SOLUTION: Find a real or imaginary solution 86. 3v^2 + 4v - 1 = 0

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Question 170965: Find a real or imaginary solution

86. 3v^2 + 4v - 1 = 0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for v:


Starting with the general quadratic


av%5E2%2Bbv%2Bc=0


the general solution using the quadratic equation is:


v+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 3%2Av%5E2%2B4%2Av-1=0 ( notice a=3, b=4, and c=-1)





v+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A3%2A-1+%29%29%2F%282%2A3%29 Plug in a=3, b=4, and c=-1




v+=+%28-4+%2B-+sqrt%28+16-4%2A3%2A-1+%29%29%2F%282%2A3%29 Square 4 to get 16




v+=+%28-4+%2B-+sqrt%28+16%2B12+%29%29%2F%282%2A3%29 Multiply -4%2A-1%2A3 to get 12




v+=+%28-4+%2B-+sqrt%28+28+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)




v+=+%28-4+%2B-+2%2Asqrt%287%29%29%2F%282%2A3%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




v+=+%28-4+%2B-+2%2Asqrt%287%29%29%2F6 Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


v+=+%28-4+%2B+2%2Asqrt%287%29%29%2F6 or v+=+%28-4+-+2%2Asqrt%287%29%29%2F6



Now break up the fraction



v=-4%2F6%2B2%2Asqrt%287%29%2F6 or v=-4%2F6-2%2Asqrt%287%29%2F6



Simplify



v=-2%2F3%2Bsqrt%287%29%2F3 or v=-2%2F3-sqrt%287%29%2F3



So the solutions are:

v=-2%2F3%2Bsqrt%287%29%2F3 or v=-2%2F3-sqrt%287%29%2F3