If the zeroes of a polynomial f(x) = ax3 +3bx2 3cx+d are in A.P. prove that 2b3-3abc+a2d=0
Answers
Answered by
384
Let the roots be p,q, and r.
We know that
1. p.q.r = -d/a
2. p+q+r = -3b/a
3. 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/a
3p=-3b/a
p=-b/a
(p-m)(p)(p+m)=-d/a
p.(p²-m²)=-d/a
since p=-b/a, we get p²-m²=d/b
Applying the third condition:
p(p-m)+p(p+m)+(p-m)(p+m)=3c/a
2p²+(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
We know that
1. p.q.r = -d/a
2. p+q+r = -3b/a
3. 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/a
3p=-3b/a
p=-b/a
(p-m)(p)(p+m)=-d/a
p.(p²-m²)=-d/a
since p=-b/a, we get p²-m²=d/b
Applying the third condition:
p(p-m)+p(p+m)+(p-m)(p+m)=3c/a
2p²+(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
Answered by
15
Answer:give me brainliest answer
Step-by-step explanation:
Attachments:
Similar questions