SOLUTION: college algebra - completing the square would you help me finish this? I wasn't sure what factors to pick out from 196 so I choose 4 times 49. How do I choose what to use? x^

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Question 405860: college algebra - completing the square
would you help me finish this? I wasn't sure what factors to pick out from 196 so I choose 4 times 49. How do I choose what to use?
x^2-10x+24=0
x=-10+-sqrt((-10)^2-4(1)(24))/2(1))
x=10 +- sqrt((100-(-96))/2)
x=10 +-sqrt((196)/(2)
2sqrt(4(49))/2))
sqrt(2^2(49))/(2)
2+-sqrt49/2
2(1+-sqrt49/2)
1sqrt49
x=1sqrt49 right?
Found 2 solutions by josmiceli, emargo19:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-10x%2B24=0
Here's how I do it:
x%5E2+-+10x+=+-24
Take 1/2 of the coefficient of x, square it, and add it
to both sides.
x%5E2+-+10x+%2B+%28-10%2F2%29%5E2+=+%28-10%2F2%29%5E2+-+24
x%5E2+-+10x+%2B+25+=+25+-+24
%28x+-+5%29%5E2+=+1
Take the square root of both sides
x+-+5+=+1
x+=+6
and, also
x+-+5+=+-1 (the other square root of 1)
x+=+4
from these, I get x-6+=+0 and x-4+=+0, so
The factors are
%28x+-+6%29%2A%28x+-+4%29

Answer by emargo19(101) About Me  (Show Source):