Reply
Tue 26 Jul, 2005 02:04 am
If a, b and c are real numbers and not all equal, how can I prove that the quadratic equation:
(c-a)x^2 - 2(a-b)x + (b-c) = 0
has unequal real roots?
My exams are approaching and I am doing some practice.
This is a question from my textbook but I don't know how to do it.
The question comes from the section (Quadratic equation & Discriminant)
What I learnt is that when the discriminant > 0, the equation has unequal real roots.
In the above equation, the discriminant is 4(a-b)(a-b) - 4(c-a)(b-c)
but I don't know how to prove this thing > 0
please give me some steps ...
Note: x^2 = x to the power 2
If this is true then
(a-b)^2>(c-a)(b-c)
is true.
multiplying this out, and doing a little rearrangement then
a^2+b^2+c^2>ab+ac+bc
now let c=a+m and c=b+n where m and n are any real number
a^2+b^2+c^2=3c^2+2cm+2cn+m^2+n^2
and
ab+ac+bc=3c^2+2mc+2nc+mn
and
3c^2+2cm+2cn+m^2+n^2>3c^2+2mc+2nc+mn
so
m^2+n^2>mn
now add (-2mn) to both sides
m^2+n^2-2mn>mn-2mn
and
(m-n)^2>-mn
since (m-n)^2 is always positive
this is true for any real m and n
so
(a-b)^2>(c-a)(b-c)
and
4(a-b)^2-4(c-a)(b-c)>0
Rap
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace