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

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hartnnBest 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

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

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

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

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

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

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

hartnnBest 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

hartnnBest 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

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

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

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

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

hartnnBest 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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