How to Solve $\frac{(x^2+1)}{x} + \frac{x}{(x^2+1)} = 2.9$?

Question

My cousin has this math homework problem she is stuck on and I have also failed at it. Maybe you guys can help?

$$\frac{x^{2}+1}{x} + \frac{x}{x^{2}+1} = 2.9$$

The answer is $0.5$ or $2$, but we're interested in how to get there.

Answer 1

Take a look at the following steps

$$\eqalign{ & {{{x^2} + 1} \over x} + {x \over {{x^2} + 1}} = 2.9 \cr & y = {{{x^2} + 1} \over x} \cr & y + {1 \over y} = 2.9 \cr & {y^2} - 2.9y + 1 = 0 \cr & y = {5 \over 2},{2 \over 5} \cr & {x^2} - yx + 1 = 0 \cr & x = {{y \pm \sqrt {{y^2} - 4} } \over 2},\,\,\,{y^2} - 4 \ge 0 \cr & y \ne {2 \over 5} \cr & y = {5 \over 2}\,\,\,\, \to \,\,\,x = 2,0.5 \cr} $$

Answer 2

We could introduce a new variable into the mix. Let $\frac{x^2+1}{x} =: z$, provided that $x\neq 0$ and $x^2+1\neq 0$ we have a new equation of the form $$z+\frac{1}{z} = 2.9 $$

EDIT: $x^2+1$ can't be zero, anyway, so that bit is redundant. Assuming we operate in $\mathbb{R}$

Answer 3

Let's consider the following problem ($b\in\mathbb{R},y\neq 0$):

$$y+\frac{1}{y}=b$$

It leads to $y^{2}-yb+1=0$, which is easy to solve: $$y=\frac{b\pm\sqrt{b^{2}-4}}{2}$$ You can see there is no real solution if $|b|<2$. Assume $|b|\geq 2$.

Suppose now that you have $y=\frac{x^{2}+1}{x}$, $b=2.9$. You have to solve

$$\frac{x^{2}+1}{x}=\frac{2.9\pm\sqrt{2.9^{2}-4}}{2}$$

Answer 4

HINT: your equation is equivalent to $$\frac{(x-2) (2 x-1) \left(5 x^2-2 x+5\right)}{10 x \left(x^2+1\right)}=0$$