Question

Find the value of $k$ if the equation $x^2-2\left(k+1\right)x+k^2=0$ has equal roots?

Answer 1

Step 1: Determine the discriminant $\left(D\right)$ of the quadratic equation

Given that, equation $x^2-2\left(k+1\right)x+k^2=0$ has real and equal roots.

The discriminant of the quadratic equation is determined by using the formula:

$D=B^2-4AC$

For equation $x^2-2\left(k+1\right)x+k^2=0$ the value $A=1,B=-2\left(k+1\right)$ and $C=k^2$.

Substituting the values of $A$, $B$ and $C$ in discriminant formula.

$\begin{array}{rcl}D& =& B^2-4AC\\ & =& {\left[-2\left(k+1\right)\right]}^2-4\left(1\right)\left(k^2\right)\\ & =& 4\left[k^2+2k+1-k^2\right]\\ & =& 4\left[2k+1\right]\end{array}$

Step 2: Determine the value of $k$

According to the nature of roots of a quadratic equation if the discriminant $D=0$, then the quadratic equation has real and equal roots.

By nature of roots of equation

$\begin{array}{rcl}D& =& 0\\ & \Rightarrow & 4\left[2k+1\right]=0\\ & \Rightarrow & 2k+1=0\\ & \Rightarrow & k=-\frac{1}{2}\end{array}$

Hence, for $k=-\frac{1}{2}$the equation $x^2-2\left(k+1\right)x+k^2=0$ has equal roots.