if a is real, what is the only real number that could be a mutiple root of $x^3 +ax+1$=0
No one in my class know how to do it, so i have to ask it here.
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$\begingroup$ Can we use calculus? $\endgroup$ – azarel Feb 19 '12 at 3:15
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Let the multiple root be $r$, and let the other root be $s$. If $r$ is to be real, then $s$ must be real also. From Vieta's formulas, we have $2r + s = 0$ and $r^2s = -1$. The first equation gives $s = -2r$, which we plug into the second equation to get $r^2s = -2r^3 = -1$, so $r = \boxed{\left(\frac12\right)^{1/3}}$.
answered Feb 19 '12 at 3:16
