Question

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

Answer 1

$\left(ii\right)kx(x-2)+6=0$
$ork{x}^2-2kx+6=0$
Comparing this equation with $a{x}^2+bx+c=0$, we get
a = k, b = - 2k and c = 6
$Discriminant=b^2-4ac$
$=(-2k{)}^2-4(k\left)\right(6)$
$=4k^2-24k$
For equal roots,
$b^2-4ac=0$
$4k^2-24k=0$
$4k(k-6)=0$
$Either4k=0ork=6$
However, if k = 0, then the equation will not have the terms '$x^2$' and 'x'. Therefore, for this equation to has two equal roots, k should be 6.