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Help me!!!!!!!!!!!!!!!!!!!!!!!! ( A.maths quadratic functions)

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1#
發表於 3/12/2005 01:45 PM | 只看該作者 回帖獎勵 |倒序瀏覽 |閱讀模式
Let f(x) = x² + kx + ( p - 2k - 4) and g(x) = x² + ( p - 2k -4)x + k
where k >0 and p is a prime no. greater than 3
It is given that the equation f(x) = 0 has two distinct real roots a and b
where a > 0 > b

a) show that (a-2)(b-2) = p.Hence, show that k = p - 3.
I dunno know how to do the red part.
Plz help me. thx
2#
發表於 3/12/2005 11:10 PM | 只看該作者
If p is prime number,
we have
a-2=±1 or b-2=±1
a=3 or a=1 or b=3(rej) or 1(rej)
since a+b=-k
(a-2)(-k-a-2) = p
when a=1 then (1-2)(-k-1-2)=p, k=p-3
when a=3 then (3-2)(-k-3-2)=p, p+k+5=0 (rej)
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3#
 樓主| 發表於 3/12/2005 11:24 PM | 只看該作者
Originally posted by 落雷 at 2005-12-3 11:10 PM:
If p is prime number,
we have
a-2=±1 or b-2=±1
a=3 or a=1 or b=3(rej) or 1(rej)
since a+b=-k
(a-2)(-k-a-2) = p
when a=1 then (1-2)(-k-1-2)=p, k=p-3
when a=3 then (3-2)(-k-3-2)=p, p+k+5=0  ...

原來係咁....
我仲以為個prime no. 冇用...
原來係個hint

thank you very much
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