Question

найти корни уравнения
$2x^{4}-19x^{2}=9=0$

Answer 1

$2x^4-19x^2+9=0\ x^2=k\ 2k^2-19k+9=0\ D=361-72=289\ sqrt{d}=17\ k1=(19+17)/4=9\ k2=(19-17)/4=2/4=1/2 \x^2=9\ x=3\ x^2=1/2\ x=sqrt{1/2}$" title="$2x^4-19x^2+9=0\ x^2=k\ 2k^2-19k+9=0\ D=361-72=289\ sqrt{d}=17\ k1=(19+17)/4=9\ k2=(19-17)/4=2/4=1/2 \x^2=9\ x=3\ x^2=1/2\ x=sqrt{1/2}$" alt="$2x^4-19x^2+9=0\ x^2=k\ 2k^2-19k+9=0\ D=361-72=289\ sqrt{d}=17\ k1=(19+17)/4=9\ k2=(19-17)/4=2/4=1/2 \x^2=9\ x=3\ x^2=1/2\ x=sqrt{1/2}$" />