Question

find the values of k for which the following equations have real and equal roots (k+1)x²-2(k-1)x+1=0

Answer 1

$\bold{\boxed{HEY!!!}}$

Here is your answer :-

Given equation => ( k + 1 )x² - 2( k - 1 )x + 1 = 0

Comparing above equation with standard form

Here , a = k + 1 b = 2( k - 1 ). c = 1

Since the roots are real and equal , descriminant is 0

b² - 4ac = 0

[2( k - 1 )]² - 4( k + 1 )( 1 ) = 0

4( k² - 2k + 1 ) - 4k - 4 = 0

4k² - 8k + 4 - 4k - 4 = 0

4k² - 12k = 0

4k ( k - 3 ) = 0

4k = 0 or. k - 3 = 0

k = 0. or. k = 3

$\bold{\boxed{\:K = 0 \:or \:K = 3 }}$

$\bold{\boxed{THANKS...}}$

Answer 2

( K + 1 )X² - 2 ( K - 1 )X + 1 = 0



Here,


a = k + 1 , b = -2(k-1 ) = -2k + 2 and c = 1



Discriminant ( D ) = 0


b² - 4ac = 0


( -2k + 2 )² - 4 × (k+1) × 1 = 0



(-2k)² + (2)² + 2•-2k•2 - 4k - 4 = 0



4k² + 4 - 8k - 4k - 4 = 0



4k² - 12k = 0



4k ( k-3)=0

4k = 0 or ( k-3) = 0


k = 0 or k = 3