# Proof using Discriminants

• Apr 24th 2013, 11:40 AM
HelenMc9
Proof using Discriminants

Prove that the equation x² + (a-2d)x + (ab - ad - b²) = 0 has real roots for all values of a, b and d.

Here's what I've done so far (correct me if I've done something wrong btw) ...

a= 1 b = a-2d c= ab - ad - b²
b² - 4ac ≥ 0

so if I'm thinking right, I have to prove the left side is positive? I've been trying different things but I'm stuck ... can somebody help me please? :)
• Apr 24th 2013, 11:57 AM
Plato
Re: Proof using Discriminants
Quote:

Originally Posted by HelenMc9
Prove that the equation x² + (a-2d)x + (ab - ad - b²) = 0 has real roots for all values of a, b and d.
Here's what I've done so far (correct me if I've done something wrong btw) ...
a= 1 b = a-2d c= ab - ad - b²
b² - 4ac ≥ 0

You can show that \$\displaystyle (a-2d)^2-4(ab-ad-b^2)=(a-2b)^2+4d^2~?\$
• Apr 24th 2013, 12:58 PM
HelenMc9
Re: Proof using Discriminants
where does the 4d^2 come from?
• Apr 24th 2013, 01:06 PM
Plato
Re: Proof using Discriminants
Quote:

Originally Posted by HelenMc9
where does the 4d^2 come from?

• Apr 24th 2013, 01:17 PM
HelenMc9
Re: Proof using Discriminants
hmm ... really sorry, but still a bit confused!
How do you get to ' (a - 2d)² - 4(ab - ad -b²) = (a -2b)² + 4d² ' ??
• Apr 24th 2013, 01:23 PM
Plato
Re: Proof using Discriminants
Quote:

Originally Posted by HelenMc9
hmm ... really sorry, but still a bit confused!
How do you get to ' (a - 2d)² - 4(ab - ad -b²) = (a -2b)² + 4d² ' ??

If one cannot do simple algebra, then this question is beyond doing.

• Apr 25th 2013, 10:16 AM
HelenMc9
Re: Proof using Discriminants
Well that's what I needed - someone to help me figure out where I was going wrong with the 'simple algebra'!! I can see now it was quite obvious but for me, a secondary school student who's not particularly mathematically minded, it's sometimes the 'obvious' stuff that catches me out! So thank-you for helping, I understand it now! :)
so now to finish off the question: I can say (a-2b)^2 - 4d^2 is greater than zero because anything squared has to be positive?
• Apr 25th 2013, 02:54 PM
Jesstess123
Re: Proof using Discriminants
Not really if 4d^2 is bigger than (a-2b)^2, then its smaller than zero.
• Apr 25th 2013, 03:08 PM
Plato
Re: Proof using Discriminants
Quote:

Originally Posted by HelenMc9
so now to finish off the question: I can say (a-2b)^2 - 4d^2 is greater than zero because anything squared has to be positive?

Quote:

Originally Posted by Jesstess123
Not really if 4d^2 is bigger than (a-2b)^2, then its smaller than zero.

The point is that you both are incorrect.
Because it is \$\displaystyle (a-2b)^2{~\color{blue}+~}4d^2\$.
The sum of two squares is non-negative.