Question

Find the value of $k$ for which the following equation has equal roots.
$x^2-2kx+7k-12=0$

Answer 1

Consider the given quadratic equation.

$x^2-2kx+7k-12=0$

We know that, if a quadratic equation $a{x}^2+bx+c=0$ has equal roots, then

$b^2-4ac=0$

Here,

$a=1$

$b=-2k$

$c=7k-12$

Therefore,

$4k^2-4(7k-12)=0$

$4k^2-28k+48=0$

$k^2-7k+12=0$

$(k-3)(k-4)=0$

$k=3,4$

Hence, these are required values of $k$.