First, let's eliminate the three by dividing both sides by that number.
#(3( 6x^4 - 5x^2 - 1) = 0)/3#
#6x^4 - 5x^2 - 1 = 0#
Now, we can factor the left side of the equation. We need to find numbers that multiply to -1, multiply to #6x^4#, and add to #-5x^2#. The first one I noticed immediately are factors #6x^2#, #x^2#, 1, and -1. We can arrange it as so:
#6x^4 - 5x^2 - 1 = 0#
#(6x^2+1)(x^2-1) = 0#
Notice that we can break down the #x^2-1# even further. Remember that a difference of squares factors into the respective sum and difference of their square roots. This is what I mean by that: #a^2-b^2=(a+b)(a-b)#.
Now we get:
#(6x^2+1)(x+1)(x-1) = 0#
Now we can solve for x. Each set of terms in the parentheses will be set to 0. So:
#6x^2+1=0#
#x+1=0#
#x-1=0#
The last two are easy to solve. You get #x=-1# and #x=1#, respectively. The first one actually deals with an imaginary root. If you have not learned that yet, you can ignore this last part. If you did learn imaginary numbers, let's solve for the third value of x:
#6x^2+1=0#
#x^2 = -1/6#
#x = ±sqrt(-1/6)#
#x = ± i sqrt(1/6)#
#x = ±i sqrt1/sqrt6#
#x = ±i 1/sqrt6#
To get the final answer, we need to eliminate the radical from the denominator.
#x = ±i1/sqrt6#
#x=±i(1sqrt6)/(sqrt6sqrt6)#
#x = ±i sqrt6/6#
Your results are:
#x = ±1, ±i sqrt6/6#
