Solve the equation y^2 - 5y + 6 = 0?
回答 (7)
2009-05-17 9:58 am
✔ 最佳答案
y² - 5y + 6 = 0Find 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}
2009-05-17 10:04 am
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-3) (y-2) = 0
Therefore y-3 = 0 or y - 2 = 0
Hence y=3 or y=2
The Solution set is {3,2}
參考: Self Knowledge
2009-05-17 10:01 am
y^2-3y-2y+6=0
y(y-3)-2(y-3)=0
(y-2)(y-3)=0
y=2
or
y=3
y(y-3)-2(y-3)=0
(y-2)(y-3)=0
y=2
or
y=3
2014-10-02 8:43 pm
i just want to say that i used these answers to answer a question in class and they are wrong. Thanks anyway!!!
2009-05-17 11:33 am
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
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
2009-05-17 11:05 am
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
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
2009-05-17 10:13 am
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..:-)
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..:-)
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