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

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Answer 1 · 2018-01-19T12:59:42

See below

$14x^2 + 5 = 3x^4$
$0 = 3x^4 - 14x^2 - 5$
$0 = 3x^4 - 15x^2 + x^2 - 5$
$0 = 3x^2(x^2-5) + 1(x^2-5)$
$0 = (3x^2+1)(x^2-5)$

$0 = 3x^2 + 1$
$-1 = 3x^2$
$-1/3 = x^2$
$+-sqrt(-1/3) = x$
These roots are not real.

$x^2 - 5 = 0$
$x^2 = 5$
$x = +-sqrt5$
These roots are real.

How do you find the real solutions of the polynomial 14x^2+5=3x^414x^2+5=3x^4?