Question

6y² + 17y + 12 = 0.

Answer 1

The given quadratic equation is 6y² + 17y + 12 = 0.
On splitting the middle term 17y as 9y + 8y, we get:
6y² + 9y + 8y + 12 = 0
=> 3y(2y + 3) + 4(2y + 3) = 0
=> (2y + 3) (3y + 4) = 0
We know that if the product of two numbers is zero, then at least one of them must be zero.
Thus,
2y + 3 = 0 or 3y + 4 = 0
=> y = –$\frac{3}{2}$ or y = –$\frac{4}{3}$
Therefore, the solution of the given equation is y = –$\frac{3}{2}$ , –$\frac{4}{3}$.