One factor of f(x) = x3 − 14x2 + 61x − 84 is (x − 7). What are the zeros of the function?
1 answer:
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Let the required point be (a,b)
The distance of (a,b) from (7,-2) is
=
But this distance needs to be betweem 50 & 60
So
Squaring all sides
2500 < (a-7)² + (b+2)² < 3600
Let a = 7
So we have
2500 < (b+2)² <3600
b+2 < 60 or b+2 > -60 => b <58 or b > -62
Also
b+2 >50 or b + 2 < -50 => b >48 or B < -52
Let us take one value of b < 58 say b = 50
So now we have the point as (7, 50)
The other point is (7,-2)
Distance between them
=
This is between 50 & 60
Hence one point which has a distance between 50 & 60 from the point (7,-2) is (7, 50)