Question

If the roots of the equations (a²+b²)x²-2b(a+c)x+(b²+c²)=0 are equal, then (a)2b = a + c (b)b² = ac (c)$b= \frac{2ac}{a+c}$ (d)b = ac​

Answer 1

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Answer 2

SOLUTION :  

Option (b) is correct :  b² = ac

Given : (a² + b²)x² - 2b(a + c )x + ( b² + c²) = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = (a² + b²) , b = - 2b( a + c)  , c = ( b² + c²)

D(discriminant) = b² – 4ac

= {- 2b(a + c)}² - 4(a² +b²)(b² + c²)  

= 4b²(a² + c² + 2ac)² - 4(a² + b²)(b²+ c²)  

[(a + b)² = a² + b² + 2ab]

= 4b²(a² + c² + 2ac)² - 4( a²b² + a²c² + b⁴ +  b²c² )

= 4(a²b² + b²c² + 2ab²c  - a²b² -  a²c² - b⁴ -  b²c² )  

= (a²b² - a²b² + b²c² - b²c² + 2ab²c  -  a²c² -  b⁴ )  

= 2ab²c  -  a²c² -  b⁴  

= -(a²c² + b⁴ - 2ab²c)  

= a²c² + b⁴ - 2ab²c  

= (ac)² + (b²)² - 2×ac × b²

= (ac - b²)²  

[(a - b)² = a² + b² - 2ab]

Given roots are equal so, D =  0

(ac - b²)² =  0

(ac - b²) =   0

ac = b²

b² = ac

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