How do you solve the equation for y: #2x^4+x^2-3=0#?

1 Answer
George C.
Jul 4, 2015

#0 = 2x^4+x^2-3#

#= (2x^2+3)(x^2-1)#

#= (2x^2+3)(x-1)(x+1)#

So #x = 1# or #x = -1#

Explanation:

#2x^4+x^2-3 = 2(x^2)^2 + (x^2) - 3#

is quadratic in #x^2#

Notice that the sum of the coefficients is #0#

#2+1-3 = 0#

So #x^2 = 1# is a root and #(x^2 - 1)# is a factor.

#x^2 - 1 = x^2 - 1^2 = (x-1)(x+1)#

using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

The other factor of #2x^4+x^2-3# must be #(2x^2+3)# in order that when multiplied by #(x^2-1)# it results in the leading #2x^4# term and the trailing #-3# term.

#2x^2+3 = 0# has no real roots, since #x^2 >= 0#, hence #2x^2+3 >= 3#

So the only real roots of the original quartic equation are #x = +-1#

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