Question

Find the value of k for the quadratic equation kx (x-2) + 6=0 so that it has two equal roots.

Answer 1

Answer:

we know that, when an equation has equal roots the value of b²-4ac = 0

So now we are going to solve this eqn and find the value of k.

Explanation:

given quadratic eqn is

kx(x -2) + 6 =0

( multiplying kx inside the bracket)

kx² -2kx +6=0

from the above eqn

a=k , b= -2k, c= +6

Now,

substituting a,b,c values in the eqn

b² -4ac=0

(-2k)² - 4 (k)(6) =0

4k²-4(k)(6) =0

taking 4 common

4(k²-6k) = 0

sending 4 to that side

k²-6k= 0/4

k²-6k = 0

taking k common

k(k-6)=0

k=0 , k-6=0

k=0 , k=6

k=0 (neglected)(because k cannot be 0)

Verification

6x(x-2)+6=0

6x²-12x+6=0

taking 6 common

6(x²-2x+1)=0

x²-2x+1=0

on factorization

we get

x=1,1 (equal roots)