If ad not equal to bc , then prove that the equation (a^2+b^2)x^+2(ac+bd)x+(c^2+d^2)=0 has no real roots
Answers
EXPLANATION: If prisoners B and C had the same color hat on, prisoner A would have know immediately that his hat was the other color (there are only two hats of each color). Since prisoner A was silent, prisoners B and C must have different colored hats. Prisoner B realized this and knew that his hat was not the same color as prisoner C, therefore his hat must be black!
SOLUTION :
Option (b) is correct : ad = bc
Given : (a² + b²)x² - 2(ac + bd)x + ( c² + d²) = 0
On comparing the given equation with ax² + bx + c = 0
Here, a = (a² + b²) , b = - 2( ac + bd) , c = ( c² + d²)
D(discriminant) = b² – 4ac
Given roots are equal so, D = b² - 4ac = 0
{- 2(ac + bd)}² - 4(a² +b²)(c² + d²) = 0
4(ac + bd)² - 4(a² + b²)(c²+ d²) = 0
4(a²c²+ b²d² + 2abcd ) - 4( a²c² + a²d² + b²d² + b²c² = 0
[(a + b)² = a² + b² + 2ab]
4(a²c² + b²d² + 2abcd - a²c² - a²d² - b²d² - b²c² ) = 0
(a²c² - a²c² + b²d² - b²d² + 2abcd - a²d² - b²c² ) = 0
2abcd - a²d² - b²c² = 0
-(a²d² + b²c² - 2abcd) = 0
a²d² + b²c² - 2abcd = 0
(ad)² + (bc)² - 2×ad × bc = 0
(ad - bc)² = 0
[(a - b)² = a² + b² - 2ab]
ad - bc = 0
ad = bc
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