Question

|x^2 - 20| = -8x решить уравнение

Answer 1

Ответ:

$|x^2-20|=-8x\\\\a)\ \ x^2-20\geq 0\ \ \to \ \ \ (x-\sqrt{20})(x+\sqrt{20})\geq 0\ \ ,\\\\x\in (-\infty ;-2\sqrt5\, ]\cup [\, 2\sqrt5;+\infty )\\\\x^2-20=-8x\ \ ,\ \ \ x^2+8x-20=0\ \ ,\ \ D/4=36\ ,\\\\x_1=-10\ \ ,\ \ \ x_2=2\notin (-\infty ;-2\sqrt5\, ]\cup [\, 2\sqrt5;+\infty )\\\\b)\ \ x^2-20<0\ \ \to \ \ x\in (-2\sqrt5\, ;\, 2\sqrt5\, )\\\\20-x^2=-8x\ \ ,\ \ x^2-8x-20=0\ \ ,\ \ D/4=36\ ,\\\\x_1=-2\notin (-\infty ;-2\sqrt5\, ]\cup [\, 2\sqrt5;+\infty )\ \ ,\ \ \ x_2=10\\\\Otvet:\ \ x=-10\ \ ,\ \ x=10\ .$