If alpha and Beta are the zeros of the qp
f(x) = x2 -p(x+1)-c , show that (alpha +1)
(ß + 1) =1-c
Answers
Step-by-step explanation:
Given :-
α,β are the zeores of the quadratic polynomial
f(x) = x²-p(x+1)-c
To find :-
Show that (α+1)(β+1) = 1-c.
Solution :-
Given quadratic polynomial f(x) = x²-p(x+1)-c
=> f(x) = x²-px-p-c
=> f(x) = x² - px - (p+c)
On comparing with the standard quadratic polynomial ax²+bx+c
We have ,
a = 1
b = -p
c = -(p+c)
Given zeroes are α and β
We know that
Sum of the zeroes = -b/a
=>α+β = -(-p)/1
=> α+β = p ------------------(1)
And
Product of the zeroes = c/a
=> α×β = -(p+c)/1
=> α×β = -(p+c)
=> αβ = -(p+c) ----------------(2)
Now,
The value of (α+1)(β+1)
=> α(β+1) + 1(β+1)
=> αβ+α + β+1
=> αβ+(α+β)+1
From (1) and (2)
On Substituting the values in the above formula then
=>-(p+c) +p + 1
=> -p-c+p+1
=>(-p+p)+1-c
=> 0+1-c
=> 1-c
=> αβ+(α+β)+1 = 1-c
=> (α+1)(β+1) = 1-c.
Hence, Proved.
Answer:-
If α,β are the zeores of the quadratic polynomial
f(x) = x²-p(x+1)-c then (α+1)(β+1) = 1-c.
Used formulae:-
- The standard quadratic polynomial is ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a