Hmmm. is the second term x^2 or not?
14x^2-5x^2=7x^2
x^2=1/7
x= +-sqrt (1/7)
Find the real solutions of the equation:
14x^2 - 5x^2 - 1 = 0
I know that you have to use au^2 + bu +c = 0
so u^2 - 5x -1 which is prime.
So would the answer be that the solution is the empty set?
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