Reducing quartic equations to quadratic

Question

I'm trying to re-learn basic math/algebra, and I can't get passed one question concerned with reducing quartic equation to a quadratic:

Find, correct to 3 significant figures, all the roots of the equation $8x^4 - 8x^2 + 1 = \frac12\sqrt 3$

I've tried substitution:

Let $y = 8x^2$, making:

$$ y^2 -y + 1 = \frac12\sqrt 3 \\ (y-1)^2 = \frac12\sqrt 3 \\ y-1 = {\sqrt{\frac12\sqrt 3}} \\ y-1 = ±0.931 \\ y = 1.931\text{ or } 0.069 $$

Therefore:

$$ x = √({1.931 \over 8}) \ or \ √({0.069 \over 8}) \\ x = 0.491 \ or \ 0.093 $$

But, as you can see on Wolfram Alpha, the roots are ±0.991 and ±0131.

What am I doing wrong?

Answer 1

$$8x^4 - 8x^2 + 1 = \frac{1}{2}\sqrt 3$$ $$16x^4-16x^2+2-\sqrt3=0$$ $$(4x^2)^2-4\cdot4x^2+2-\sqrt3=0$$ $4x^2=y,x=\pm\frac{\sqrt y}{2}$ $$y^2-4y+2-\sqrt3=0$$ $$(y-2)^2=2+\sqrt3$$ $$y=2\pm\sqrt{2+\sqrt3}$$ $$x=\pm\frac{\sqrt{2\pm\sqrt{2+\sqrt3}}}{2}$$

Answer 2

Note that $$y = 8x^2\;\implies\;y^2 = 64 x^4$$

In addition, putting aside that error, you incorrectly factored $$y^2 - y +1 \neq (y - 1)^2 = y^2 - 2x + 1$$

Try substituting $y = x^2$ and then divide each side of the resulting equation by $8$.