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jimthompson5910 (jim_thompson5910):
x-16x simplifies to _____
OpenStudy (anonymous):
(x-8)^2?
jimthompson5910 (jim_thompson5910):
oh is there a square in there?
jimthompson5910 (jim_thompson5910):
is the first x supposed to be squared?
OpenStudy (anonymous):
yes it's x^2-16x+64=81
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jimthompson5910 (jim_thompson5910):
oh ok
jimthompson5910 (jim_thompson5910):
so yes, the left side factors to (x-8)^2
jimthompson5910 (jim_thompson5910):
x^2-16x+64=81
turns into
(x-8)^2 = 81
jimthompson5910 (jim_thompson5910):
then you take the square root of both sides (don't forget the plus/minus) to get
x-8 = sqrt(81) or x-8 = -sqrt(81)
what's next?
OpenStudy (anonymous):
x=89,x=-77?
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jimthompson5910 (jim_thompson5910):
sqrt(81) = 9
jimthompson5910 (jim_thompson5910):
so
x-8 = sqrt(81) or x-8 = -sqrt(81)
then turns into
x-8 = 9 or x-8 = -9
OpenStudy (anonymous):
so i have to square root the answer first then solve?
jimthompson5910 (jim_thompson5910):
correct
OpenStudy (anonymous):
so x=17, x=-1?
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OpenStudy (anonymous):
?
jimthompson5910 (jim_thompson5910):
yep, you got it
OpenStudy (anonymous):
k what about this problem.... x^2-18x+81=100...big numbers are intimidating to me
OpenStudy (anonymous):
would it be (x-9)^2
jimthompson5910 (jim_thompson5910):
so you would get (x-9)^2 = 100
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jimthompson5910 (jim_thompson5910):
take the square root of both sides
OpenStudy (anonymous):
so does it look like x-9=sqrt100
>>x-9=10>> x-9=-10
x=-19,
x=-1?
OpenStudy (anonymous):
\[x-9 = \pm 10\]
jimthompson5910 (jim_thompson5910):
x-9 = 10
x = 10+9
x = 19 is your first solution
jimthompson5910 (jim_thompson5910):
x = -1 is the second, so you got that part right
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OpenStudy (anonymous):
ok thanks ...x^2+4x+_?=(x+_?)^2
jimthompson5910 (jim_thompson5910):
take half of 4 and square it to fill in the first blank
jimthompson5910 (jim_thompson5910):
the second blank is just half of the x coefficient
OpenStudy (anonymous):
show me?
jimthompson5910 (jim_thompson5910):
what's half of 4
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OpenStudy (anonymous):
2
jimthompson5910 (jim_thompson5910):
square that to get ____
OpenStudy (anonymous):
4
OpenStudy (anonymous):
now what
jimthompson5910 (jim_thompson5910):
so that goes in the first blank
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jimthompson5910 (jim_thompson5910):
the second blank is 2 because that's half of 4 (the x coefficient)
OpenStudy (anonymous):
k got it and 4 in the next ?
jimthompson5910 (jim_thompson5910):
2 in the next
jimthompson5910 (jim_thompson5910):
because (x+2)^2 = x^2 + 4x + 4
OpenStudy (anonymous):
oh ok gotcha!
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OpenStudy (anonymous):
what if the equation becomes negative such as x^2-4x+_?=(x-_?)^2 is this like the difference of squares
jimthompson5910 (jim_thompson5910):
half of -4 is ____
OpenStudy (anonymous):
-2
jimthompson5910 (jim_thompson5910):
square that to get ____
OpenStudy (anonymous):
4
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jimthompson5910 (jim_thompson5910):
so that goes in the first blank
jimthompson5910 (jim_thompson5910):
now you just factor x^2 - 4x + 4 to find the value that goes in the second blank
OpenStudy (anonymous):
ok great complete the square problem:
x^2+4x+3=0
jimthompson5910 (jim_thompson5910):
last one...
take half of 4 to get 2
then square it to get 4
add this to both sides to get
x^2 + 4x + 3 + 4 = 4
x^2 + 4x + 4 + 3 = 4
(x^2 + 4x + 4) + 3 = 4
(x+2)^2 + 3 = 4
(x+2)^2 = 4-3
(x+2)^2 = 1
I'll let you finish up
OpenStudy (anonymous):
where did the 4's go though is it subtracted?
on the other side
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