# [SOLVED]140.56 finding zero of order 4 polynominial

#### karush

##### Well-known member
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

Staff member
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
Start by dividing by 2: 8x^4 + (3/2)x^2 - 1 = 0

#### karush

##### Well-known member
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 #### Wilmer

##### In Memoriam
The square root of that changes the 1/16 to 1/4