if zeroes of polynonial f(x) =x³-3x²+x+1 are (a-b), a, (a+b). Find a&b.
Answers
Answer :
Given :-
Zeroes of polynonial f(x) =x³-3x²+x+1 are (a-b), a, (a+b).
To find :-
Find a&b?
Solution :-
Given Polynomial = f(x) =x³-3x²+x+1
On Comparing this with the standard Cubic Polynomial ax³+bx²+cx+d then
We have ,
a = 1
b=-3
c=1
d=1
Given zeroes are (a-b) , a ,(a+b)
We know that
Sum of the zeroes = -b/a
=> (a-b)+a+(a+b)= -(-3)/1
=> a-b+a+a+b = 3
=> a+a+a = 3
=> 3a = 3
=> a = 3/3
=> a = 1
The value of a = 1 -----------(1)
Product of the zeroes = -d/a
=> (a-b)(a)(a+b) = -1/1
=> (a-b)(a)(a+b) = -1
On Substituting the value of a
From (1)
=> (1-b)(1)(1+b) = -1
=> (1-b)(1+b) = -1
=> 1²-b² = -1
Since (a+b)(a-b) = a²-b²
Where , a = 1 and b=b
=> 1-b² = -1
=> -b² = -1-1
=> -b² = -2
=> b² = 2
=> b = ±√2
The value of b = √2 or -√2
Answer :-
The value of a = 1
The value of b = √2 or -√2
Check :-
If a = 1 and b =√2 then
1+√2 , 1 , 1-√2 are the zeroes then
The Cubic Polynomial is
x³-(1+√2+1+1-√2)x²+(1+√2+1-√2+1-2)x- (1-√2)(1+√2)
=> x³-3x²+x+1
If a = 1 and b =-√2 then
1+√2 , 1 , 1-√2 are the zeroes then
The Cubic Polynomial is
x³-(1+√2+1+1-√2)x²+(1+√2+1-√2+1-2)x- (1-√2)(1+√2)
=> x³-3x²+x+1
Verified the given relations in the given problem.
Used formulae:-
- (a+b)(a-b) = a²-b²
- The standard Cubic Polynomial is ax³+bx²+cx+d.
- Sum of the zeroes = -b/a
- Product of the zeroes = -d/a
- Sum of the product of two zeroes taken at a time = c/a