3y^2-5y+2=0
3y^2-5y+2=0
(3y + 1)(y - 2) = 0
you know what to do from here.
3y^2 - 5y + 2 = 0.
A*C = 3*2 = 6 = (-2)*(-3). sum = -2 + (-3) = -5 = B.
3y^2 + (-2y-3y) + 2 = 0,
(3y^2 - 2y) + (-3y + 2) = 0,
y(3y - 2) - (3y - 2) = 0.
(3y-2)(y - 1) = 0,
3y-2 = 0, Y = 2/3.
y-1 = 0, Y = 1.
To solve the quadratic equation 3y^2 - 5y + 2 = 0, you can either factorize it or use the quadratic formula.
First, let's try factoring:
The equation is in the form ax^2 + bx + c = 0, where a = 3, b = -5, and c = 2.
We need to find two numbers whose product is ac (3 * 2 = 6) and whose sum is b (-5).
Upon inspection, we see that the numbers are -3 and -2:
3y^2 - 3y - 2y + 2 = 0
3y(y - 1) - 2(y - 1) = 0
(3y - 2)(y - 1) = 0
Now, using the zero-product property, we set each of the factors equal to zero:
3y - 2 = 0 or y - 1 = 0
Solving these two equations, we get:
3y = 2 or y = 1
Dividing the first equation by 3, we find:
y = 2/3
Therefore, the solutions to the quadratic equation 3y^2 - 5y + 2 = 0 are y = 2/3 and y = 1.
If factoring the equation is difficult or impossible, you can always use the quadratic formula to find the solutions.
The quadratic formula states that for any quadratic equation ax^2 + bx + c = 0, the solutions can be found with the following formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation 3y^2 - 5y + 2 = 0, a = 3, b = -5, and c = 2. Plugging these values into the quadratic formula, we get:
y = (-(-5) ± √((-5)^2 - 4 * 3 * 2)) / (2 * 3)
y = (5 ± √(25 - 24)) / 6
y = (5 ± √(1)) / 6
Simplifying further:
y = (5 ± 1) / 6
This gives us two possible solutions:
y = (5 + 1) / 6 = 6 / 6 = 1
y = (5 - 1) / 6 = 4 / 6 = 2/3
Again, we find that the solutions to the quadratic equation 3y^2 - 5y + 2 = 0 are y = 1 and y = 2/3.