Question

Find the value of k such that the quadratic equation x(x-2k)+6=0 has real and equal roots

Answer 1

k=
`(x^(2)+6 )/(2x)`

Explanation:

x(x−2k)+6=0

Step 1: Add -x^2 to both sides.

−2kx+x2+6+−x2=0+−x2

−2kx+6=−x2

Step 2: Add -6 to both sides.

−2kx+6+−6=−x2+−6

−2kx=−x2−6

Step 3: Divide both sides by -2x.

−2kx /−2x = −x2−6 /−2x

k =