polynomial

Leong13

show that the roots of the quadratic equation
ax2+bx+c=0 are given by X=-b+/-[b2-4ac]/2a(formula).Deduce that both roots are real b2-ac>0 and both roots are complex if b2-4ac< 0 .Find all possible values k such that the equation x2-(k-3)x+k2+2k+5= 0 has real roots.If A and B are real roots of this quadratic equation,show that A2+B2=-(k+5)2+24.Hence,find the maximum value .

蝎杰

对不起，借用楼住的贴！！
有几题polynomial不会做！！

if 2y-5x+1<0 and 3y-2x-4>0 , show that x>1
这种问题到底是用什么原理解答呢？？
我没头绪....

prove that without refering to mathematical tables,that
a)2<(5^.5)<3 and deduce that 0<(5^.5)<1
b)[(5^.5)+2]^4+[(5^.5)-2]^4=322
c)321<[(5^.5)+2]^4<322
这题的a 我会做，不过接下来的，我就不会做了....
有谁可以告诉我怎么做吗？？

dunwan2tellu

if 2y-5x+1<0 and 3y-2x-4>0 , show that x>1
这种问题到底是用什么原理解答呢？？
我没头绪....

解方程组 2y-5x+1=0 和 3y-2x-4=0 得 x=1 , y=2 . 带入 y=2 进任何不等试 得 x>1 .

dunwan2tellu

prove that without refering to mathematical tables,that
a)2<(5^.5)<3 and deduce that 0<(5^.5)-2<1
b)[(5^.5)+2]^4+[(5^.5)-2]^4=322
c)321<[(5^.5)+2]^4<322

a) 的少了东西-2

b)设a=5^.5 ,
(a+2)^4 + (a-2)^4 = (a^2 +4a+4)^2 + (a^2-4a+4)^2
=(9+4a)^2 + (9-4a)^2 = 2(81+16a^2)=2(81+80)=322

c)
从a和b得知 321=322-1<322-[(5^.5)-2]^4 <322

[(5^.5)+2]^4 = 322-[(5^.5)-2]^4

321<[(5^.5)+2]^4 <322   QED

[ 本帖最后由 dunwan2tellu 于 20-8-2005 04:35 PM 编辑 ]

dunwan2tellu

show that the roots of the quadratic equation
ax2+bx+c=0 are given by X=-b+/-[b2-4ac]/2a(formula).Deduce that both roots are real b2-ac>0 and both roots are complex if b2-4ac< 0 .Find all possible values k such that the equation x2-(k-3)x+k2+2k+5= 0 has real roots.If A and B are real roots of this quadratic equation,show that A2+B2=-(k+5)2+24.Hence,find the maximum value .

顺手解了第一题

前部分就省略了..

x2-(k-3)x+k2+2k+5= 0 有实根，所以
判别试＝(k-3)^2-4(k2+2k+5)=-(3k+11)(k+1)>0 解得

-11/3 < k <-1

A2+B2=(A+B)^2 - 2AB = (k-3)^2 - 2(k2+2k+5)=-k2-10k-1
=-(k+5)^2+24  

To maximise , k+5 = minimum

from 4/3 < k+5 < 4 , k_min = 4/3 so mx = -(4/3)^2+24 = 200/9 when k=-11/3

[ 本帖最后由 dunwan2tellu 于 20-8-2005 04:47 PM 编辑 ]

蝎杰

抱歉，又要麻烦大家了....这题又怎么解呢？？
b)show that,if n is a positive integer,then the value of (n+2)n is between n^2 and (n+1)^2
c)if n is a positive integer,use your results from a)and b) to find 2 consecutive integers,in terms of n,such that
   
(n+2)^.5+n^.5
(n+2)^.5-n^.5

lies between these 2 integers.

*the result of a)is a+b+2(ab)^.5
                     a-b

無聊人

回第1帖：formula题，用completing the square 方法

dunwan2tellu

b)show that,if n is a positive integer,then the value of (n+2)n is between n^2 and (n+1)^2
c)if n is a positive integer,use your results from a)and b) to find 2 consecutive integers,in terms of n,such that
   
(n+2)^.5+n^.5
(n+2)^.5-n^.5

lies between these 2 integers.

*the result of a)is a+b+2(ab)^.5
                     a-b

b)n^2<n(n+2)=n^2 +2n <n^2+2n+1=(n+1)^2

c)由a)的result,或称为有理化分母，可得

n+1+sqrt{n(n+1)}

再用b)的result看得出

n+1+n=2n+1<n+1+sqrt{n(n+1)}<n+1+(n+1)=2n+2

蝎杰

？？？？不明白，一点都不明白b)和c)的解答，可以再更详细一点吗？

dunwan2tellu

b) n(n+2)=n2 +2n < n2 +2n +1 = (n+1)^2 ------(i)

   n(n+2)=n2 + 2n > n2  -------(ii)

(i),(ii)--> n2 < n(n+2) <(n+1)^2

c)
(sqrt[n+2]+sqrt[n])/(sqrt[n+2]-sqrt[n])

=(sqrt[n+2]+sqrt[n])(sqrt[n+2]+sqrt[n])/(sqrt[n+2]-sqrt[n])(sqrt[n+2]+sqrt[n])

=n+1+sqrt[n(n+2)]

But from b) n2 < n(n+2) <(n+1)^2  ---> n < sqrt[n(n+2)] < n+1

So  
n+1+sqrt[n(n+2)] < n+1 + n+1 = 2n+2 ------(i)
n+1+sqrt[n(n+2)] > n+1 n = 2n+1 ------(ii)

(i),(ii) --> 2n+1 < n+1+sqrt[n(n+2)] < 2n+2