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?
Best Answer
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 $8$.