If 3 and (-2) are the zeroes of the polynomial x²+(a+1)x-b, find the value of a and b
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Step-by-step explanation:
Given:-
3 and (-2) are the zeroes of the polynomial x^2+(a+1)x-b.
To find:-
Find the values of a and b ?
Solution:-
Given Polynomial = x^2+(a+1)x-b.
On Comparing this with the standard quadratic Polynomial ax^2+bx+c
a = 1
b= (a+1)
c = -b
Given zeroes are 3 and -2
We know that
Sum of the zeroes (α+β)= -b/a
=> 3+(-2) = - (a+1)/1
=> 3-2 = -(a+1)
=> 1= -a-1
=> -a-1 = 1
=> -a = 1+1
=> -a = 2
=> a = -2
The value of a = -2
and
Product of the zeroes (αβ) = c/a
=> 3×(-2) = -b/1
=> -6 = -b
=> -b = -6
=> b = 6
The Value of b = 6
Answer:-
The value of a = -2
The Value of b = 6
Used formulae:-
- The standard quadratic Polynomial ax^2+bx+c
- Sum of the zeroes (α+β)= -b/a
- Product of the zeroes (αβ) = c/a
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