Question

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

Answer 1

Dear student,

Following is the asnwer:

x(x - 2k) + 6 = 0
$$\begin{gathered} x^2-2kx+6=0 \\ ForrealequalrootsD=0 \\ D=b^2-4ac=4k^2-24=0 \\ {k}^2=6 \\ k=\pm \sqrt{6} \end{gathered}$$