S sithdas New member Joined Nov 16, 2010 Messages 1 Nov 16, 2010 #1 10y^2=-5y. Not sure how to do this one. Please help? I know the answers are ,0 and 1/2 but I need to know how to do them.

10y^2=-5y. Not sure how to do this one. Please help? I know the answers are ,0 and 1/2 but I need to know how to do them.

M masters Full Member Joined Mar 30, 2007 Messages 378 Nov 16, 2010 #2 sithdas said: 10y^2=-5y. Not sure how to do this one. Please help? I know the answers are ,0 and 1/2 ?? but I need to know how to do them. Click to expand... Hi sithdas, \(\displaystyle 10y^2=-5y\) First thing, transpose the -5y to the left side setting the equation to zero. \(\displaystyle 10y^2+5y=0\) Next, look for common monomial factors. Do you see that 5y is a common factor? \(\displaystyle 5y(2y+1)=0\) Now use the zero product property to set each factor to 0 and solve each of them. You'll discover that one of the solutions is \(\displaystyle -\frac{1}{2}\: \text{not}\: \frac{1}{2}\)

sithdas said: 10y^2=-5y. Not sure how to do this one. Please help? I know the answers are ,0 and 1/2 ?? but I need to know how to do them. Click to expand... Hi sithdas, \(\displaystyle 10y^2=-5y\) First thing, transpose the -5y to the left side setting the equation to zero. \(\displaystyle 10y^2+5y=0\) Next, look for common monomial factors. Do you see that 5y is a common factor? \(\displaystyle 5y(2y+1)=0\) Now use the zero product property to set each factor to 0 and solve each of them. You'll discover that one of the solutions is \(\displaystyle -\frac{1}{2}\: \text{not}\: \frac{1}{2}\)