Question

If a is real, what is the only real number that could be a multiple root of x^3 + ax + 1 = 0?

Answer 1

\[x^3+ax+1=0\]\[x^3=-ax-1\quad;\quad f(x)=x^3\quad,\quad g(x)=-ax-1\]\[f'(x)=3x^2\quad\Rightarrow\quad y-y_0=f'(x_0)(x-x_0)\]\[y-x_0^3=3x_0^2(x-x_0)\quad\Rightarrow\quad y=3x_0^2x-2x_0^3\]\[-2x_0^3=-1\quad\Rightarrow\quad x_0=\left(\frac{1}{2}\right)^{1/3}\]\[a=-3x_0^2=-3\left(\frac{1}{2}\right)^{2/3}=-\frac{3}{\sqrt[3]{4}}\]

Answer 2

I don't know what this concatenation of symbols and operations has as its answer.

Please humor me and provide the answer in form that someone like me might recognize as a cube root of something or other. Thanks.

I am not one of the Luddites but I do lean that way. :) Thanks.

http://www.merriam-webster.com/dictionary/luddite

Answer 3

I'm wondering about the -3. See, I got the cube root of a single fraction. I worked with it and cannot get it to equal your answer. The -3 portion of your answer doesn't fit my work but I may be wrong.

Answer 4

looks like Directrix might not be studying calculus

Answer 5

I wonder what level of math Directrix is studying?

Answer 6

Vieta's formulas:\[x_1=x_2=u\quad,\quad x_3=v\]\[x_1+x_2+x_3=0=2u+v\]\[x_1x_2+x_2x_3+x_3x_1=a=u^2+2uv\]\[x_1x_2x_3=-1=u^2v\]\[u^2(-2u)=-1\quad\Rightarrow\quad u=\left(\frac{1}{2}\right)^{1/3}\quad,\quad v=-2\left(\frac{1}{2}\right)^{1/3}\]\[a=u^2+2uv=\left(\frac{1}{2}\right)^{2/3}-4\left(\frac{1}{2}\right)^{2/3}=-3\left(\frac{1}{2}\right)^{2/3}=-\frac{3}{\sqrt[3]{4}}\]

Answer 7

Great Job!
I believe the student can follow that one.
I like both methods.