Question

if x= root p+q + root p-q / root p+q - root p-q then prove that qx^2-2px + q = 0

Answer 1

Answer:

0

Step-by-step explanation:

Property: If    

b

a

​

=  

d

c

​

  then,

a−b

a+b

​

=  

c−d

c+d

​

 

Now,

1

x

​

=  

p+q

​

−  

p−q

​

 

p+q

​

+  

p−q

​

 

​

 

Using the property,

x−1

x+1

​

=  

p+q

​

+  

p−q

​

−  

p+q

​

+  

p−q

​

 

p+q

​

+  

p−q

​

+  

p+q

​

−  

p−q

​

 

​

 

x−1

x+1

​

=  

p−q

​

 

p+q

​

 

​

 

Squaring both sides,

(  

x−1

x+1

​

)  

2

=  

p−q

p+q

​

 

(x  

2

+1)−2x

(x  

2

+1)+2x

​

=  

p−q

p+q

​

 

⇒  

2x

x  

2

+1

​

=  

q

p

​

 

(x  

2

+1)q−2px=0

qx  

2

+q−2px=0