SOLUTION: How do I solve 3x(to the 4th power)-2x(to the 2nd power)=16

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Question 1149771: How do I solve 3x(to the 4th power)-2x(to the 2nd power)=16
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.

    3x^4 - 2x^2 = 16.


Introduce new variable t = x^2.

Then your equation takes the form


    3t^2 - 2t - 16 = 0.


Solve it using the quadratic formula


    t%5B1%2C2%5D = %282+%2B-+sqrt%282%5E2+%2B+4%2A3%2A16%29%29%2F%282%2A3%29 = %282+%2B-+sqrt%28196%29%29%2F6 = %282+%2B-+14%29%2F6.


Only positive root is acceptable (since  t = x^2),  t = %282+%2B+14%29%2F6 = 16%2F6 = 8%2F3.


It implies  x^2 = 8%2F3;  hence,  x = +/- sqrt%288%2F3%29 = +/- %282%2Asqrt%286%29%29%2F3.    ANSWER

Solved.

Introducing new variable is the STANDARD WAY solving such problems.