Question

Solve the quadratic equation $3x^4-13x^2+10=0$.

Answer 1

The given quadratic equation is,
$3x^4-13x^2+10=0$
$3{\left(x^2\right)}^2-13\left(x^2\right)+10=0$
Substituting $x^2=m$ we get
$3m^2-13m+10=0$
$3m^2-3m-10m+10=0$
$3m(m-1)-10(m-1)=0$
$(m-1)(3m-10)=0$
$m-1=0$ or $3m-10=0$
$\therefore$ $m=1$ or $3m=10$
$\therefore$ $m=1$ or $m=\frac{10}{3}$
Substituting $m=x^2$, we get
$x^2=1$ or $x^2=\frac{10}{3}$
$x=\pm 1$ or $x=\pm \sqrt{\frac{10}{3}}$
$\therefore$ $x=1,-1,\sqrt{\frac{10}{3}},-\sqrt{\frac{10}{3}}$
$1,-1,\sqrt{\frac{10}{3}}$ and $-\sqrt{\frac{10}{3}}$ are the roots of the given quadratic equation.