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

(a-2d)² - 4(1)(ab-ad-b²) ≥ 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? :)

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

(a-2d)² - 4(1)(ab-ad-b²) ≥ 0

You can show that $\displaystyle (a-2d)^2-4(ab-ad-b^2)=(a-2b)^2+4d^2~?$

Re: Proof using Discriminants

where does the 4d^2 come from?

Re: Proof using Discriminants

Quote:

Originally Posted by

**HelenMc9** where does the 4d^2 come from?

Simple algebra: $\displaystyle (a-2d)^2=a^2-4ad+4d^2~.$

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 ' ??

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.

$\displaystyle (a-2d)^2-4(ab-ad-b^2)=a^2-4ad+4d^2-4ab+4ad+4b^2.$

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?

Re: Proof using Discriminants

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

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.