(1)
In the following quadratic equations, k represents a non-zero constant. State whether they have " no real root ", " two equal real roots" or " two distinct real roots".
(a) x^2 - kx + k - 2 = 0
(b) k(x^2) - 6x - 2k = 0
(2)
In a party, each person shook hands with everyone else. If the total number of handshakes was 780, how many people were there in the party?
f4 maths -
2007-07-03 5:42 am
回答 (4)
2007-07-04 12:41 am
✔ 最佳答案
1a) △ = b^2 - 4ac= ( - k )^2 - 4 ( 1 )( k - 2 )
= k^2-4k+8
=k^2-4k+4+4
=(k^2-4k+4)+4
=(k-2)^2+4
( k - 2 )^2 ≧ 0 , so ( k - 2 )^2 + 4 > 0
Therefore,the equation has two distinct real roots.
1b) △ = b^2 - 4ac
= ( - 6 )^2 - 4 ( k )( -2k)
= 36+8k^2
8k^2 ≧ 0 , so 36 + 8k^2 > 0
Therefore,the equation has two distinct real roots.
2)Let the total numbers of people be x
since one person's total number of handshakes with others is (x-1)
so the total number of handshakes is
(x(x-1))/2
(because A handshakes with B is equal to B handshakes with A...So the number is
divided by 2)
(x(x-1))/2 =780
x^2-x=1560
x^2-x-1560=0
(x-40)(x+39)=0
x=40 or x=-39(rejected)
Therefore ,the total numbers of people is 40
chocolate328154 is wrong because no people will use left hand handshaking
with one person ,then use right hand with him again...
So you should count the number of handshakes but no need to find whether he uses
left or right hand...
2007-07-08 8:18 am
(2)Consider:
No, of person no, of handshakes
2 1
3 3 ( AB, AC, BC)
4 6 (AB,AC,AD.BC.BD.CD)
. .
. .
. .
n n(n-1)/2 ( ie the (n-1)th triangular number)
Hence, we can set up the equation n(n-1)/2 = 780
n(n-1) = 1560
n = 40 ( n > 0)
There are 40 people in the party.
No, of person no, of handshakes
2 1
3 3 ( AB, AC, BC)
4 6 (AB,AC,AD.BC.BD.CD)
. .
. .
. .
n n(n-1)/2 ( ie the (n-1)th triangular number)
Hence, we can set up the equation n(n-1)/2 = 780
n(n-1) = 1560
n = 40 ( n > 0)
There are 40 people in the party.
2007-07-03 6:12 am
I'm afraid the above answers have some problems.
( 1 a ) △ = b^2 - 4ac
= ( - k )^2 - 4 ( 1 )( k - 2 )
= k^2 - 4k + 8
= k^2 - 4k + 4 + 4
= ( k - 2 )^2 + 4 ( by completeing square )
As you can see, ( k - 2 )^2 ≧ 0 , so ( k - 2 )^2 + 4 > 0
Hence △ > 0 for all real values of k and the equation should have two distinct real roots.
b ) △ = ( - 6 )^2 - 4 ( k )( - 2k )
= 36 + 8k^2
Again, 8k^2 ≧ 0 , so 36 + 8k^2 > 0
Hence △ > 0 for all real values of k and the equation should have two distinct real roots.
2 ) Let the number of people be n.
As you know, the total number of hands is 2n but the total number of handshakes of each person should be (2n - 1).
Just imagine you're one of them and you shake others' hands with your right hand, surely you will not shake your right hand with your left hand!
So,
n ( 2n - 1 ) = 780
2n^2 - n - 780 = 0
( 2n + 39 )( n - 20 ) = 0
n = -39 / 2 ( rejected ) or n = 20
Therefore the total number of people is 20.
( 1 a ) △ = b^2 - 4ac
= ( - k )^2 - 4 ( 1 )( k - 2 )
= k^2 - 4k + 8
= k^2 - 4k + 4 + 4
= ( k - 2 )^2 + 4 ( by completeing square )
As you can see, ( k - 2 )^2 ≧ 0 , so ( k - 2 )^2 + 4 > 0
Hence △ > 0 for all real values of k and the equation should have two distinct real roots.
b ) △ = ( - 6 )^2 - 4 ( k )( - 2k )
= 36 + 8k^2
Again, 8k^2 ≧ 0 , so 36 + 8k^2 > 0
Hence △ > 0 for all real values of k and the equation should have two distinct real roots.
2 ) Let the number of people be n.
As you know, the total number of hands is 2n but the total number of handshakes of each person should be (2n - 1).
Just imagine you're one of them and you shake others' hands with your right hand, surely you will not shake your right hand with your left hand!
So,
n ( 2n - 1 ) = 780
2n^2 - n - 780 = 0
( 2n + 39 )( n - 20 ) = 0
n = -39 / 2 ( rejected ) or n = 20
Therefore the total number of people is 20.
參考: My Maths Knowledge
2007-07-03 5:52 am
b^2 - 4(a)(c) >/=/< 0
<<你識呢條式嫁啦?
so,
(a) (-k)^2 -4(1)(k-2) = k^2 -4(k-2) = k^2 -4k+8
k= no real roots
(b) 你應該識嫁啦...唔想打呀,好麻煩-_-
需要再問我啦
第2題唔識orz
死啦,考完個會考乜都唔記得
<<你識呢條式嫁啦?
so,
(a) (-k)^2 -4(1)(k-2) = k^2 -4(k-2) = k^2 -4k+8
k= no real roots
(b) 你應該識嫁啦...唔想打呀,好麻煩-_-
需要再問我啦
第2題唔識orz
死啦,考完個會考乜都唔記得
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