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Mathematics 19 Online
OpenStudy (anonymous):

x^2+23x+k what makes this a perfect square ?

OpenStudy (anonymous):

you are trying to find the value of k such that this would be a perfect square?

OpenStudy (anonymous):

469/4

OpenStudy (anonymous):

- Yes .

OpenStudy (anonymous):

sorry 529/4

OpenStudy (anonymous):

OK, that would mean you want this equation to be in the form \[(x+b)^2 = x^2 + 23x + k\]

OpenStudy (anonymous):

\[x^2 + 2bx + b^2\] so actually 23 is 2*k

OpenStudy (anonymous):

so \[k = 23/2, k^2 = (23/2)^2\] as spetzz stated aboved

OpenStudy (anonymous):

therefore the equation is \[x^2 + 23x + (23/2)^2\] = \[(x+23/2)^2\]

OpenStudy (anonymous):

sorry should be \[b\] in the post above the final equation post

OpenStudy (anonymous):

not k

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