equation reducible to quadratic (1 Viewer)

Green Yoda · Mar 23, 2016

(x^2-1/x) - (18x/x^2-1)=3

Green Yoda · Mar 23, 2016

another one: x^2-8x-2=12/x^2-8x-1

integral95 · Mar 23, 2016

Rathin said:
another one: x^2-8x-2=12/x^2-8x-1

Assuming it's

$x^2-8x-2 = \frac{12}{x^2} -8x-1$

You eliminate the -8x from both sides and multiply both sides by x^2 to get

$x^4-x^2-12 = 0 \\ \\ \Rightarrow (x^2-4)(x^2+3) = 0$

Well i think you can solve it from here.

kawaiipotato · Mar 24, 2016

edit

Rathin said:
(x^2-1/x) - (18x/x^2-1)=3

$\frac{x^2 - 1}{x} - \frac{18x}{x^2 - 1} = 3$

$x \neq 0, \pm 1$

$multiplying by$ x(x^2 - 1) $throughout$

$\Rightarrow (x^2 - 1)^2 - 18x^2 = 3x(x^2 - 1)$

$(x^2 - 1)^2 - 18x^2 - 3x(x^2 - 1) = 0$

$x^4 - 2x^2 + 1 - 18x^2 - 3x^3 + 3x = 0$

$x^4 - 3x^3 - 20x^2 + 3x + 1$

$$Notice the symmetry, so divide by$ x^2 ($using the fact$ x \neq 0)$

$x^2 - 3x - 20 + \frac{3}{x} + \frac{1}{x^2}$

$\left(x^2 + \frac{1}{x^2}\right) - 3\left(x - \frac{1}{x}\right) - 20 = 0$

$$let$ y = x-\frac{1}{x}$

$y^2 = \left(x^2 + \frac{1}{x^2}\right) - 2$

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

$Reduces down to$

$y^2 + 2 - 3y - 20 = 0$

$y^2 - 3y - 18 = 0$

equation reducible to quadratic (1 Viewer)

Green Yoda

Hi Φ

Green Yoda

Hi Φ

integral95

Well-Known Member

kawaiipotato

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