document.write( "Question 168786: I asked the same question ealier and got an answer to try it instead of helping. Can someone please help me find what I am doing wrong. \r
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\n" ); document.write( "\n" ); document.write( "I got(y-3)^2 but I dont come out to the same thing when I plug in the answer I dont get the +9. I also came up with(y+3)^2 but still dont get -6y.
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Algebra.Com's Answer #124437 by Electrified_Levi(103)\"\" \"About 
You can put this solution on YOUR website!
Hi, Hope I can help,
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\n" ); document.write( "If you are wanting us to solve this quadratic equation, here is how you do it.
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\n" ); document.write( "\"+y%5E2-6y%2B+9+\"
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\n" ); document.write( "The easiest way you can solve this is by factoring, in this case we can
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\n" ); document.write( "First we need to write the \"y\"'s in parentheses
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\n" ); document.write( "It is only \"y\" squared so there will be two \"y\"'s
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\n" ); document.write( "( y )(y )
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\n" ); document.write( "Next, we need to find the rest of the factors of this quadratic
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\n" ); document.write( "If the factors of the third term, in this case \"9\", add up to the middle term( in this case (-6)), it can be factored
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\n" ); document.write( "factors of \"9\" added together
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\n" ); document.write( "(9) + (1) = 10
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\n" ); document.write( "(-9) + (-1) = (-10)
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\n" ); document.write( "(3) + (3) = 6
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\n" ); document.write( "(-3) + (-3) = (-6)
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\n" ); document.write( "The last pair of factors add up to the middle term
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\n" ); document.write( "Now we will put the factors in the parentheses with the \"y\"'s
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\n" ); document.write( "( y )(y ) = (y - 3 )(y - 3 )
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\n" ); document.write( "This is the factored quadratic
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\n" ); document.write( "( y - 3 )(y - 3 ) = 0, you can check by using the FOIL method ( First, Inner, Outer, Last )
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\n" ); document.write( "\"+%28+y+-+3+%29%28y+-+3+%29+\" = (F) \"+%28+highlight%28y%29+-+3+%29%28highlight%28y%29+-+3+%29+\" = (O) \"+%28+highlight%28y%29+-+3+%29%28y+-+highlight%283%29+%29+\" = (I) \"+%28+y+-+highlight%283%29+%29%28highlight%28y%29+-+3+%29+\" = (L) \"+%28+y+-+highlight%283%29+%29%28y+-+highlight%283%29+%29+\", Remember the negative signs on the \"3's\"
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\n" ); document.write( "\"+y%5E2+-+3y+-+3y+%2B+9+\", combine the like terms, \"+%28y%5E2%29+highlight%28-+3y+-+3y%29+%2B+9+\" (forget about the multiplication sign, it is just being weird) = \"+y%5E2+-+6y+%2B+9+\" ( True )
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\n" ); document.write( " now we take each of the factors and have them equal \"0\", then solve for \"y\"
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\n" ); document.write( "First factor = ( y - 3 )
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\n" ); document.write( "\"+%28+y+-+3+%29+=+0+\", removing the parentheses
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\n" ); document.write( "\"++y+-+3++=+0+\", moving (-3) to the right side
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\n" ); document.write( "\"++y+-+3+%2B+3++=+0+%2B+3+\" = \"+y+=+3+\"
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\n" ); document.write( "\"+y+=+3+\", the other factor is the same, so we would get the same answer.
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\n" ); document.write( "your answer is \"+y+=+3+\", you can check by replacing \"y\" with \"3\" in the original equation
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\n" ); document.write( "\"+y%5E2-6y%2B+9+=+0+\" = \"+%283%29%5E2-6%283%29%2B+9+=+0+\" = \"+%289%29-%2818%29%2B+9+=+0+\" = \"+%28-9%29+%2B+9+=+0+\" = \"+0+=+0+\" ( True )
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\n" ); document.write( "You can also use the quadratic formula, for every quadratic equation, make sure your equation is in the form \"+ax%5E2+%2B+bx+%2B+c+=+0+\" ( in this case \"y\" ) it is
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\n" ); document.write( "\"+y%5E2-6y%2B+9+\"
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\n" ); document.write( "a = 1
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\n" ); document.write( "b = -6
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\n" ); document.write( "c = 9
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\n" ); document.write( "\"y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" = \"y+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A%281%29%2A%289%29+%29%29%2F%282%2A%281%29%29+\" = \"y+=+%28+6+%2B-+sqrt%28+%2836%29-36+%29%29%2F%282%29%29+\" = \"y+=+%28+6+%2B-+sqrt+%280%29%29%2F%282%29%29+\" = \"y+=+%28+6+%2B-+0%29%2F%282%29%29+\" = \"y+=+%28+6+%2F2%29+\" = \"y+=+3+\"
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\n" ); document.write( "Your answer is \"+y+=+3+\"
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\n" ); document.write( "Hope I helped, Levi
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