Question

If a and B are the zeros of the quadratic polynomial f(x) = x+ - p(x + 1) - c, show that
(a + 1) (B + 1) = 1-c.




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Answer 1

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Hey!

___________________

Given,

Alpha and beta are the zeroes of the quadratic polynomial x^2 - p (x + 1) - c

Alpha ( @ )

Beta ( ß )

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

x^2 - px - p - c

x^2 - px - (p+c)

Comparing with ax^2 + bx + c

a = 1

b = - p

c = - (p+c)

We know,

Alpha + Beta = - b/a = - (-p) = p

Alpha × Beta = c/a = - ( p+c )

Thus,

( Alpha + 1 ) ( Beta + 1 )

= @ß + @ + ß + 1

= - (p+c) + p + 1

= - p - c + p + 1

= 1 - c

Thus, ( Alpha + 1 ) ( Beta + 1 ) = 1 - c

___________________

Hope it helps...!!!

Answer 2

Answer:

1

Step-by-step explanation:

ANSWER

(i) Given that alpha and beta are roots of quadratic equation

f(x)=x

2

−p(x+1)−c=x

2

−px−p−c=x

2

−px−(p+c)

Comparing with ax

2

+bx+c,

we have, a=1, b=−p and c=−(p+c)

∴α+β=−b/a=

1

−(−p)

=p and α×β=

a

c

=

1

−(p+c)

=−(p+c)

=(α+1)×(β+1)=(α×β)+α+β+1=−(p+c)+p+1

=−p−c+p+1

=1−c

The given equation is x

2

−p(x+1)−q=0 or x

2

−px−(p+q)=0

Therefore the sum of the roots α+β=p and product of the roots αβ=−(p+q)

Therefore we have,

=

α

2

+2α+q

α

2

+2α+1

+

β

2

+2β+q

β

2

+2β+1

=

α

2

+2α−α−β−αβ

(α+1)

2

+

β

2

+2β−α−β−αβ)

(β+1)

2

(substituting the value of q)

=

(α−β)(α+1)

(α+1)

2

+

(β−α)(β+1)

(β+1)

2

=

α−β

α+1

−

α−β

β+1

=1