Question

The roots of the equation ax²-bx+(c+k)=0 are reciprocal to each other, then find k.​

Answer 1

Answer:

The roots are reciprocal to each other.therfore

Let P and 1 / P be the roots of the equation ax²-bx+(c+k)=0

P× 1/ P = c+k / a

1 = c+k / a

c+k = a

k = a-c

Answer 2

Answer:

Step-by-step explanation:

Given equation is ax² - bx + ( c + k ) = 0

Comparing the given equation with ax²+ bx + c = 0,

we have a = a , b = - b and c = ( c + k ).

Sum of the zeroes = - b / a = ( - ( - b ) / a = b / a. And

Product of tthe zeroes = c / a = ( c + k ) / a.

∴ ( b / 2a - x ) ( b / 2a + x ) = ( c +k ) / a ( ∵ b / a × 1/ 2 = b / 2a )

⇒ ( b / 2a )² - x² = ( c + k ) / a

⇒( b² / 4a² ) - x² = ( c + k ) / a

⇒x² = ( b² / 4a² ) - ( c + k ) / a

⇒ x² = [ b² - 4a (c + k ) ] / 4a²

⇒ x = ± √ [ b² - 4a ( c + k ) / 4a²

⇒ x = ± √[ b² - 4a ( c + k )] / 2a

Given that, the equation of the roots are reciprocal to each other.

So, √[b² - 4a ( c + k ) ] / 2a = 2a / - √[ b ² - 4a (c + k ) ]

⇒ [ √[ b² - 4a(c + k )]² = 4a²

⇒ b² - 4a ( c + k ) = 4a²

⇒b² - 4ac - 4ak = 4a²

⇒4ak = b² - 4a² - 4ac

⇒k = ( b² - 4a² - 4ac ) / 4a is the answer.