Question

If the roots of $(k+1)x^2-2(k+1)x+1=0$ are real and equal, then find the value of $k$.

Answer 1

Step 1: Find the discriminant of given equation.

Given: The roots of $(k+1)x^2-2(k+1)x+1=0$ are real and equal.

We know that discriminant of a quadratic equation $a{x}^2+bx+c=0$ is $\left(D\right)=b^2-4ac$

On comparing the given equation with the general quadratic equation, we get;

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

So, discriminant of given equation is ${[-2(k+1\left)\right]}^2-4(k+1)\left(1\right)$

Step 2: Find the value of $k$.

As the roots of the equation are real and equal.

Therefore, discriminant $\left(D\right)=0$

$$\begin{gathered} \Rightarrow {[-2(k+1\left)\right]}^2-4(k+1)\left(1\right)=0 \\ \Rightarrow 4{(k+1)}^2-4(k+1)=0 \\ \Rightarrow 4(k+1)[k+1-1]=0 \\ \Rightarrow (k+1)k=0 \\ \Rightarrow k=0\mathrm{or}-1 \end{gathered}$$

Hence the value of $k$ is $0$ or $-1$.