Math, asked by pvpavan9, 8 months ago

if a, b are the roots of x^2-p(x-1), show that (a-1)(b-1)=1-c. Hence prove that
a^2+2a+1/​

Answers

Answered by kalpanagoyal903
0

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

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