? – Amirali Jul 15 '22 at 14:48
  • Actually it wasn't necessary to find $a+b$, we could use $ab=\frac{x^2}{x+1}=\frac12$....Thanks a lot @labbhattacharjee! It was very interesting! – Amirali Jul 15 '22 at 14:56
  • Use substitution $t=\frac{x^2}{x+1}$, then $t^2=\frac54-2t$. – Ivan Kaznacheyeu Jul 15 '22 at 15:38

4 Answers4

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Simplify as $$4x^2[(x+1)^2+1]=5(x+1)^2 \implies 4x^4+8x^3+3x^2-10x-5=0$$ Denoting LHS as $f(x)$, note that $f(1)=0 ~\& ~f(-1/2)=0.$ This means that by remainder theorem $$f(x) ~\text{is divisible by}~ (2x+1)(x-1)=2x^2-x-1=g(x)$$ dividing $f(x)$ by $g(x)$ we see that $$f(x)=(2x^2-x-1)(2x^2+5x+5)$$ By solving $2x^2+5x+5=0$ we get $$x=\frac{-5\pm \sqrt{-15}}{4}=\frac{-5\pm i\sqrt{15}}{4}.$$

Finally, the equation has 4 roots $$x=-1/2,1,\frac{-5\pm i\sqrt{15}}{4}.$$

Z Ahmed
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$x^2 + \left(\frac{x}{x+1}\right)^2 = \frac{5}{4}$

Let $y=-x\qquad ⇒ y^2 + \left(\frac{y}{1-y}\right)^2 = \frac{5}{4}$

Let $z=\frac{x}{x+1}\qquad ⇒ z^2 + \left(\frac{z}{1-z}\right)^2 = \frac{5}{4}$

By inspection, $(x=1)$ is a solution.

$(x=1) → (y=-1) → (z=-1)$

Solve $z=-1$, back for x, we have $\;\left(x=-\frac{1}{2}\right)$

Original formula can be rewritten as quartic polynomial.
With 2 known roots, it reduced to simple quadratics.

albert chan
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To solve $4x^3+12x^2+15x+5=0$, start by applying the rational root theorem. In this case, potential rational roots are $\pm \lbrace \frac{1}{4}, \frac{1}{2}, 1, \frac{5}{4}, \frac{5}{2}, 5 \rbrace$. Trying them out, we see that $x=\frac{-1}{2}$ satsfies the equation. So $x + \frac{1}{2}$ (or for the sake of avoiding fractions, $2x + 1$) is a factor.

$$4x^3+12x^2+15x+5 = (2x + 1)(ax^2 + bx + c)$$

The quadratic term works out to $2x^2 + 5x + 5$. By the quadratic formula, $x = \frac{-5 \pm i\sqrt{15}}{4}$.

Dan
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Let $t=x+1$

$\displaystyle (t-1)^2 + \left( \frac{t-1}{t} \right)^2 = \frac{5}{4}$

$(t^2 + 2 + \frac{1}{t^2}) - 2\left(t + \frac{1}{t} \right) = \frac{5}{4}$

Let $s = \left(t + \frac{1}{t} \right)$

$s^2 - 2s - \frac{5}{4} = (s+\frac{1}{2})(s-\frac{5}{2}) = 0$

We reduced solving quartic for x to quadratic for s, then quadratic for t

$\displaystyle s=-\frac{1}{2}\qquad ⇒ x = \frac{-5 ± i \sqrt{15}}{4}$

$\displaystyle s=+\frac{5}{2}\qquad ⇒ x = \frac{1 ± 3}{4}$

albert chan
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