Can a fourth-order equation be solved like a quadratic equation?

Question

I was asked to find the zeros of $y = x^4 + 5x^2 +6$.

I tried to turn this into a quadratic to factor it as follows:

$y = x^4 + 5x^2 +6 = {(x^2)}^2 + 5{(x^2)}^1 + 6$

Put another way:

Let $t = {x^2}$

$y = t^2 + 5t +6$

$y = (t + 6)(t - 1)$

$t =$ {$-6, 1$}

I threw out the -6 solution, but thought that if $t = x^2 = 1$, then $x = 1$.

I can see that when I plug 1 back into the original formula, $1^4 + 5(1^2) + 6 \ne 0$, but I don't know where I went wrong conceptually.

How embarrassing. So does that mean $\sqrt{-2}$ and $\sqrt{-3}$ are solutions? And the reason it says "no zeros" is because solutions with imaginary numbers are not zeros? I'm still unclear if this is an OK approach and I just can't add/multiply. — dumb question, Oct 03 '13 at 22:40
The roots are $i \sqrt 2, -i \sqrt 2, i \sqrt 3, -i \sqrt 3.$ — Will Jagy, Oct 03 '13 at 22:41

score 2 · Answer 1 · answered Oct 03 '13 at 22:34

2

You factored incorrectly. It factors into $y = (t+2)(t+3)$.

answered Oct 03 '13 at 22:34

Joe

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score 2 · Accepted Answer · answered Oct 03 '13 at 22:35

2

Yes, you can do what you are trying. The only problem is you did not factor it correctly, as the constant term would be $-6$ instead of $+6$

answered Oct 03 '13 at 22:35

Ross Millikan

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score 1 · Answer 3 · answered Oct 03 '13 at 22:59

Yes that is a valid method. In reply to your comment about the answer being "no zeros," that comes from the fact that there are no real zeros, which is what that means.

If you were looking for only real roots, you could have also deduced that from looking at the original equation. $x^4$ and $x^2$ are always positive, thus for real $x, x^4 + 5x^2 + 6 > 0$

Can a fourth-order equation be solved like a quadratic equation?

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