Question

Find the value of k for which the Q.E $k{x}^2+1-2(k-1)x+x^2=0$ has equal roots.

Answer 1

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

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

$\Rightarrow x^2(k+1)-2(k-1)+1=0$

Here It is in the form ob $a{x}^2+bx+c=0$

$a=(k+1)b=-2(k-1)c=1$

for real and equal roots;

$D=0$

$\Rightarrow b^2-4ac=0$

$\Rightarrow [-2(k-1){]}^2-4(k+1).1=0$

$\Rightarrow 4(k^2+1-2k)-4k-4=0$

$\Rightarrow 4k^2+4-8k-4k-4=0$

$\Rightarrow 4k^2-12k=0$

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

$4k=0$ or $k-3=0$

$k=0/4$ $k=3$

$k=0$

$\therefore$ $k=0,3.$