Question

Find the value of $k$ for which the quadratic equation $x^2-2kx+(7k-12)=0$ has equal roots.

Answer 1

Step 1: Using the discriminant, compare the general quadratic equation

Discriminant $D=B^2-4AC$ compared with the equation $A{x}^2+Bx+C=0$.

So, $A=1,B=-2k$ and, $7k-12$ for the equation $x^2-2kx+(7k-12)=0$.

For roots to be equal, $D=0$.

Step 2: Substitute the values of $A,B$ and $C$ in the equation $D=B^2-4AC$

After simplifying,

$$\begin{gathered} \Rightarrow {(-2k)}^2-4\left(1\right)(7k-12)=0 \\ \Rightarrow 4k^2-28k+48=0 \\ \Rightarrow k^2-7k+12=0 \\ \Rightarrow k^2-4k-3k+12=0 \end{gathered}$$

Simplifying further,

$$\begin{gathered} \Rightarrow k(k-4)-3(k-4)=0 \\ \Rightarrow (k-3)(k-4)=0 \end{gathered}$$

Step 3: Set factors equal to $0$

$k-3=0$ or,

$k-4=0$

$k=3$ or,

$k=4$

Therefore, the value of $k$ are $3$ and $4$.