Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Quadratic Function proof?
Given a quadratic of the form f(x) = x^2 + bx + c, prove that the equation f(x) - f(a) = 0 will always have 2 integer solutions for all a,b,c ϵ ℝ, or provide a counter example to disprove the statement.
3 Answers
- PhilipLv 65 years agoFavorite Answer
Consider eqn f(x)-f(a)=0. Now f(x)-f(a)=x^2+bx+c - (a^2+ab+c)
= (x^2-a^2) + (bx-ba) + (c-c) = (x-a)(x+a+b) & eqn f(x) - f(a) = 0
has solutions x=a & x=-(a+b). Suppose (a,b,c)=([1/2],[1/3],[1/4]).
Then solutions to f(x) - f(a) = 0 are x = [1/2] & x = [-5/6]. These
solutions are non-integer even though a,b,c are all real. This
counter-example disproves the statement.
- PopeLv 75 years ago
Let a = b = c = 0.
f(x) = x² + (0)x + (0) = x²
Let f(x) - f(a) = 0.
x² - 0² = 0
x² = 0
This equation has only one integer solution, 0.
- Elizabeth MLv 75 years ago
f(x)-f(a)=(x^2+bx+c)-(a^2+ab+c)
=x^2-a^2+b(x-a)
=(x-a)(x+a+b)
= 0 when x=a or x=-a-b and if a and b are real
and both non-zero integers then there are
two integer solutions.
