Question

Помогите пожалуйста, нужно решить через дискриминант.​

Answer 1

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

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

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

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

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

$4x^4-13x^2+3=0\x^2=t\4t^2-13t+3=0\D=(-13)^2-4*4*3=169-48=121=11^2\t_1=frac{13+11}{2*4}=3\ t_2=frac{13-11}{2*4} =0,25\x^2=3\x_{1,2}=pmsqrt{3} \x^2=0,25\x_{1,2}=pmsqrt{0,25} =pm5sqrt{0,01}$