SOLUTION: Could someone help me solve this problem using quadratic formula step by step. I am lost and do not understand. x^2+4x+3=0 Thank you

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Question 387141: Could someone help me solve this problem using quadratic formula step by step. I am lost and do not understand.
x^2+4x+3=0
Thank you
Found 2 solutions by richard1234, jim_thompson5910:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula states that, for any second-degree polynomial ax%5E2+%2B+bx+%2B+c+=+0, where a, b, and c are coefficients, then

x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F2a

In your provided quadratic, a = 1, b = 4, c = 3. Now we just plug in those numbers and simplify:

x+=+%28-4+%2B-+sqrt%284%5E2+-+4%283%29%281%29%29%29%2F2

x+=+%28-4+%2B-+2%29%2F2 --> x = -3 or -1. You can check each solution and see that they satisfy the equation.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2%2B4x%2B3=0 Start with the given equation.


Notice that the quadratic x%5E2%2B4x%2B3 is in the form of Ax%5E2%2BBx%2BC where A=1, B=4, and C=3


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%284%29+%2B-+sqrt%28+%284%29%5E2-4%281%29%283%29+%29%29%2F%282%281%29%29 Plug in A=1, B=4, and C=3


x+=+%28-4+%2B-+sqrt%28+16-4%281%29%283%29+%29%29%2F%282%281%29%29 Square 4 to get 16.


x+=+%28-4+%2B-+sqrt%28+16-12+%29%29%2F%282%281%29%29 Multiply 4%281%29%283%29 to get 12


x+=+%28-4+%2B-+sqrt%28+4+%29%29%2F%282%281%29%29 Subtract 12 from 16 to get 4


x+=+%28-4+%2B-+sqrt%28+4+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-4+%2B-+2%29%2F%282%29 Take the square root of 4 to get 2.


x+=+%28-4+%2B+2%29%2F%282%29 or x+=+%28-4+-+2%29%2F%282%29 Break up the expression.


x+=+%28-2%29%2F%282%29 or x+=++%28-6%29%2F%282%29 Combine like terms.


x+=+-1 or x+=+-3 Simplify.


So the solutions are x+=+-1 or x+=+-3


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim