Question

Find the value of k if x= 1÷2-√3 is azero of polynomial 2kx^2 - 7x+ k

Answer 1

Hi ,

x = 1/( 2 - √3 )

rationalize the denominator ,

x = ( 2 + √3 ) / ( 2 - √3 ) ( 2 + √3 )

x = ( 2 + √3 ) / [ 2² - ( √3 )² ]

x = ( 2 + √3 ) / ( 4 - 3 )

x = 2+ √3 -----( 1 )

according to the problem given,

let p( x ) = 2kx² -7x +k

If x = 2 + √3 is a zero of the polynomial then

p( x ) = 0

p( 2 + √3 ) = 0

2k ( 2 + √3 )² -7(2+√3) + k = 0

2k[ 2² + (√3)² + 2×2×√3 ] -14 - 7√3 + k =0

2k( 7 + 4√3 ) - 14 - 8√3 + k = 0

k(14 + 8√3 +1 ) = 14 + 8√3

k = ( 14 + 8√3 )/(15+8√3)

I hope this helps you.

:)