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OpenStudy (anonymous):
The goal of this problem is to solve the equation by completing the square.
x^2 - 13 x + 36 = 0
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OpenStudy (helder_edwin):
first add -36 to both sides
OpenStudy (anonymous):
Right, that I understand. I'm just finding it hard to think of a perfect square which would involve -13.
OpenStudy (helder_edwin):
then u got
\[ \large x^2-13x+\qquad=-36 \]
OpenStudy (helder_edwin):
in the blank: put the coefficient of x (-13) halved and then squared and add the same to the right hand side
OpenStudy (anonymous):
x^2-13x+169/4=25/4
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OpenStudy (anonymous):
Right?
OpenStudy (helder_edwin):
u forgot the -36 on the RHS
\[ \large x^2-13x+\color{red}{\frac{169}{4}}=-36+\color{red}{\frac{169}{4}}=
\frac{25}{4} \]
OpenStudy (helder_edwin):
sorry u forgot nothing.
OpenStudy (helder_edwin):
factor the LHS
OpenStudy (anonymous):
Alright. So, (x-13/2) = 25/4?
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OpenStudy (anonymous):
Sorry...the LHS should be squared.
OpenStudy (helder_edwin):
yes. u r right
OpenStudy (anonymous):
Okay, wonderful. So...is the sol'n 13/2 plus or minus sqrt(25/4)?
OpenStudy (helder_edwin):
yes
\[ \large x=13/2\pm\sqrt{25/4} \]
OpenStudy (anonymous):
Okay, wonderful. Thanks so much!!
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OpenStudy (helder_edwin):
u r welcome
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