Question

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

Answer 1

$2x^{4}-19x^{2}+9=0$

$x^{2}=t$

$2t^{2}-19t+9=0$

$x_{1}=0,5$

$x_{2}=9$

$x^{2}=9$

$x=+-3$

$x^{2}=0,5$

$x=+-sqrt{0,5}=+-frac{1}{sqrt{2} }$

Ответ: $x_{1}=3, x_{2}=-3, x_{3}= frac{1}{sqrt{2}}, x_{4}=-frac{1}{sqrt{2}}$