√(3x+4) + 4 = 2x
√(3x+4) = 2x-4
3x+4 = (2x-4)^2
3x+4 = 4x^2 - 16x + 16
4x^2 - 19x + 12 = 0
(4x-3)(x-4) = 0
x = 3/4 or 4
But, squaring sometimes introduces extraneous roots. In this case, x = 3/4 does not work.
√(3*3/4 + 4) + 4 = 2(3/4)
√(9/4 + 4) + 4 = 3/2
√(25/4) + 4 = 3/2
5/2 + 4 = 3/2
NO
So, how did we get 3/4 as a solution?
Because
-√(25/4) + 4 = 3/2
-5/2 + 4 = 3/2
3/2 = 3/2
when you square both sides, both the +√ and the -√ work. But the original equation only used the +√.
sqrt(3x+4) + 4 = 2x
can you solve this with steps to help me please?
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