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/
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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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