SOLUTION: Please help with this problem: {{{2x^2+8x+1=0}}} Thanks

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Question 144457This question is from textbook
: Please help with this problem:
2x%5E2%2B8x%2B1=0
Thanks This question is from textbook

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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 2%2Ax%5E2%2B8%2Ax%2B1=0 ( notice a=2, b=8, and c=1)




x+=+%28-8+%2B-+sqrt%28+%288%29%5E2-4%2A2%2A1+%29%29%2F%282%2A2%29 Plug in a=2, b=8, and c=1



x+=+%28-8+%2B-+sqrt%28+64-4%2A2%2A1+%29%29%2F%282%2A2%29 Square 8 to get 64



x+=+%28-8+%2B-+sqrt%28+64%2B-8+%29%29%2F%282%2A2%29 Multiply -4%2A1%2A2 to get -8



x+=+%28-8+%2B-+sqrt%28+56+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-8+%2B-+2%2Asqrt%2814%29%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-8+%2B-+2%2Asqrt%2814%29%29%2F4 Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

x+=+%28-8+%2B+2%2Asqrt%2814%29%29%2F4 or x+=+%28-8+-+2%2Asqrt%2814%29%29%2F4


Now break up the fraction


x=-8%2F4%2B2%2Asqrt%2814%29%2F4 or x=-8%2F4-2%2Asqrt%2814%29%2F4


Simplify


x=-2%2Bsqrt%2814%29%2F2 or x=-2-sqrt%2814%29%2F2



So our answers are


x=-2%2Bsqrt%2814%29%2F2 or x=-2-sqrt%2814%29%2F2


which approximate to


x=-0.129171306613029 or x=-3.87082869338697