i think the answer is 121? not sure how to answer this will post screenshot
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OpenStudy (anonymous):
zepdrix (zepdrix):
121? hehe that seems way too large :D
zepdrix (zepdrix):
So like..
zepdrix (zepdrix):
\[\large\rm y=x^2+bx\]To complete the square, Toby adds 121 to each side,\[\large\rm y+121=x^2+bx+121\]
zepdrix (zepdrix):
121 is a perfect square, do you know which one? :)
Hint 12*12 = 144
9*9 = 81
it's somewhere between there, ya?
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OpenStudy (anonymous):
9?
OpenStudy (anonymous):
so thats the coefficent?
OpenStudy (anonymous):
oh you mean for 121!!
zepdrix (zepdrix):
9*9=81, not 121 0_o
OpenStudy (anonymous):
wait so 9 or11?
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zepdrix (zepdrix):
We're not to the answer yet...
I'm just asking if you can tell me the square root of 121.
OpenStudy (anonymous):
oh okay that is 11
zepdrix (zepdrix):
Ok good :)
Let's rewrite the 121 as 11*11 then.\[\large\rm y+121=x^2+bx+11\cdot11\]
When completing the square, we have this squared number on the end,
and we have TWO of these x's in the middle.
So the completed square should look like this:\[\large\rm x^2+11x+11x+11\cdot11\]
zepdrix (zepdrix):
So what is the coefficient on your x term? :)
Combine like-terms first!!
zepdrix (zepdrix):
Or if this easier for you,
square root the 121,
then double that number.
That's how you find the middle coefficient ^^
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