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?

You did not substitute $y = 8x^2$ correctly. See if you can figure out the error. — Jim Belk, Jan 24 '14 at 21:42
Try writing $8x^4-8x^2+1 = 2(4x^4-4x^2+1)-1 = 2(y-1)^2-1$ where $y = 2x^2$ to see if matters improve any. — Dilip Sarwate, Jan 24 '14 at 22:03

score 6 · Answer 1 · answered Jan 24 '14 at 21:50

6

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 $8$.

answered Jan 24 '14 at 21:50

amWhy

209,954

score 1 · Accepted Answer · answered Jan 24 '14 at 22:16

1

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

answered Jan 24 '14 at 22:16

Adi Dani

16,949

Reducing quartic equations to quadratic

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