Answer by MT_ for Can a fourth-order equation be solved like a quadratic...
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...
View ArticleAnswer by Ross Millikan for Can a fourth-order equation be solved like a...
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 $+6$
View ArticleAnswer by Joe for Can a fourth-order equation be solved like a quadratic...
You factored incorrectly. It factors into $y = (t+2)(t+3)$.
View ArticleCan a fourth-order equation be solved like a quadratic equation?
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 =...
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