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

1 Answer
Oct 1, 2015

Answer:

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

Explanation:

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

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

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

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

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

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

If #q=1#
then
#color(white)("XXX")x^2=1#

#rarrcolor(white)("X")x=+-1#