N nae New member Joined Jun 18, 2006 Messages 33 Jun 28, 2006 #1 5y^2-16y=-3 i know it has to be in standard form which is 5y^2-16+-3=0 5y^2-15+1y+-3 This is where i get lost?
5y^2-16y=-3 i know it has to be in standard form which is 5y^2-16+-3=0 5y^2-15+1y+-3 This is where i get lost?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,337 Jun 28, 2006 #2 nae said: 5y^2-16y=-3 i know it has to be in standard form which is 5y^2-16+-3=0 5y^2-15+1y+-3 This is where i get lost? Click to expand... 1) Where did the equal sign go? 2) Where did the "y" go? 3) Why didn't the "-3" change signs? 5y^2 - 16y = -3 Add 3 5y^2 - 16y + 3 = -3 + 3 Simplify 5y^2 - 16y + 3 = 0 Are we factorign by grouping, or what? 5y^2 - 15y - y + 3 = 0 <== This is almost magical. Why would you do this unless you knew where it was going? Common 5 in the first two terms. Watch the signes on the second two terms. 5(y^2 - 3y) - (y - 3) = 0 Common y in the first two terms. 5y(y - 3) - (y - 3) = 0 Common (y-3) in each term -- Watch very carefully. (5y - 1)(y - 3) = 0 Now what? Are we solving or just factoring?
nae said: 5y^2-16y=-3 i know it has to be in standard form which is 5y^2-16+-3=0 5y^2-15+1y+-3 This is where i get lost? Click to expand... 1) Where did the equal sign go? 2) Where did the "y" go? 3) Why didn't the "-3" change signs? 5y^2 - 16y = -3 Add 3 5y^2 - 16y + 3 = -3 + 3 Simplify 5y^2 - 16y + 3 = 0 Are we factorign by grouping, or what? 5y^2 - 15y - y + 3 = 0 <== This is almost magical. Why would you do this unless you knew where it was going? Common 5 in the first two terms. Watch the signes on the second two terms. 5(y^2 - 3y) - (y - 3) = 0 Common y in the first two terms. 5y(y - 3) - (y - 3) = 0 Common (y-3) in each term -- Watch very carefully. (5y - 1)(y - 3) = 0 Now what? Are we solving or just factoring?