Is it true that, if \(b>a+c>0\), then the quadratic equation \(ax^2 +bx+c=0\) has two roots?
One of the roots of the equation \(x^2 + ax + b = 0\) is \(1 + \sqrt 3\). Find \(a\) and \(b\) if you know that they are rational.
Author: A. Khrabrov
Do there exist integers \(a\) and \(b\) such that
a) the equation \(x^2 + ax + b = 0\) does not have roots, and the equation \(\lfloor x^2\rfloor + ax + b = 0\) does have roots?
b) the equation \(x^2 + 2ax + b = 0\) does not have roots, and the equation \(\lfloor x^2\rfloor + 2ax + b = 0\) does have roots?
Note that here, square brackets represent integers and curly brackets represent non-integer values or 0.
The equations \[ax^2 + bx + c = 0 \tag{1}\] and \[- ax^2 + bx + c \tag{2}\] are given. Prove that if \(x_1\) and \(x_2\) are, respectively, any roots of the equations (1) and (2), then there is a root \(x_3\) of the equation \(\frac 12 ax^2 + bx + c\) such that either \(x_1 \leq x_3 \leq x_2\) or \(x_1 \geq x_3 \geq x_2\).
Solving the problem: “What is the solution of the expression \(x^{2000} + x^{1999} + x^{1998} + 1000x^{1000} + 1000x^{999} + 1000x^{998} + 2000x^3 + 2000x^2 + 2000x + 3000\) (\(x\) is a real number) if \(x^2 + x + 1 = 0\)?”, Vasya got the answer of 3000. Is Vasya right?