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Mathematics 7 Online
OpenStudy (iwanttogotostanford):

Using the completing-the-square method, rewrite f(x) = x^2 − 8x + 3 in vertex form.

OpenStudy (iwanttogotostanford):

@urbanmorgans @jjwalton04 @Rishi_K

OpenStudy (iwanttogotostanford):

WOULD IT BE: f(x) = (x − 4)2 + 3??

OpenStudy (jjwalton04):

do they have the same product?

OpenStudy (iwanttogotostanford):

these are my choices: f(x) = (x − 8)2 f(x) = (x − 4)2 − 13 f(x) = (x − 4)2 + 3 f(x) = (x − 4)2 + 16

OpenStudy (iwanttogotostanford):

yes they do @jjwalton04

OpenStudy (anonymous):

x23456y−9−12−13−12−9

OpenStudy (iwanttogotostanford):

so, would it be C???

OpenStudy (jjwalton04):

is that the only one with the same product?

OpenStudy (iwanttogotostanford):

not sure...

OpenStudy (iwanttogotostanford):

yes?

OpenStudy (jjwalton04):

then your right'

OpenStudy (iwanttogotostanford):

thanks!

OpenStudy (jjwalton04):

No prob bob

OpenStudy (iwanttogotostanford):

is this answer right> Which of the following values "completes the square," or creates a perfect square trinomial, for x2 − 12x + ___? –6 12 36 –36 i think it is C

OpenStudy (johnweldon1993):

\[\large (x-4)^2 + 3 = x^2 - 8x + 16 + 3 = x^2 - 8x + 19\] Doesn't look the same to me \[\large x^2 - 8x + 3\] Complete the square...subtract 3 from both sides of the equation \[\large x^2 - 8x = -3\] Take half the coefficient of the 'x'...and square it...add that to both sides \[\large x^2 - 8x + 16 = -3 + 16\] Rewrite as sum of squares \[\large (x-4)^2 = 13\] Finally subtract 13 from both sides to get vertex form \[\large (x-4)^2 - 13\]

OpenStudy (iwanttogotostanford):

@johnweldon1993 ok, thank you !!!! i would've gotten that wrong then...

OpenStudy (jjwalton04):

i think the answer for your second question is 36 though

OpenStudy (jjwalton04):

@johnweldon1993 am i right?

OpenStudy (johnweldon1993):

Correct

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