Question

show that : alpha^2+2alpha+1/alpha^2+2alpha+c + beta^2+2beta+1/beta^2+2beta+c = 1 ​

Answer 1

Answer:

Step-by-step explanation:

(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