Question

Помогите пожалуйста, нужно решить через дискриминант.​

Answer 1

$4x^4-9x^2+2=0\\x^2-t\\4t^2-9t+2=0\\D=(-9)^2-4*4*2=81-32=49=7^2\\t_1=\frac{9+7}{2*4} =2\\t_2=\frac{9-7}{2*4} =0,75\\x^2=2\\x_{1,2}=\pm\sqrt{2} \\x^2=0,75\\x_{3,4}=\pm\sqrt{0,75} =\pm 5\sqrt{0,03}$

$2x^4-5x^2+2=0\\x^2=t\\2t^2-5t+2=0\\D=(-5)^2-4*2*2=25-16=9=3^2\\t_1=\frac{5+3}{2*2}=2\\ t_2=\frac{5-3}{2*2} =0,5\\x^2=2\\x_{1,2}=\pm\sqrt{2} \\x^2=0,5\\x_{3,4}=\pm\sqrt{0,5}$

$x^4+3x^2-70=0\\x^2=t\\t^2+3t-70=0\\D=3^2-4*(-70)=9+280=289=17^2\\t_1=\frac{-3+17}{2} =7\\t_2=\frac{-3-17}{2} =-10 \\x^2=7\\x_{1,2}=\pm\sqrt{7} \\x^2\neq -10$

$x^4-50x^2+49=0\\x^2=t\\t^2-50t+49=0\\D=(-50)^2-4*49=2500-196=2304=48^2\\t_1=\frac{50+48}{2} =49\\t_2=\frac{50-48}{2}=1\\ x^2=49\\x_{1,2}=\pm7\\x^2=1\\x_{3,4}=\pm1$

$x^4-5x^2-36=0\\x^2=t\\t^2-5t-36=0\\D=(-5)^2-4*(-36)=25+144=169=13^2\\t_1=\frac{5+13}{2} =9\\t_2=\frac{5-13}{2} =-4\\x^2=9\\x_{1,2}=\pm3\\x^2\neq -4$

$4x^4-13x^2+3=0\\x^2=t\\4t^2-13t+3=0\\D=(-13)^2-4*4*3=169-48=121=11^2\\t_1=\frac{13+11}{2*4}=3\\ t_2=\frac{13-11}{2*4} =0,25\\x^2=3\\x_{1,2}=\pm\sqrt{3} \\x^2=0,25\\x_{1,2}=\pm\sqrt{0,25} =\pm5\sqrt{0,01}$