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Mathematics 20 Online
OpenStudy (anonymous):

x-16x+64=81

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

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?

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?

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

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

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

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

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!

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

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

jimthompson5910 (jim_thompson5910):

what do you mean

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