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Question 1042585: Hey, can someone please help?
Solve by completing the square: 9x=5x^2-1
I can normally figure out complete the square problems, but the leading coefficient in this is throwing me off! Please help! thank you!
Found 3 solutions by Fombitz, Boreal, rothauserc: Answer by Fombitz(32388) (Show Source): Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! rewrite as 5x^2-9x=1
factor out a 5 a
5[x^2-(9/5)x]=1
divide by 5
(x-(9/5)x=1/5
now complete the square
x-(9/5)x+81/100=(1/5)+(81/100); divide the middle term by 2, to make 9/10, then square it and add to both sides.
(x-(9/10)^2=(101/100), common denominator of 100.
x=(9/10)+/-sqrt(101)/10, since square root of 100 is 10
numerically, this is 0.9+/-1, or very close to it. Answer should be 1.9 and -0.1.

Quadratic formula:
x=(1/10)(9+/-sqrt(81+20)) and that is (9/10)+/-sqrt (101/100)
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! 9x = 5x^2 - 1
:
subtract 9x from both sides of =
:
5x^2 -9x -1 = 0
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add 1 to both sides of =
:
5x^2 -9x = 1
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divide both sides of = by 5
:
x^2 -9x/5 = 1/5
:
x^2 -9x/5 +81/100 = 1/5 +81/100
:
(x - 9/10)^2 = 101/100
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x - 9/10 = square root(101) / 10
:
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x = (9/10) + (square root(101) / 10) = 1.905
x = (9/10) - (square root(101) / 10) = -0.105
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