If the
the roots
roots of the equation
(a²+b²)x²+2(bc-ad)x+c²+d²=0 are equal
show that ac+bd=0
Answers
Answer:
the answer
Step-by-step explanation:
Well , we know the condition for real and equal roots.
i.e. D = 0
= > b² - 4 ac = 0
From question we have given :
b = 2 ( b c - a d )
a = a² + b²
c = c² - d²
Now put all value in condition.
= > ( 2 ( b c - a d )² - 4 * ( a² + b² ) ( c² + d² ) = 0
= > 4 b² c² +4 a² d² - 8 b c a d - 4 * ( a² c² + a² d²+ b² c² + b² d² ) = 0
= >4 b² c² +4 a² d² - 8 b c a d - 4 a² c² - 4 a² d² - 4 b² c² - b² d² = 0
= > Clearly 4 b² c² & 4 a² d² cancel out .
= > - 8 b c a d - 4 a² c² - b² d² = 0
= > - 4 ( a² c² + b² d² + 2 a c b d ) = 0
= > ( a c + b d )² = 0
= > a c + b d = 0
Hence proved.
Answer:
ac + bd = 0
Step-by-step explanation:
Given,
Roots of (a²+b²)x²+2(bc-ad)x+c²+d²=0 are equal, which means discriminant is 0.
= > discriminant = 0
= > [ 2( bc - ad ) ]² - 4( a² + b² )( c² + d² ) = 0
= > [ 4( bc - ad )² ] - 4[ a²c² + a²d² + b²c² + b²d² ] = 0
= > 4[ ( b²c² + a²d² - 2abcd ) - ( a²c² + a²d² + b²c² + b²d² ) ] = 0
= > [ b²c² + a²d² - 2abcd - a²c² - a²d² - b²c² - b²d²] = 0
= > [ - 2abcd - a²c² - b²d² ] = 0
= > a²c² + b²d² + 2ac.bd = 0
= > ( ac + bd )² = 0
= > ac + bd = 0
Hence proved.