Question

If alfa,beta are the zeroes of the polynomial p(x) = x2 – p(x+1) – c such that (alfa+ 1) (bita+ 1) = 0. What is the value of c?​

Answer 1

Step-by-step explanation:

Given :-

α, β are the zeroes of the polynomial

p(x) = x²-p(x+1)-c such that (α+ 1)(β+ 1) = 0

To find :-

What is the value of c?

Solution :-

Given quadratic polynomial is p(x) = x²-p(x+1) - c

=> p(x) = x²-px-p-c

=> p(x) = x²-px-(p+c)

On comparing this with the standard quadratic polynomial ax²+bx+c

a = 1

b = -p

c = -(p+c)

We know that

Sum of the zeroes = -b/a

=> α+ β = -b/a

=> -(-p)/1

=> p

α+ β = p ---------------(1)

Product of the zeroes = c/a

=> αβ = -(p+c)/1

=> αβ = -(p+c) ---------(2)

We have ,

(1+α)(1+ β) = 0

=> 1+α+ β+ αβ = 0

=> 1+p+(-(p+c)) = 0

=> 1+p-p-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 is ax²+bx+c
• Sum of the zeroes = -b/a
• Product of the zeroes = c/a