Question

решите целое уравнение
18x^4-20x^2+2=0

Answer 1

$18x^{4}-20x^{2} +2=0$

Пусть $x^{2} =t$

Тогда:

$18t^{2}-20t+2=0$

$D=b^{2}-4ac=20^{2}-4*18*2=400-4*36=400-144=256\\\\t1=\frac{-b+\sqrt{D} }{2a}=\frac{20+16}{2*18}=\frac{36}{36}=1\\t2=\frac{-b-\sqrt{D} }{2a}=\frac{20-16}{2*18}=\frac{4}{36}=\frac{1}{9}$

$x^{2} =1\\x=1; \\x =-1$

$x^{2} =\frac{1}{9}\\x=\frac{1}{3};\\x=-\frac{1}{3}$

Ответ:

$x=1, \\x=-1,\\x=\frac{1}{3}\\x=-\frac{1}{3}$.