SOLUTION: solve by completing the square 2x^2+5x-1=0 the solutions are x=??????????? HELP ME!!! PLEASE

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Question 155267: solve by completing the square
2x^2+5x-1=0
the solutions are x=???????????

HELP ME!!! PLEASE
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve by completing the square:
2x%5E2%2B5x-1+=+0 First, divide through by 2 to get the coefficient of x%5E2 equal to 1.
x%5E2%2B%285%2F2%29x-1%2F2+=+0 Now add 1%2F2 to both sides.
x%5E2%2B%285%2F2%29x+=+1%2F2 Complete the square in the x-terms by add the square of half the x-coefficient (%28%281%2F2%29%285%2F2%29%29%5E2+=+highlight%2825%2F16%29) to both sides of the equation.
x%5E2%2B%285%2F2%29x%2Bhighlight%2825%2F16%29+=+1%2F2+%2B+highlight%2825%2F16%29 Simplify, where possible.
x%5E2%2B%285%2F2%29x+%2B+25%2F16+=+33%2F16 Now factor the left side.
%28x%2B5%2F4%29%28x%2B5%2F4%29+=+33%2F16 Simplify.
%28x%2B5%2F4%29%5E2+=+33%2F16 Take the square root of both sides.
%28x%2B5%2F4%29+=+sqrt%2833%2F16%29 Simplify.
%28x%2B5%2F4%29+=+0%2B-sqrt%2833%29%2F4 Now subtract 5%2F4 from both sides.
x+=+%285%2F4%29%2B-sqrt%2833%29%2F4
x+=+%285%2Bsqrt%2833%29%29%2F4 or x+=+%285-sqrt%2833%29%29%2F4