1. ## non-whole roots

How could I know that 4x^4 + 3^2 - 3 has two roots?

2. ## Re: non-whole roots

I've noticed it would have a parabola shape and subtracting 3 creates a y-axis crossing point at y=-3.

3. ## Re: non-whole roots

Originally Posted by Stuck Man
How could I know that 4x^4 + 3^2 - 3 has two roots?
You don't. It's not true. Or are you asking about real roots?

Is that really the problem? $4x^4+ 3^2- 3= 4x^4+ 9- 3= 4x^4+ 6= 0$ so $x^4= -6/4$ so that there are four roots, the four fourth roots of -3/2, none of which are real.

If the equation is $4x^4+ x^2- 3= 0$, we can let $y= x^2$ so we are dealing with the quadratic equation $4y^2+ y- 3= 0$. By the quadratic formula,
$y= \frac{-1\pm\sqrt{1+ 48}}{8}= \frac{-1\pm 7}{8}$
which has the two roots y= 6/8= 3/4 and y= -8/8= -1.

Now, If $y= x^2= 3/4$, then $x= \pm\sqrt{3}/2$, two real roots. If $y= x^2= -1$, then $x= \pm i$, two imaginary roots.

The crucial point was that, because this had only even factors, we could reduce to a second order equation. Positive roots of that equation give real roots of the original equation.

4. ## Re: non-whole roots

I typed it wrong, its 4x^4+3x^2-3.

5. ## Re: non-whole roots

Use the substitution, let $x^2=t$ so you get a quadratic equation in $t$ which you can solve, afterwards do the back-substitution. Let see how many roots you get.

Thanks.

7. ## Re: non-whole roots

You're welcome! But, what's your result? ...

8. ## Re: non-whole roots

There is 2 real roots. I have exact values but there is quite a lot to type. They are about +/- 0.75.