Question

Find k, if the roots of x²–(3k–2)x+2k=0 are equal and real.

Answer 1

$\mathbf{ {x}^{2} - ( 3k - 2 ) + 2k = 0}</p><p> \\ \\ \\ \bold{On \: comparing \: the \: given \: equation \: with \: ( a {x}^{2} + bx + c = 0 ), \: we \: get </p><p>} \\ \\ </p><p>a = 1 \\ </p><p>b = - ( 3k - 2 ) \\ </p><p>c = 2k</p><p>$

We know, discriminant = b² - 4ac

=> [ - ( 3k - 2 ) ]² - 4( 1 × 2k )

=> 9k² + 4 - 12k - 8k

=> 9k² + 4 - 20 k

=> 9k² - 20k + 4

=> 9k² - 18k - 2k + 4

=> 9k( k - 2 ) - 2( k - 2 )

=> ( k - 2 ) ( 9k - 2 )

$\bold{We \: know, \: For \: equal \: roots, \: discriminant = 0}$

=> ( k - 2 ) ( 9k - 2 ) = 0

$\mathbf{By \: Zero \: Product \: Rule} \\ \\ \\ k = 2 \: \: \: \: \: or \: \: \: \: k = \frac{2}{9}$

Answer 2

HOPE IT WILL HELP YOU