SOLUTION: I need to complete the square: {{{2x^2+10x-3 =0}}}
I've been using {{{2x^2+4x-1=0}}} to figure out the formula since I know the answer to this second one.
I used the Complete th
Question 288289: I need to complete the square:
I've been using to figure out the formula since I know the answer to this second one.
I used the Complete the Square Solver on here, but it comes up with a different answer than the book.
The formula I'm using is: x=-4 +or- square root of (4)^2-4(2)(-1) over 2(2)
so, -4+or- square root of 16 + 8 over 4
then -4+or- square root of 24 over 4
then -4+or- 2 times square root of 6 over 4
after simplifying, i get -1+or- square root of 6 over 2.
My book says the answer is -2+or- square root of 6 over 2.
Where did I go wrong?? If I can't solve THIS one, then I know I can't solve the others.
Could somebody please explain where I messed up? I'm assuming I should have multiplied something by 2 somewhere along the way to come up with the book answer, but I can't figure out where or more importantly, WHY.
Thanks for any help given, I appreciate it! Found 2 solutions by stanbon, richwmiller:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! I need to complete the square: 2x^2+10x-3 =0
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2x^2 + 10x = 3
2(x^2 + 5x + ? = 3 + 2*?
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2(x^2 + 5x + (5/2)^2) = 3 + 2(5/2)^2
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2(x+(5/2))^2 = 3 + 25/2
2(x+(5/2))^2 = 31/2
(x+(5/2))^2 = 31/4
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x+(5/2) = +-sqrt(31)/2
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x = [-5+-sqrt(31)]/2
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I'm not sure where you got your "answers"
but they are wrong.
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Cheers,
Stan H.
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You can put this solution on YOUR website! 2x^2+10x-3 =0
2x^2+10x=3
2*(x^2+5x)=3
x^2+5x+25/4=6/4+25/4
x^2+5x+25/4=31/4
(x+5/2)^2=(31/4)
x = 1/2 (sqrt(31)-5)
x = 1/2 (-5-sqrt(31))
2x^2+4x-1=0
2x^2+4x=1
2(x^2+2x+1)=3
2(x+1)^2=3
x = 1/2 (sqrt(6)-2)
x = 1/2 (-2-sqrt(6))
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