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Mathematics 9 Online
OpenStudy (anonymous):

Please help: x^2+13x

OpenStudy (anonymous):

Are we looking for solutions?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Start by removing the x from both terms:\[x(x+13)=0\] Do you know how to proceed?

OpenStudy (anonymous):

subtract 13 from both sides?

OpenStudy (anonymous):

first you can't find the solution without a left side, did it ask for the roots/zeroes of the function?

OpenStudy (anonymous):

right* side

OpenStudy (anonymous):

if it's asking for roots or zeroes first start with \[x^2+13x=0\] because when it asks for zeroes it's asking for x intercepts we know that y = 0, in otherwords y=f(x)=x^2+13x in this case where y = 0

OpenStudy (anonymous):

it's asking me to write the resulting perfect square in trinomial form

OpenStudy (anonymous):

I came up with x^2+13x + 169/4= (x + 13/2)^2

OpenStudy (anonymous):

first take half of 13

OpenStudy (anonymous):

6.5

OpenStudy (anonymous):

then square it it

OpenStudy (anonymous):

42.25

OpenStudy (anonymous):

now since you added 42.25 to one side you have to add 42.25 on the other side

OpenStudy (anonymous):

\[x^2+13x+42.25=42.25\]

OpenStudy (anonymous):

your perfect squre is the left side

OpenStudy (anonymous):

x^2 +13x is the final answer?

OpenStudy (anonymous):

no your trinomial is \[Ax^2+Bx+C=x^2+13x+42.25\]

OpenStudy (anonymous):

I see it now

OpenStudy (anonymous):

this question is really weird because trinomial form is Ax^2+Bx+C however when you do complete the square you get Ax^2+Bx+C-(B/2)^2

OpenStudy (anonymous):

im lost

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