Question

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2+kx+3=0$
(ii) $kx(x-2)+6=0$

Answer 1

i)

If $2x^2+kx+3=0$ has equal roots, then $b^2-4ac=0$.

Here, $a=2,b=k,c=3$

$x^2+kx+3=0b^2-4ac=0k^2-4\left(2\right)\left(3\right)=0k^2=24k=\pm \sqrt{24}=\pm 2\sqrt{6}$

ii)

If $k{x}^2-2kx+6=0$ has equal roots, then $b^2-4ac=0$.

Here, $a=k,b=-2k,c=6$

$kx(x-2)+6=0k{x}^2-2kx+6=04k^2-4\left(k\right)\left(6\right)=04k^2-24k=04k(k-6)=0$

For $k=0,$ equation is not quadratic.

$k=6$