Question

Разность корней квадратного уравнения х^2-4х+q=0 равна 20.найдите q.

Answer 1

X^2-4x+q=0
x1-x2=20 условие
x1+x2=4 терема виета
x1x2=q

2x1=24
x1=12
x2=-8
x1*x2=-96
q=-96

Answer 2

$x^2 - 4x + q = 0\\\\ D = 16 - 4q = 4(4 - q)\\\\ x_1 = \frac{4 + 2\sqrt{4 - q}}{2}\\\\ x_2 = \frac{4 - 2\sqrt{4 - q}}{2}\\\\ |\frac{4 + 2\sqrt{4 - q}}{2} - \frac{4 - 2\sqrt{4 - q}}{2}| = 20\\\\ |2 + \sqrt{4 - q} - 2 + \sqrt{4 - q}| = 20\\\\ |2\sqrt{4 - q}| = 20\\\\ \sqrt{4 - q} = 10\\\\ 4 - q = 100\\\\ \boxed{q = -96}$