Question

Find the set of values of $k$ for which the given equation has and real and equal roots.
$k^2{x}^2-2(k-1)x+4=0$

Answer 1

$k^2{x}^2-2(k-1)x+4=0$

for the roots to be real and equal, condition is, $b^2-4ac=0$

$[-2(k-1){]}^2-4(k^2\left)\right(4)=0$

$4(k-1{)}^2=4(k^2\left)\right(4)$

$(k-1{)}^2=4k^2$

$k^2+1-2k=4k^2$

$3k^2+2k-1=0$

$(k+1)(3k-1)=0$

$k=-1,\frac{1}{3}$