Real Roots

• Nov 12th 2007, 05:18 AM
PvtBillPilgrim
Real Roots
I'm stuck on this problem. Let a and b be fixed non-zero real numbers and let f(x) = 2(b^2)x^3 + abx^2 - [(a^2) + (3b^2) + 1] - 3ab. Prove that f(x) = 0 has three real roots and one of these roots must lie in the interval [-1,1]. Thank you for your help.
• Nov 12th 2007, 05:49 AM
janvdl
Quote:

Originally Posted by PvtBillPilgrim
I'm stuck on this problem. Let a and b be fixed non-zero real numbers and let f(x) = 2(b^2)x^3 + abx^2 - [(a^2) + (3b^2) + 1] - 3ab. Prove that f(x) = 0 has three real roots and one of these roots must lie in the interval [-1,1]. Thank you for your help.

I've factorised it to this:

\$\displaystyle b^2 (2x^3 - 3) + ab(x^2 - 3) - a^2 - 1 = 0 \$
• Nov 12th 2007, 09:03 AM
PvtBillPilgrim
Thank you very much for the response, but I'm not really sure what you're doing. Could you explain it a little bit further? What are you doing with the delta? How do you reach the conclusions?
• Nov 12th 2007, 09:43 AM
ThePerfectHacker
You can show that f(0)>0 and f(-1)<0. So there is a zero on [-1,1].
• Nov 12th 2007, 03:06 PM
PvtBillPilgrim
What exactly is this "Delta" you're referring to and how did you pick a,b,c in b^2 - 4ac?
• Nov 12th 2007, 03:28 PM
PvtBillPilgrim
I'm pretty sure I know how to prove that one root lies in the interval [-1,1]. I find f(-1) and f(1) and then use the Intermediate Value Theorem since f(-1) is less than zero and f(1) is greater than zero. However, how do I show that there are three roots? Can someone just show me this?

By the way, there should be an x after the brackets in my original function f(x).
• Nov 12th 2007, 04:45 PM
ThePerfectHacker
You can try to graph the function to see if it intersects x-axis three times. But since we do not know exactly the values of a,b it is not that easy to do. However, we can approximate how it looks like. For example, find f'(x) and set it as zero so x=-1/3. That is a relative max/min which is inside the interval. Now try the argument that says it is negative or something like that.
• Nov 12th 2007, 09:24 PM
janvdl
Quote:

Originally Posted by PvtBillPilgrim
What exactly is this "Delta" you're referring to and how did you pick a,b,c in b^2 - 4ac?

Delta is how we determine the type of roots.

\$\displaystyle Delta = b^2 - 4ac\$

The b, a, and c refers to: \$\displaystyle ax^2 + bx + c\$

If Delta is:
1. Greater than zero and a perfect square, then the roots are Real, Rational, and not equal.

2. Greater than zero but not a perfect square, then the roots are Real, Irrational, and not equal.

3. Equal to zero. The roots are Real, Rational, and Equal.

4. Smaller than zero. The roots do not exist.
• Nov 13th 2007, 06:41 AM
ThePerfectHacker
Quote:

Originally Posted by janvdl
Delta is how we determine the type of roots.

....

But we are not solving for a,b,c we are solving for x. Those coefficients are already given to us.
• Nov 13th 2007, 07:27 AM
janvdl
Quote:

Originally Posted by ThePerfectHacker
But we are not solving for a,b,c we are solving for x. Those coefficients are already given to us.

Heck yes, you're right... What a major mistake i made there! :eek:

I'm scaring myself... I need to start thinking properly...