f.4 maths 急!!!! quadratic equation(20分)

a) 我根據你修正後的 equation o黎做:
3x2-2(3k+1)x+(3k2+k+4) = 0
For it to have unequal real roots, △ > 0, i.e.
[-2(3k+1)]2 - 4(3)(3k2+k+4) > 0
4(9k2+6k+1) - 12(3k2+k+4) > 0
(9k2+6k+1) - 3(3k2+k+4) > 0
3k - 11 > 0
k > 11/3
b) i) Sub x = 4 into the equation:
3(4)2-2(3k+1)(4)+(3k2+k+4) = 0
3(16) - 24k - 8 + 3k2+k+4 = 0
3k2- 23k + 44 = 0
k = 4 or 11/3 (rejected according to the result of a)
b) ii) Sub k = 4 into the equation, we have the original equation is:
3x2-2(3k+1)x+(3k2+k+4) = 0
3x2 - 26x + 56 = 0
(3x - 14)(x - 4) = 0
∴x = 4 or x = 14/3 (the other root)

唉…唔好成日都直接出答案啦~~害人0架~

首先 你條式後面少左加減號.... solve 唔到 sorry!
a) 佢話有 two unequal real roots<<<<<
je係話 △(deta)＞0 (你識呢part唔駛再講啦)
b)i) 佢話 given that 4 is a root of the equation<<<<<
咁你就用 4 put 入去條式個 x 到，就搵到 the value of k<<<<
b)ii) the other root of the equation<<<<<<
因為上一part你搵到 the value of k<<<< je係你知 b同c既actual value
就可以用sum of roots = -b/a 同 product of roots = c/a 既 formula 計
(因為你已經知道一個root既value 同b and c既value)

希望我講既野可以幫到你少少啦~~!!

sorry,你條qudratic equation少左d加減號,只可以用concept答你,

a)deta>=0

b^2-4ac>=0

and hence u can find range of k in this part.


b)given that 4 is a root of the equation, find

i)the value of k

因為你知道x=4是其中一個root,所以put x=4 into the equation:

3x^2-2(3k-1)x-(3k^2 k 4)=0

you will find the value of k in this equation.


ii)既然你知道k的value,咁你重新寫番條equation

之後就係一條好簡單ge qudratic equation 啦!用factor method ,qudratic formula or even

newton's method to solve it!

希望幫到你啦:)

a) b^2-4ac >0

b)3x^2-2(3k-1)x-(3k^2+ k+ 4)=0
3x^2-6kx+2x-3k^2-k-4=0
3x^2+(-6k+2)x-3k^2-k-4=0
a=3 b=(-6k+2) c=3k^2-k-4

c) find k first n sub back into the equation and put x=4 in the equtaion