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mskyeg
Group Title
What are the real or imaginary solutions of the polynomial equation?
x3 = 216
 one year ago
 one year ago
mskyeg Group Title
What are the real or imaginary solutions of the polynomial equation? x3 = 216
 one year ago
 one year ago

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mskyeg Group TitleBest ResponseYou've already chosen the best response.0
Possible answers
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(x^3216=0 \\ x^36^3=0 \\ \text{now apply }a^3b^3 formula\)
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
how do i do that
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
like what does it mean
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
know the formula ?
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
like what do i plug intohe formula
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
how does that help all i get is x^3  216
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
thats because u don't know that formula
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
oooh i thought that WAS the formula
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\large a^3b^3=(ab)(a^2+ab+b^2)\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
now put a=x,b=6
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
and then equate it to 0
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\((x6)(x^2+6x+36)=0 \\ (x6)=0,(x^2+6x+36)=0 \\ x=6,\text{use quadratic formula for x^2+6x+36=0}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\)
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
I tried but i got a really weird answer
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
what u got, a,b,c as ?
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
a=6 b=3+i square root 54 c=3i square root 54
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
oh a=1 b=6 and c=36
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
so two other values of x are ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
one obviously is x=6
 one year ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
what about the other two
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
thats what u have u work out using \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\) u have a,b,c values correct.
 one year ago
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