maths @!!! ergent

a) Do u mean x [ ( 2k - 1 )x + 3 ]( 3k + 1 ) + 9 = 0?

Rearranging,

( 2k - 1 )( 3k + 1 )x^2 + 3 ( 3k + 1 )x + 9 = 0

△ = 0 for the equation has 2 equal roots, so

[ 3 ( 3k + 1 )]^2 - 4 ( 2k - 1 )( 3k + 1 )( 9 ) = 0

( 3k + 1 )^2 - 4 ( 2k - 1 )( 3k + 1 ) = 0

( 3k + 1 )[ ( 3k + 1 ) - 4 ( 2k - 1 )] = 0

( 3k + 1 )( 3k + 1 - 8k + 4 ) = 0

( 3k + 1 )( -5k + 5 ) = 0

k = - 1 / 3 ( rejected ) or k = 1

k = -1 / 3 has to be rejected because the coefficient of x^2, ( 2k - 1 )( 3k + 1 ) will become 0 and this will not be a quadratic equation.

b) Then,

( 2 - 1 )( 3 + 1 )x^2 + 3 ( 3 + 1 )x + 9 = 0

4x^2 + 12x + 9 = 0

( 2x + 3 )^2 = 0

x = - 3 / 2

k = 1
x = - 3 / 2