If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0
Answers
Step-by-step explanation:
Given:-
The zeroes of polynomial x3 – ax2 + bx – c are in AP.
To find:-
Show that 2a3 – 9ab + 27c = 0
Solution:-
Given cubic polynomial is x^3 - ax^2 + bx - c
Let P(x) = x^3 - ax^2 + bx - c
On comparing with the standard cubic polynomial ax^3 + bx^2 + cx + d then
we have
a = 1
b= -a
c = b
d= -c
Since it is a cubic polynomial it has three zeroes
If they are in the AP then
they are t-d , t , t+d
Sum of the zeroes = -b/a
=> t-d+t+t+d = -(-a)/1
=>3t = a
Therefore, a = 3t ( or) t = a/3 -----(1)
Product of the zeroes = -d/a
=>(t-d)(t)(t+d)=-(-c)/1
=>t(t^2-d^2) = c
=>t^2 - d^2 = c/t -----(2)
=>Sum of the product of the two zeroes taken at time = c/a
=>(t-d)(t)+(t)(t+d)+(t+d)(t-d) = b/1
=>t^2-td +t^2+td+t^2-d^2 = b
=>t^2+t^2+t^2-d^2 = b
=>2t^2+c/t = b
=>(2t^3 +c)/t = b
=>2t^3 + c = bt
=>2(a/3)^3 + c = b(a/3)
=>2(a^3/27) + c = ab/3
=>(2a^3+27c)/27 = ab/3
=>2a^3 + 27c = 27×(ab/3)
=>2a^3 +27c = 9ab
=>2a^3 +27c -9ab = 0
=>2a^3 -9ab +27c = 0
=>RHS
LHS = RHS
Hence, proved
Answer:-
If the zeroes of polynomial x^3 – ax^2 + bx – c are in AP then show that 2a^3 – 9ab + 27c = 0
Used formulae:-
ax^3 +bx^2+cx+d is a cubic polynomial then
- Sum of the zeroes = -b/a
- Product of the zeroes = -d/a
- Sum of the product of the two zeroes taken at time = c/a