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[SOLVED] 140.56 finding zero of order 4 polynominial

karush

Well-known member
Jan 31, 2012
2,657
Find the zeros
$$f(x)=16x^4+3x^2-2$$
ok I presume we can solve this with the quadratic formula even with powers of 4 and 2
by setting $u=x^2$ I was able to get


$$x=\pm\sqrt{-\frac{3}{2}+\frac{\sqrt{137}}{2}}$$


but $W\vert A$ says the answer is


$$x=\pm\frac{1}{4}\sqrt{-\frac{3}{2}+\frac{\sqrt{137}}{2}}$$


where does the $\frac{1}{4}$ come from?

$$\tiny{140.56}$$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
You should get (when applying the quadratic formula and discarding the complex roots):

\(\displaystyle x=\pm\sqrt{-\frac{3}{32}+\frac{\sqrt{137}}{32}}\)

The 32 comes from twice the coefficient of the 4th degree term. And this is equivalent to W|A's result.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Start by dividing by 2: 8x^4 + (3/2)x^2 - 1 = 0
 

karush

Well-known member
Jan 31, 2012
2,657
Start by dividing by 2: 8x^4 + (3/2)x^2 - 1 = 0
\begin{align}\displaystyle
y&=8x^4 + (3/2)x^2 - 1 \\
&=\frac{-(3/2)\pm\sqrt{(3/2)^2-4(8)(-1)}}{2(8)}\\
&=\frac{1}{16} \left(-\frac{3}{2} \pm \frac{\sqrt{137}}{2}\right)
\end{align}

maybe:rolleyes:
 

Wilmer

In Memoriam
Mar 19, 2012
376
The square root of that changes the 1/16 to 1/4