Question

If α and β are the zeroes of the quadratic polynomial p(x)=x²-p(x+1)-c such that (α+1)(β+1)=0
WHAT IS THE VALUE OF C?


best answer gets brainliest

Answer 1

x^2-p(x+1)-c

x^2-px-p-c

x^2-px-(p+c)

now comparing this equation with

ax^2+bx+c

we get,

a=1

b=-p

c=-(p+c)

now, alpha+beta=-b/a=

-(-p)/1

alpha+beta=p

alpha*beta=c/a=

-(p+c)/1

alpha*beta=-(p+c)

now, multiply (alpha+1)(beta+1)

we get,

alpha*beta+alpha+beta+1

putting up the value

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

-p-c+p+1=0

-c+1=0

-c=-1

c=1

hope you will get!!

Answer 2

Answer:

(α+1)(β+1)=0

=> αβ + α + β + 1 = 0

We know that:

αβ = c/a

α + β = -b/a

p(x) = x² - p ( x + 1 ) - c = 0

=> p(x) = x² - px - p - c = 0

α + β = p

αβ = - p - c

Put the values:

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

=> c = -1

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