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?
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?