Solve the equation y^2 - 5y + 6 = 0?

y² - 5y + 6 = 0

Find two numbers whose sum is -5 and whose product is 6
(-2)+(-3) = -5
(-2)(-3) = 6

y² - 2y - 3y + 6 = 0
y(y-2)-3(y-2) = 0
(y-2)(y-3) = 0

y-2 = 0
y = 2

y-3 = 0
y = 3


y = {2, 3}

y^2 - 5y + 6 = 0
y^2 - 3y - 2y + 6 = 0
y(y-3) - 2(y-3) = 0
(y-3) (y-2) = 0

Therefore y-3 = 0 or y - 2 = 0
Hence y=3 or y=2

The Solution set is {3,2}

y^2-3y-2y+6=0

y(y-3)-2(y-3)=0

(y-2)(y-3)=0

y=2

or

y=3

i just want to say that i used these answers to answer a question in class and they are wrong. Thanks anyway!!!

y^2 - 5y + 6 = 0

Factor a trinomial

(y - 2)(y - 3) = 0

We can multiply using the FOIL Method to confirm (check and test) the answer.

y * y = y^2

y * -3 = -3y

-2 * y = -2y

-2 * -3 = 6

y^2 - 3y - 2y + 6

y^2 - 5y + 6 = 0

Subproblem 1

y -2 = 0

Add 2 to each side of the equation

y - 2 + 2 = 0 + 2

y = 2

Subproblem 2

y - 3 = 0

Add 3 to each side of the equation

y - 3 + 3 = 0 + 3

y = 3

Solution:

y = 2, 3

y^2 - 5y + 6 = 0
y^2 - 2y - 3y + 6 = 0
(y^2 - 2y) - (3y - 6) = 0
y(y - 2) - 3(y - 2) = 0
(y - 2)(y - 3) = 0

y - 2 = 0
y = 2

y - 3 = 0
y = 3

∴ y = 2, 3

We can write the above equation as

y^2 - 3y - 2y + 6 = 0

Take 'y' common on y^2 and -3y we'll get

y(y - 3) - 2y + 6 = 0

take ' 2 ' as common in -2y and 6 we'll get

y(y - 3) - 2(y - 3) = 0

Take (y-3) common in y(y-3) and -2(y-3) we'll get

(y - 3)(y - 2) = 0 so we can equate

y - 3 = 0 and

y - 2 = 0

so we get y = 3 and y = 2 that's the solution.

Verification

if y = 3

3^2 - 5(3) + 6 = 0
9 - 15 + 6 = 0
-6 + 6 =0
0 = 0
so correct

if y = 2
2^2 - 5(2) + 6 = 0
4 - 10 + 6 = 0
-6 + 6 = 0
0 = 0

so correct

so solution is y = 2 and y = 3..:-)