Question

Решите уравнение: (x+3)(х-5)(х+2)(х-4)=60

Answer 1

$\underbrace {(x+3)(x-5)}\underbrace {(x+2)(x-4)}=60\\\\(x^2-2x-15)(x^2-2x-8)=60\\\\t=x^2-2x-15\; \; \Rightarrow \; \; x^2-2x-8=t+7\\\\t(t+7)=60\\\\t^2+7t-60=0\\\\t_1=-12\; ,\; \; t_2=5\\\\a)\; x^2-2x-15=-12\; ,\; \; x^2-2x-3=0\\\\x_1=-1,\; x_2=3\\\\b)\; \; x^2-2x-15=5\; ,\; \; x^2-2x-20=0\\\\\frac{D}{4}=1+20=21\; ,\; \; \; x_{3,4}=1\pm \sqrt{21}\\\\Otvet:\; -1;\; \; 3;\; \; 1-\sqrt{21};\; \; 1+\sqrt{21}\; .$