Предмет: Алгебра, автор: Knyazevamoya

Помогите решить пожалуйста!
(х+5)^2 = 4
( x-2)^2 = 3
(x+7)^2 = 0
(x-6)^2 = -9

Ответы

Автор ответа: nedenispolybomy

0

(х+5)^2 = 4;
$x^2+10x+25=4; \ x_1+x_2=-10; x_1*x_2=21; x_1=-7; x_2=-3; \ ( x-2)^2 = 3; x^2-6x+4=3; x_{1,2}= frac{4+- sqrt{12} }{2}; x_1=2- sqrt{3}; x_2= sqrt{3}+2$
$(x+7)^2 = 0; x^2+14x+49=0; \ x_1+x_2=-14; x_1*x_2=49; x=-7$
$(x-6)^2 = -9; x^2-12x+45=0; D=-36; D<0$ - корней нет.

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