Finding the range of $x$ such that $x^2+\frac{x^{2}}{(x+1)^2}<\frac54$

Question

Find the range of $x$: $$x^2+\frac{x^{2}}{(x+1)^2}<\frac54$$

I am confused about a good approach to this question. My teacher says a graphical approach is the most convenient one, but isn't there any other method to find the range?

Answer 1

$$ x^2 +\frac{x^2}{(x+1)^2} \lt \frac 54 \\ \iff x^2(x+1)^2 +x^2 \lt \frac 54 (x+1)^2 \\ \iff x^4 +2x^3 +2x^2 \lt \frac 54 (x^2 +2x +1) \\ \iff x^4 +2x^3 +\frac 34 x^2 -\frac 52 x -\frac 54 \lt 0 \\ \iff 4x^4 +8x^3 +3x^2 -10x -5 \lt 0$$ $1$ is a root, so factor. $$\iff (x-1)(4x^3 +12x^2+15x+5) \lt 0 $$ $-\frac 12$ is a root of the cubic, so factor. $$\iff (x-1)(x+\frac 12) (4x^2+10x+10)\lt 0$$ The quadratic has a negative discriminant and hence is always positive. $$\iff (x-1)(x+\frac 12) \lt 0 \\ \iff -\frac 12 \lt x \lt 1$$

Answer 2

Hint:

Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$,

$$x^2+\left(\dfrac x{x+1}\right)^2=\left(x-\dfrac x{x+1}\right)^2+2\cdot\dfrac{x^2}{x+1}$$

If $\dfrac{x^2}{x+1}=a,$ we have $$a^2+2a<\dfrac54\iff(a+1)^2<\dfrac94\iff-\dfrac32<a+1<\dfrac32$$

Answer 3

but isn't there any other method to find the range?

First I derived the minimums of $f(x)$. Very easy task finding $x=-2$ and $x=0$

Then by simple inspection (I spent 30'' to do that) immediately find that the range such that $f(x)<1.25$ is given by

$$x \in (-0.5;1)$$

The suggested graphical method is easy too...

the red line is $y=5/4$

the blue line is the graphics of your $f(x)$

to calculate the two bounds I think it is evident that

$$f(1)=1+\frac{1}{4}=\frac{5}{4}$$

$$f(-0.5)=\frac{1}{4}+1=\frac{5}{4}$$