SOLUTION: point of intersection 5x+4y-32=0 and x^2+y^2=25 How>?

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Question 660257: point of intersection 5x+4y-32=0 and x^2+y^2=25
How>?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
point of intersection
x^2+y^2=25
y^2=25-x^2
y=√(25-x^2)
..
5x+4y-32=0
4√(25-x^2)=32-5x
square both sides
16(25-x^2)=1024-320x+25x^2
400-16x^2=1024-320x+25x^2
41x^2-320x+624=0
solve by following quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=41, b=-320, c=624
ans:
x=4
y=√(25-x^2)=√(25-16)=√9=3
or
x≈3.805
y=√(25-x^2)=√(25-14.478)=√10.522≈3.244
2 points of intersection: (4,3) and (3.805,3.244)