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How do you solve $3 x^{4} + 12 x^{2} - 15 = 0$ $3x^4+12x^2-15=0$ ?

1 Answer

Oct 1, 2015

Assuming we are restricted to Real (non-Complex) solutions:
$XXX x = \pm 1$ $color(white)("XXX")x=+-1$

Explanation:

Given
$XXX 3 x^{4} + 12 x^{2} - 15 = 0$ $color(white)("XXX")3x^4+12x^2-15=0$

Dividing both sides by $3$ $3$
$XXX x^{4} + 4 x^{2} - 5 = 0$ $color(white)("XXX")x^4+4x^2-5=0$

Temporarily substituting $q$ $q$ for $x^{2}$ $x^2$
$XXX q^{2} + 4 q - 5 = 0$ $color(white)("XXX")q^2+4q-5=0$

Factoring the left side
$XXX (q + 5) (q - 1) = 0$ $color(white)("XXX")(q+5)(q-1) = 0$

With potential solutions
$XXX q = - 5$ $color(white)("XXX")q=-5$ and $q = 1$ $q=1$

If $q = - 5$ $q=-5$
then
$XXX x^{2} = - 5$ $color(white)("XXX")x^2=-5$
which is impossible for Real values of $x$ $x$

If $q = 1$ $q=1$
then
$XXX x^{2} = 1$ $color(white)("XXX")x^2=1$

$\to X x = \pm 1$ $rarrcolor(white)("X")x=+-1$

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