Question

Question: 15
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.

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Answer 1

Step-by-step explanation:

x² - p ( x + 1 ) + c

⇒ x² - p x - p + c

⇒ x² - p x + ( c - p )

Comparing with ax² + bx + c, we get :

a = 1

b = - p

c = c - p .

Given  :

( α + 1 )( β + 1 ) = 0

⇒ αβ + α + β + 1 = 0

Note that, sum of roots = - b/a

α + β = - b / a

But b = - p

a = 1

So α + β = - ( - p ) / 1 = p

Product of roots = αβ = c / a

⇒ αβ = ( c - p )

Hence write this as :

αβ + α + β + 1 = 0

⇒ c - p + p + 1 = 0

⇒ c + 1 = 0

⇒ c = -1

Hope it will help you

Answer 2

Polynomial f(x) = x² – p(x + 1) – c

___________ [ GIVEN ]

• We have to show that (α + 1)(β + 1) = 1 – c.

_____________________________

=> f(x) = x² - p(x + 1) - c

=> x² - px - p - c

=> x² - px -(p + c)

Here ..

a = 1

b = - p

c = -(p + c)

Now ..

• Sum of zeros = -b/a

→ α + β = -b/a

→ α + β = -(-p)/1 = + p

• Product of zeros = c/a

→ α β = c/a

→ α β = -(p + c)/1 = -(p + c)

_____________________________

(α + 1)(β + 1) = 1 – c.

Take L.H.S.

=> (α + 1)(β + 1)

=> α(β + 1) + 1(β + 1)

=> αβ + α + β + 1

Put the know values in above equation

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

=> - p - c + p + 1

=> - c + 1

=> 1 - c

L.H.S. = R.H.S.

_______ [ HENCE PROVED ]

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