# How do you find the real solutions of the polynomial 14x^2+5=3x^4?

Jan 19, 2018

See below

#### Explanation:

$14 {x}^{2} + 5 = 3 {x}^{4}$
$0 = 3 {x}^{4} - 14 {x}^{2} - 5$
$0 = 3 {x}^{4} - 15 {x}^{2} + {x}^{2} - 5$
$0 = 3 {x}^{2} \left({x}^{2} - 5\right) + 1 \left({x}^{2} - 5\right)$
$0 = \left(3 {x}^{2} + 1\right) \left({x}^{2} - 5\right)$

$0 = 3 {x}^{2} + 1$
$- 1 = 3 {x}^{2}$
$- \frac{1}{3} = {x}^{2}$
$\pm \sqrt{- \frac{1}{3}} = x$
These roots are not real.

${x}^{2} - 5 = 0$
${x}^{2} = 5$
$x = \pm \sqrt{5}$
These roots are real.