F.4 Quardratic Equations

[匿名]

2013-10-01 8:00 pm

1. Suppose k and k-3 are the roots of the quardratic equation x^2 - 3kx + h = 0. fiond the values of h and k.

2.Solve the following equation 3(2x + 3)^2 - 18x = 3(x - 1)^2 + 8

更新1:

Q1. answer: h=18, k=-3

回答 (1)

Stardust

2013-10-01 8:14 pm

✔ 最佳答案

1) k+k-3=-(-3k)/1
2k+3=3k
3=k
k=3 3(3-3)=h/1
0=h
h=0 2) 3(2x + 3)^2 - 18x = 3(x - 1)^2 + 8
3[(2x)²+2(2x)(3)+3²]-18x=3[x²-2(x)(1)+1²]+8
3(4x²+12x+9)-18x=3(x²-2x+1)+8
12x²+36x+27-18x=3x²-6x+3+8
9x²+24x+16=0
(3x+4)(3x+4)=0
x=-4/3

2013-10-01 12:16:10 補充：
Q1用sum of roots & product of roots

2013-10-01 12:51:51 補充：
打錯正負號－　－

1) k+k-3=-(-3k)/1
2k-3=3k
-3=k
k=-3

-3(-3-3)=h/1
-3(-6)=h
18=h
h=18

2013-10-01 12:54:31 補充：
remember：

sum of roots =-b/a
product of roots =c/a

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