SOLUTION: if the roots of x^2-ax+b=0 are real & differ by a quantity which is less than c(c>0),prove that b lies between (1/4)(a^2-c^2)& (1/4)a^2

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Question 629643: if the roots of x^2-ax+b=0 are real & differ by a quantity which is less than c(c>0),prove that b lies between (1/4)(a^2-c^2)& (1/4)a^2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Using the Quadratic Formula on your equation we get:
x+=+%28-%28-a%29+%2B-+sqrt%28%28-a%29%5E2-4%281%29%28b%29%29%29%2F2%281%29
which simplifies as follows:
x+=+%28-%28-a%29+%2B-+sqrt%28a%5E2-4%281%29%28b%29%29%29%2F2%281%29
x+=+%28-%28-a%29+%2B-+sqrt%28a%5E2-4b%29%29%2F2%281%29
x+=+%28a+%2B-+sqrt%28a%5E2-4b%29%29%2F2
which is short for:
x+=+%28a+%2B+sqrt%28a%5E2-4b%29%29%2F2 or x+=+%28a+-+sqrt%28a%5E2-4b%29%29%2F2

The difference of these is:
%28a+%2B+sqrt%28a%5E2-4b%29%29%2F2+-+%28a+-+sqrt%28a%5E2-4b%29%29%2F2
%282sqrt%28a%5E2-4b%29%29%2F2
sqrt%28a%5E2-4b%29

We are told that this difference is less than c, and as is true of any square root, it is not negative. So
sqrt%28a%5E2-4b%29+%3C+c and sqrt%28a%5E2-4b%29+%3E=+0
Squaring both sides of both we get:
a%5E2-4b+%3C+c%5E2 and a%5E2-4b+%3E=+0
Now we solve both for b:
a%5E2-c%5E2+%3C+4b and a%5E2+%3E=+4b
%281%2F4%29%28a%5E2-c%5E2%29+%3C+b and %281%2F4%29a%5E2+%3E=+b

P.S. To me, "between p and q" means "between p and q but not equal to p or q". But I don't think there is any way to eliminate the "or equal to" part of %281%2F4%29a%5E2+%3E=+b. So the largest possible b would be equal to %281%2F4%29a%5E2 not between that and %281%2F4%29%28a%5E2-c%5E2%29.