Question 113952

{{{9 x^ 2 + 45 x = 0}}}........ each term in the expression has a common factor of {{{9x}}}, and we can simply factor out an {{{9x}}}: 

{{{9 x(x + 5) = 0}}}........ ....now we have a product of an monomial and an binomial equal to {{{0}}}

the product is equal to {{{0}}} if one or both factors are equal to {{{0}}}

so

if {{{9x=0}}} then {{{9 x(x + 5) = 0}}}

it will be true only if {{{x=0}}}; notice that {{{9}}} is not equal to {{{0}}}

so  {{{x=0}}} is first solution

then

if {{{(x + 5) = 0}}},{{{9 x(x + 5)}}} will be equal to {{{ 0}}} too

{{{x + 5 = 0}}} only if {{{x = -5}}}

so

{{{x = -5}}}..........another solution


check both solutions:

{{{9 x^ 2 + 45 x = 0}}}........ plug in {{{x=0}}} 

{{{9 *0^ 2 + 45 *0 = 0}}}

{{{0 + 0 = 0}}}

{{{0  = 0}}}


{{{9 x^ 2 + 45 x = 0}}}........ plug in {{{x = -5}}}

{{{9 *(-5)^ 2 + 45 *(-5) = 0}}}

{{{9*25 - 225= 0}}}

{{{225 - 225 = 0}}}

{{{0 = 0}}}