Question

If the roots of the equation
* (a+b)x2 +2 (bc-ad) x+c+d = 0 are equal,
show that ac + bd = 0.​

Answer 1

Answer:

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.

Step-by-step explanation:

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