How to solve this bi-quadratic example?

Question

$$5\left(\frac{x+1}{x}\right)^2 - 16\left( \frac{x-1}{x}\right) - 52 = 0$$

I learned how to use a swap for instance I'd say that

$$\frac{x+1}{x}=t$$ and then swap throughout the equation, but I'm stuck here, I can't get rid of x completely.

Someone edited my question, I don't know how to fix it, I had x + 1/x not (x+1)/x

Answer 1

You can write $(x + \frac{1}{x})^2$ as $(x - \frac{1}{x})^2 + 4$.

Answer 2

$$5(\frac{x+1}{x})^2 - 16( \frac{x-1}{x}) - 52 = 0$$

$$5(1+\frac{1}{x})^2 - 16( 1-\frac{1}{x}) - 52 = 0$$

$$5+\frac{10}{x}+\frac{5}{x^2} - 16+\frac{16}{x} - 52 = 0$$

$$\frac{5}{x^2}+\frac{26}{x} - 63 = 0$$

$$(\frac{5}{x}-9)(\frac{1}{x}+7)=0$$

OR

$$5t^2+26t-63=0$$

where $t=\frac{1}{x}$

Answer 3

Let $\frac 1 x=a$

we get:

$5a^2+26a-63=0$

$a=\frac 95$, so $x=\frac 59$

$a=-9$, $x=-\frac 1 9$