factoring a quatratic equation

sithdas

New member
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.

masters

Full Member
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.
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 $$\displaystyle -\frac{1}{2}\: \text{not}\: \frac{1}{2}$$