證明二次方程....

Consider the discriminant of kx^2-x-k=0

i.e.

D= (-1)^2 - 4(k)(-k) = 1+4k^2 > 0

So , the equation has two distinct real roots.

問題:已知k為一非零實數......證明二次方程kx^2-x-k=0有兩個不等實根...

解:
∆=(-1)-4(k)(-k)
=1+4k
>0
因此,方程有兩個不等實根。

2008-02-28 22:48:14 補充：
∆=(-1)-4(k)(-k)

=1+4k^2

>0

因此,方程有兩個不等實根。

If kx-x-k=0 has two real roots, delta > 0
Delta = b-4ac
= (-1)-4(k)(-k)
= 1+4k
As k is a non-zero real number, k must be positive, therefore 1+4k is also positive
i.e. delta > 0 which means kx-x-k = 0 has two real roots.

2008-02-28 12:18:15 補充：
Pls. amend the last sentence as follows:


i.e. delta > 0 which means kx²-x-k = 0 has two different real roots.