SOLUTION: Please help with this problem: Find all zeros of f: f(x) = 3x^2 + 5x - 2 I believe you use the quadratic formula to solve or do I need to factor?

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Question 110872: Please help with this problem:
Find all zeros of f: f(x) = 3x^2 + 5x - 2
I believe you use the quadratic formula to solve or do I need to factor?

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 x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


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




So lets solve 3%2Ax%5E2%2B5%2Ax-2=0 ( notice a=3, b=5, and c=-2)





x+=+%28-5+%2B-+sqrt%28+%285%29%5E2-4%2A3%2A-2+%29%29%2F%282%2A3%29 Plug in a=3, b=5, and c=-2




x+=+%28-5+%2B-+sqrt%28+25-4%2A3%2A-2+%29%29%2F%282%2A3%29 Square 5 to get 25




x+=+%28-5+%2B-+sqrt%28+25%2B24+%29%29%2F%282%2A3%29 Multiply -4%2A-2%2A3 to get 24




x+=+%28-5+%2B-+sqrt%28+49+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-5+%2B-+7%29%2F%282%2A3%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-5+%2B-+7%29%2F6 Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


x+=+%28-5+%2B+7%29%2F6 or x+=+%28-5+-+7%29%2F6


Lets look at the first part:


x=%28-5+%2B+7%29%2F6


x=2%2F6 Add the terms in the numerator

x=1%2F3 Divide


So one answer is

x=1%2F3




Now lets look at the second part:


x=%28-5+-+7%29%2F6


x=-12%2F6 Subtract the terms in the numerator

x=-2 Divide


So another answer is

x=-2


So our solutions are:

x=1%2F3 or x=-2