SOLUTION: x^2+4x+3=0, how do i solve this

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Question 295717: x^2+4x+3=0, how do i solve this
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 x%5E2%2B4%2Ax%2B3=0 ( notice a=1, b=4, and c=3)





x+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29 Plug in a=1, b=4, and c=3




x+=+%28-4+%2B-+sqrt%28+16-4%2A1%2A3+%29%29%2F%282%2A1%29 Square 4 to get 16




x+=+%28-4+%2B-+sqrt%28+16%2B-12+%29%29%2F%282%2A1%29 Multiply -4%2A3%2A1 to get -12




x+=+%28-4+%2B-+sqrt%28+4+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-4+%2B-+2%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-4+%2B-+2%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%28-4+%2B+2%29%2F2 or x+=+%28-4+-+2%29%2F2


Lets look at the first part:


x=%28-4+%2B+2%29%2F2


x=-2%2F2 Add the terms in the numerator

x=-1 Divide


So one answer is

x=-1




Now lets look at the second part:


x=%28-4+-+2%29%2F2


x=-6%2F2 Subtract the terms in the numerator

x=-3 Divide


So another answer is

x=-3


So our solutions are:

x=-1 or x=-3