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9x-2-41x-204-0-solved-it-




Question Number 26642 by Sr@2004 last updated on 27/Dec/17
9x^2 +41x−204=0. solved it.
$$\mathrm{9}{x}^{\mathrm{2}} +\mathrm{41}{x}−\mathrm{204}=\mathrm{0}.\:{solved}\:{it}. \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/17
x=((−41±(√(41^2 −4(9)(−204))))/(2(9)))  x=((−41±95)/(18))=3,−7(5/9)
$$\mathrm{x}=\frac{−\mathrm{41}\pm\sqrt{\mathrm{41}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{9}\right)\left(−\mathrm{204}\right)}}{\mathrm{2}\left(\mathrm{9}\right)} \\ $$$$\mathrm{x}=\frac{−\mathrm{41}\pm\mathrm{95}}{\mathrm{18}}=\mathrm{3},−\mathrm{7}\frac{\mathrm{5}}{\mathrm{9}} \\ $$
Answered by shiv15031973@gmail.com last updated on 28/Dec/17
9×204=1836  9x^2 +68x−27x−204  =x(9x+68)−3(9x+68)  =(9x+68)(x−3)^
$$\mathrm{9}×\mathrm{204}=\mathrm{1836} \\ $$$$\mathrm{9}{x}^{\mathrm{2}} +\mathrm{68}{x}−\mathrm{27}{x}−\mathrm{204} \\ $$$$={x}\left(\mathrm{9}{x}+\mathrm{68}\right)−\mathrm{3}\left(\mathrm{9}{x}+\mathrm{68}\right) \\ $$$$=\left(\mathrm{9}{x}+\mathrm{68}\right)\left({x}−\mathrm{3}\overset{} {\right)} \\ $$
Commented by shiv15031973@gmail.com last updated on 28/Dec/17
what is the amswer?is it right?^
$${what}\:{is}\:{the}\:{amswer}?{is}\:{it}\:{right}\overset{} {?} \\ $$
Commented by Sr@2004 last updated on 28/Dec/17
yes the answear is right
$${yes}\:{the}\:{answear}\:{is}\:{right} \\ $$

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