If a,B are the zeros of polynomial f(x)=x^2-p(x+1)-c such that (a+1)(B+1)=0 then c is
Answers
Step-by-step explanation:
Given :-
a,B are the zeros of polynomial f(x)=x^2-p(x+1)-c
such that (a+1)(B+1)=0
To find :-
Find the value of c ?
Solution :-
Given Polynomial = f(x)=x^2-p(x+1)-c
=>f(x) = x^2-px-p-c
=>f(x) = x^2-px -(p+c)
On comparing with the standard quadratic Polynomial ax^2+bx+c
We have
a = 1
b = -p
c = -(p+c)
Given that
f(x) has a ,B are the zeroes then
We know that
Sum of the zeroes = -b/a
=> a+B = -(-p)/1
=> a+B = p -------(1)
and
Product of the zeroes = c/a
=> aB = -(p+c)/1
=> aB = -(p+c)-------(2)
Given that
(a+1)(B+1) = 0
=>a(B+1)+1(B+1) = 0
=> aB + a + B + 1 = 0
=>(aB) + (a+B) + 1 = 0
From (1)&(2)
=> -(p+c) +p +1 = 0
=> -p-c+p+1 = 0
=> (p-p)+1-c = 0
=>0+1-c = 0
=>>1-c = 0
=>1= c
Therefore, c = 1
Answer:-
The value of c for the given problem is 1
Used formulae :-
- The standard quadratic Polynomial ax^2+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a
Answer:
We have,
f(x)=x
2
−p(x+1)−c=0
f(x)=x
2
−px−(p+c)=0 ...(1)
Since,
α,β are the zeroes of the above polynomial.
So,
α+β=p
αβ=−(p+c)
Since,
(α+1)(β+1)=0
αβ+α+β+1=0
−p−c+p+1=0
−c+1=0
c=1
Step-by-step explanation:
Correct answer is -1