All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Reasonable Root ? (Posted on 2013-02-17)
p is a nonzero real root of the equation: ax2+ bx + c = 0 and, q is a nonzero real root of the equation: -ax2+ bx + c = 0.

Does the equation ax2/2 + bx + c = 0 always have a root between p and q?

If so, prove it. Otherwise, give a counter example.

Note: Each of a, b and c is a nonzero real number.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 2 of 2 |
A variant of the quadratic formula is x = (2c)/(-b +/- sqrt[b^2-4ac])

From this it is easy to see that for a variable a that one root is always greater than -c/b and the other is always less than -c/b.

Also note that from this form of the quadratic equation, lim {a -> 0} (2c)/(-b +/- sqrt[b^2-4ac]) is equal to -c/b or diverges to infinity, depending on which root is examined.

Let p1 and p2 be the roots of ax^2+ bx + c = 0.
Let q1 and q2 be the roots of -ax^2+ bx + c = 0.
Let r1 and r2 be the roots of (a/2)x^2+ bx + c = 0.

Without loss of generality, p2 < -c/b < p1q2 < -c/b < q1; and r2 < -c/b < r1.

From the limit earlier, exactly one of r1 and r2 approaches -c/b.  If that root is r1 then q2 < -c/b < r1 < p1.  Also, if that root is r2 then p2 < r2 < -c/b < q1.

Therefore (a/2)x^2+ bx + c = 0 has root between some p and q.  More generally this is true for any k with abs(a)>abs(k)>0.

Edited on July 14, 2016, 1:56 pm
 Posted by Brian Smith on 2016-07-14 13:45:22

 Search: Search body:
Forums (0)