Question

9y² – 3y – 2 = 0

Answer 1

Given: 9y² – 3y – 2 = 0
On splitting the middle term –3y as 3y – 6y, we get:
9y² + 3y – 6y – 2 = 0
=> 3y(3y + 1) – 2(3y + 1) = 0
=> (3y + 1) (3y – 2) = 0
We know that if the product of two numbers is zero, then at least one of them must be zero.
Thus, 3y + 1 = 0 or 3y - 2 = 0
=> y = –$\frac{-1}{3}$ or y = $\frac{2}{3}$
Therefore, the solution of the given equation is y = $\frac{-1}{3}$,$\frac{2}{3}$ .