Question

Find the value(s) of $k$ if the equation $(k+1)x^2-2(k-1)x+1=0$ has real and equal roots.

• A. $k=0$
• B. $k=2$
• C. $k=3$
• D. $k=-2$

Answer 1

Answer: (A), (C)

The correct options are
A $k=3$
C $k=0$

The given equation is
$(k+1)x^2-2(k-1)x+1=0$
comparing it with $a{x}^2+bx+c=0$ we get
$a=(k+1),b=-2(k-1)$ and $c=1$
$\therefore$ Discriminant,
$D=b^2-4ac=4(k-1{)}^2-4(k+1)\times 1$
$=4(k^2-2k+1)-4k-4$
$\Rightarrow 4k^2-8k+4-4k-4=4k^2-12k$
Since roots are real and equal, so
$D=0\Rightarrow 4k^2-12k=0\Rightarrow 4k(k-3)=0$
$\Rightarrow$ either $k=0$ or $k-3=0\Rightarrow 4k(k-3)=0$
Hence, $k=0,3.$