solve for Y. 5y^2 - 11y +2 = 0?
solve for Y.
5y^2 - 11y +2 = 0
回答 (7)
✔ 最佳答案
5y^2 - 11y + 2 = 0
(5y - 1)(y - 2) = 0
y = 1/5 or 2
*βω*
y = ~0.1429
______________________
5y^2-11y+2 = 0
-2 -2
5y*5y-11y = -2
25y-11y = -2
14y = -2
----- -----
14 14
y=1429
It's probably wrong because i suck at math, but it was fun working out... and it did equal 0... albiet 0.95, it's a zero whole number i guess... i don't know...
5y^2 - 11y + 2 = 0
(5y - 1)(5y - 2) = 0
5y - 1 - 0
5y = 1
y = 1/5 (0.2)
5y - 2 = 0
5y = 2
y = 2/5 (0.4)
â´ y = 1/5 (0.2) , 2/5 (0.4)
Factor it!
Note the coefficient 5, in the 5y^2 term, and the constant 2.
The 5 gives you something like:
(5y - ____)(y - ____)
The fact that the 2 is positive means either that you will have two postive or two negative constant terms, like this:
(____ + 1)(____ + 2) or (____ - 1)(____ - 2)
Since the middle term of your equation is -11y, we must use the two negatives.
So the pieces of our puzzle are:
(5y - ____)(y - ____) and (____ - 1)(____ - 2)
Putting it all together, we get either:
(5y - 1)(y - 2) or (5y - 2)(y - 1)
Now you have to decide which one it is--I won't finish it for you!
ax^2 + bx +c = 0
first find the discriminant:
d = b^2 - 4*a*c
d = (-11)^2 - 4*5*2 = 121-40 = 81
the 2 solutions is then:
x = (-b +/- sqrt(d)) /(2*a)
x = (11 +/- 9) /(10)
x = 2 and x = 2/10 = 1/5 = 0.2
5y^2 - 11y +2 = 0
5y^2-10y-y+2=0
5y(y-2)-1(y-2)=0
(y-2)(5y-1)=0
y-2=0 means y=2
5y-1=0 means 5y=1 or y=1/5
5y^2 - 11y + 2 = 5y^2 - 10y - y + 2 = 5y(y-2) - (y-2) = (5y-1)(y-2) = 0 when y=2 or y=1/5.
收錄日期: 2021-05-01 10:35:05
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