5x^{2}-3x+12=3(x+4) There's supposed to be two different possible solutions for X, please help me find them.
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Originally Posted by ceedee 5x^{2}-3x+12=3(x+4) There's supposed to be two different possible solutions for X, please help me find them. $ \begin{align*}5x^2-3x+12&=3(x+4) \\&=3x+12\\5x^2-6x&=0\\\text{Now you solve} \end{align*}$
Yes I also get that far, but I don't get further...help please and can you explain your answer then also please?
Originally Posted by ceedee Yes I also get that far, but I don't get further...help please and can you explain your answer then also please? $5x^2 - 6x = 0$ $x(5x-6) = 0$ can you finish it now? If the product of two numbers is zero what can you say about at least one of those numbers?
Thank you for the answer, is it X=2 ?
Last edited by ceedee; Jan 27th 2018 at 10:52 AM.
$5x^2 - 6x = 0$ $\large ax^2+bx+c=0\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=5,~b=-6,~\&~c=0$
Originally Posted by ceedee Thank you for the answer, is it X=2 ? no... $x(5x-6) = 0$ in order for the product to equal zero at least one of the factors must equal zero. So either $x=0$ or $5x-6=0 \rightarrow x = \dfrac 6 5$ so $x = 0$ OR $x=\dfrac 6 5$
Thank you very much
Thank you!!