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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 
: 
add 1 to both sides of = 
: 
5x^2 -9x = 1 
: 
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 
: 
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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