1
Question
If α, β are roots of x2−p(x+1)−c=0. Show that
i) (α+1)(β+1)=1−c &
ii) α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1
i) (α+1)(β+1)=1−c &
ii) α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1
Open in App
Solution
(i) Given that alpha and beta are roots of quadratic equation
f(x)=x2−p(x+1)−c=x2−px−p−c=x2−px−(p+c)
Comparing with ax2+bx+c,
we have, a=1, b=−p and c=−(p+c)
∴α+β=−b/a=−(−p)1=p and α×β=ca=−(p+c)1=−(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 x2−p(x+1)−q=0
or x2−px−(p+q)=0
Therefore the sum of the roots
α+β=p and product of the roots αβ=−(p+q)
Therefore we have,
=α2+2α+1α2+2α+q+β2+2β+1β2+2β+q
=(α+1)2α2+2α−α−β−αβ+(β+1)2β2+2β−α−β−αβ) (substituting the value of q)
=(α+1)2(α−β)(α+1)+(β+1)2(β−α)(β+1)
=α+1α−β−β+1α−β
=1
(i) Given that alpha and beta are roots of quadratic equation
f(x)=x2−p(x+1)−c=x2−px−p−c=x2−px−(p+c)
Comparing with ax2+bx+c,
we have, a=1, b=−p and c=−(p+c)
∴α+β=−b/a=−(−p)1=p and α×β=ca=−(p+c)1=−(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 x2−p(x+1)−q=0
or x2−px−(p+q)=0
Therefore the sum of the roots
α+β=p and product of the roots αβ=−(p+q)
Therefore we have,
=α2+2α+1α2+2α+q+β2+2β+1β2+2β+q
=(α+1)2α2+2α−α−β−αβ+(β+1)2β2+2β−α−β−αβ) (substituting the value of q)
=(α+1)2(α−β)(α+1)+(β+1)2(β−α)(β+1)
=α+1α−β−β+1α−β
=1
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
