Question

find the value of K for which one root of the equation kx square - 14 x + 8 = 0 is 6 times of the other

Answer 1

Solution :

Given quadratic equation kx²-14x+8=0

Compare above equation with

ax²+bx+c=0 we get,

a = k , b = -14 , c = 8

Let One root = m

Second root = 6m

i ) Sum of the tee roots = -b/a

=> m + 6m = - (-14 )/k

=> 7m = 14/k

=> m = 14/7k

=> m = 2/k ---( 1 )

ii ) product of the roots = c/a

=> m × 6m = 8/k

=> 6m² = 8/k

=> 6( 2/k )² = 8/k [ From ( 1 ) ]

=> ( 6 ×4 )/k²= 8/k

=> 24/8 = k²/k

=> 3 = k

Therefore ,

k = 3

•••••