# Thread: I need help finding the zeros of a function.

1. ## I need help finding the zeros of a function.

The function is:

$\displaystyle 4x^4+17x^2+4 = 0$

I have tried to do the following:
$\displaystyle x^2(4x^2+17) + 4 = 0$
$\displaystyle x^2(4x^2+17) = -4$
$\displaystyle x^2 = -4$
$\displaystyle x = 2i$

$\displaystyle 4x^2+17 = -4$
$\displaystyle 4x^2 = -21$
$\displaystyle x^2 = -21/4$
$\displaystyle x^2 = sqrt(-21)/2$

But I know I do not have the right answer for there is not sqrt(-21)/2 as one of the answers in the back of the book. Any help would be appreciated.

2. let x^2 = w then x^4 = w^2

$\displaystyle 4w^2+17w+4=0$

3. Originally Posted by thyrgle
The function is:

$\displaystyle 4x^4+17x^2+4 = 0$

I have tried to do the following:
$\displaystyle x^2(4x^2+17) + 4 = 0$
$\displaystyle x^2(4x^2+17) = -4$
$\displaystyle x^2 = -4$
$\displaystyle x = 2i$

$\displaystyle 4x^2+17 = -4$
$\displaystyle 4x^2 = -21$
$\displaystyle x^2 = -21/4$
$\displaystyle x^2 = sqrt(-21)/2$

But I know I do not have the right answer for there is not sqrt(-21)/2 as one of the answers in the back of the book. Any help would be appreciated.
just after the line $\displaystyle x=2i$ how did you get $\displaystyle 4x^2+17 = -4$??

4. $\displaystyle \displaystyle 4w^2+17w+4=0\Rightarrow w=-4, \ -\frac{1}{4}\Rightarrow x^2=-4, \ -\frac{1}{4}\Rightarrow x=\pm 2i, \ \pm\frac{i}{2}$

5. I get it know:

$\displaystyle (w+4)(4w+1)$
Then
$\displaystyle (x^2+4) = 0 or (4x^2+1 = 0)$
$\displaystyle -2i, 2i, -1/2i, 1/2i$

Thanks dwsmith!

6. It is true that if ab= 0, then either a= 0 or b= 0 but that is a property of 0 only!
Saying that $\displaystyle x^2(4x^2+17) = -4$ does NOT imply that "$\displaystyle x^2= -4$" or that "$\displaystyle 4x^2+ 17= -4$".

As dwsmith said, you original equation is a quadratic in $\displaystyle w= x^2$. By the way, I would be inclined to write those last two roots as (1/2)i and (-1/2)i or even -i/2, i/2. -1/2i and 1/2i are too likely to be interpreted as -1/(2i) and 1/(2i) (which, it suddenly occurs to me, are exactly (1/2)i and (-1/2)i!)

7. Just following up on what HallsofIvy said, here is a specific counterexample:

$\displaystyle ab=-4$ has $\displaystyle a=2$, $\displaystyle b=-2$ as a solution. Notice that neither $\displaystyle a$ nor $\displaystyle b$ is $\displaystyle -4$.

Of course, there are infinitely many other solutions to this equation.