Solve the following quadratic equation by factorization method: y2-5y+6=0
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Answered by
8
y2-3y-2y+6=0
y(y-3)-2(y-3)
(y-3)y-2)
y=2 or 3
y(y-3)-2(y-3)
(y-3)y-2)
y=2 or 3
sokomukanya:
well done panashe
kkkk thank you
Answered by
6
Hi !
p(y) = y² - 5y + 6
y² - 5y + 6 = 0
sum = -5
product = 6
no:s = -2,-3
y² - 2y - 3y + 6 = 0
y(y - 2) - 3 ( y - 2) = 0
(y - 2) (y - 3) = 0
y = 2
y = 3
Hope this helps you !
p(y) = y² - 5y + 6
y² - 5y + 6 = 0
sum = -5
product = 6
no:s = -2,-3
y² - 2y - 3y + 6 = 0
y(y - 2) - 3 ( y - 2) = 0
(y - 2) (y - 3) = 0
y = 2
y = 3
Hope this helps you !
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