Question

(a – b)^2x^2 + 2 (a^2 – b^2 ) x + (a + b)^2 = 0 solve for x.Warning: iit jee level.

Answer 1

Answer:The roots of (a  

2

+b  

2

)x  

2

−2b(a+c)x+(b  

2

+c  

2

)=0 are equal.

i.e. D=b  

2

−4ac=0

⇒4b  

2

(a+c)  

2

=4(a  

2

+b  

2

)(b  

2

+c  

2

)

⇒b  

2

(a  

2

+c  

2

+2ac)=a  

2

b  

2

+a  

2

c  

2

+b  

4

+b  

2

c  

2

 

⇒b  

2

a  

2

+b  

2

a  

2

+2ab  

2

c=a  

2

b  

2

+a  

2

c  

2

+b  

4

+b  

2

c  

2

 

⇒b  

4

−2b  

2

ac+a  

2

c  

2

=(b  

2

−2ac)  

2

=0

∴b  

2

=ac

i.e. a,b,c are in GP.

Step-by-step explanation: