if the roots of the equation (a² +b²)x-2b(a+c)x+(b²+c²)=0 are equal then b² = ??
Answers
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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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