Anyone know this one?
$\displaystyle f(x)=9x^3-48x^2+64x$
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Anyone know this one?
$\displaystyle f(x)=9x^3-48x^2+64x$
helloQuote:
Originally Posted by EmG
factorize with x.
$\displaystyle x(9x^{2}-48x+64)=x(8-3x)^{2}$
I did and when I did it I get stuck at $\displaystyle f(x)=x(9x^2-48x+64)$
$\displaystyle 9x^2-48x+64=(3x)^2-48x+8^2$Quote:
Originally Posted by EmG
is it clear now ?
(Happy)
Remember !
$\displaystyle (a-b)^2=a^2-2ab+b^2$
(Happy)
Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.
okay..Quote:
Originally Posted by EmG
do you agree that $\displaystyle f(x)=x(8-3x)^{2}$
if yes,all you have to do is to put f(x)=0 and find the solutions.
When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?
$\displaystyle f(x)=0\Leftrightarrow x(8-3x^2)=0$Quote:
Originally Posted by EmG
therefore the solutions are $\displaystyle \left [ x=0,x=\frac{8}{3} \right ]$.
hope that helps.
Go back and review the basic theory that is undoubtedly given in your classnotes or textbook. You're expected to know thatQuote:
Originally Posted by EmG
$\displaystyle f(x)=0 \Rightarrow x(8-3x^2)=0 \Rightarrow x = 0 \, \text{or} \, 8 - 3x^2 = 0 \Rightarrow x = ....$
