Lthe zeroes of the polynomial f(x) = axt36x2+3 c&td are in AP. show th

1.	Lthe zeroes of the polynomial f(x) = axt36x2+3 c&td are in AP. show that 26-Bana 2020.
Answer» the correct answer was 485 Let the roots be p, q and r.We know that 1.p.q.r = -d/a2. p+q+r = -3b/a3. pq+qr+Rp = 3c/a Since the roots are in AP, we can assume the roots to p-m, p and p+m (where m is the common difference)∴ p-m+p+p+m = -3b/a3p = -3b/ap = -b/a (p-m)(p)(p+m) = -d/ap.(p²-m²) = -d/asince p = -b/a, we get p²-m² = d/b Applying the third condition:p(p-m)+p(p+m)+(p-m)(p+m) = 3c/a2p²+(p²-m²) = 3c/a substituting for p and p²-m² 2b²/a² + d/b = 3c/a Simplifying this gives:2b³+a²d = 3abc 2b³-3abc+a²d = 0

Lthe zeroes of the polynomial f(x) = axt36x2+3 c&td are in AP. show that 26-Bana 2020.

Answer»

the correct answer was 485

Let the roots be p, q and r.We know that 1.p.q.r = -d/a2. p+q+r = -3b/a3. pq+qr+Rp = 3c/a

Since the roots are in AP, we can assume the roots to p-m, p and p+m (where m is the common difference)∴ p-m+p+p+m = -3b/a3p = -3b/ap = -b/a

(p-m)(p)(p+m) = -d/ap.(p²-m²) = -d/asince p = -b/a, we get p²-m² = d/b

Applying the third condition:p(p-m)+p(p+m)+(p-m)(p+m) = 3c/a2p²+(p²-m²) = 3c/a

substituting for p and p²-m²

2b²/a² + d/b = 3c/a

Simplifying this gives:2b³+a²d = 3abc

2b³-3abc+a²d = 0