Question

Find the roots of the following equation

$3x^4-13x^2+10=0$, then

• A. $x=\pm \sqrt{\frac{10}{3}},\pm 3$
• B. $x=\pm \sqrt{\frac{10}{7}},\pm 1$
• C. $x=\pm \sqrt{\frac{10}{3}},\pm 1$
• D. $x=\pm \sqrt{\frac{10}{3}},\pm 2$

Answer 1

The correct option is C $x=\pm \sqrt{\frac{10}{3}},\pm 1$

Given equation is

$3x^4-13x^2+10=0$
$3^4-10x^2-3x^2+10=0$
$(3x^2-10)(x^2-1)=0$
$3x^2=10$

$\therefore x=\pm \sqrt{\frac{10}{3}}$

And $x^2=1$

$\therefore x=\pm 1$

$\therefore x=\pm \sqrt{\frac{10}{3}},\pm 1$