if the arithmetic mean of x and x^2 is 6,
(1) form a quadratic equation in x
(2) Hence find the possible value(s) of x
第(1)我就知道係
( x^2 + x ) / 2 = 6
x^2 + x + 12 = 0
但係我唔識做(2)...請幫忙 steps by steps pls~
P.S 第(2)條的HENCE字係咪要一定用返(1)個條式搵X咁解呢?
THX VERY MUCH~
請求幫手解arithmetic mean~(5分)
2007-04-20 12:42 am
回答 (2)
2007-04-20 1:06 am
✔ 最佳答案
( x^2 + x ) / 2 = 6x^2 + x - 12 = 0
HENCE字係要用返(1)個條式搵x
x^2 + x - 12 = 0
--> a^2 + bx + c = 0
use [-b (+/-) sqrt(b^2-4ac)] / (2a)
(+/-) means one root is +, & the other root is -
sqrt = square root
The roots are
[ (-1) + sqrt((1^2)-4(1)(-12)) ] / (2(1)) & [ (-1) - sqrt((1^2)-4(1)(-12)) ] / (2(1))
x = -4 or x = 3
2007-04-20 1:03 am
Hence 姐係一定要用1)的答案來做
Hence or otherwise就唔一定
No.2 is only equation solving.
but the discriminant is less than 0.
So I think your answer in 1) is wrong.
Hence or otherwise就唔一定
No.2 is only equation solving.
but the discriminant is less than 0.
So I think your answer in 1) is wrong.
參考: ME
收錄日期: 2021-04-12 19:19:15
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