Question

Find the value of k in which x=3 is a root of equation( (kx)2)+2x-3=0

Answer 1

Substitute x = 3 in given

quadratic equation( (kx)2)+2x-3=0

(k×3)²+2×3 - 3 = 0

=> 9k² + 6 - 3 = 0

=> 9k² + 3 = 0

=> 9k² = -3

=> k² = (-3/9)

=> k² = (-1/3 )

=> k = ± √(-1/3)

Or

if the equation is kx²+2x-3=0

then

k×3² + 2×3 - 3 = 0

=> 9k + 6 - 3 = 0

=> 9k + 3 = 0

=> 9k = -3

=> k = (-3/9)

=> k = -1/3

••••

Answer 2

((k*3)2)+2*3-3=0

((3k)2)+6-3=0

6k+3=0

6k=-3

k=-3/6

k=-1/2