Question

4x^4-12x^2+1=0 решите биквадратные уравнение подробно

Answer 1

4x⁴ - 12x² + 1 = 0

Пусть t = x²
4t² - 12t + 1 = 0
$D = 144 - 4*4 = 128 \\ \sqrt{D} = \sqrt{128} = \sqrt{16*4*2} = 8 \sqrt{2} \\ t_1 = \frac{12-8 \sqrt{2} }{8} = \frac{6-4 \sqrt{2} }{4} \\ t_2 =\frac{12+8 \sqrt{2} }{8} = \frac{6+4 \sqrt{2} }{4}$

Проведём обратную замену

$x_1^2=\frac{6-4 \sqrt{2} }{4} \\ x_2^2=\frac{6+4 \sqrt{2} }{4} \\ \\ x_1= \pm \sqrt{\frac{6-4 \sqrt{2} }{4} }\\ x_2= \pm \sqrt{\frac{6+4 \sqrt{2} }{4}} \\ \\ x_1= \pm \sqrt{\frac{4-4 \sqrt{2}+2 }{4} }\\ x_2= \pm \sqrt{\frac{4+4 \sqrt{2}+2 }{4}}\\ \\ x_1= \pm \sqrt{\frac{(2- \sqrt{2} ) ^2}{4} }\\ x_2=\pm \sqrt{\frac{(2+ \sqrt{2})^2 }{4}}$

$x_1= \pm (\frac{2- \sqrt{2}}{2} )\\ x_2=\pm (\frac{2+ \sqrt{2} }{2}) \\ \\ x_1= \pm (1-\frac{\sqrt{2}}{2} )\\ x_2=\pm (1+\frac{ \sqrt{2} }{2}) \\ \\ x_1= 1-\frac{\sqrt{2}}{2} \\ x_2=\frac{\sqrt{2}}{2} -1\\ x_3=1+\frac{ \sqrt{2} }{2} \\ x_4=-1-\frac{\sqrt{2}}{2} \\ \\ \\ \\ OTBET: 1-\frac{\sqrt{2}}{2}; \frac{\sqrt{2}}{2} -1;1+\frac{ \sqrt{2} }{2};-1-\frac{\sqrt{2}}{2}$