Find $k$ for which the equation $x^{2} - 2 k x + 7 k - 12 = 0$ has real and equal roots.

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Solution

Step 1: Finding values of $a, b$ and $c$
Given equation is $x^{2} - 2 k x + 7 k - 12 = 0$ .
When compared with the general form of quadratic equation $a x^{2} + b x + c = 0$ ,
$a = 1 b = - 2 k c = 7 k - 12$
Step 2: Finding the discriminant
Discriminant is given by $D = b^{2} - 4 a c$ .
Substitute the value of $a, b$ and $c$ ,
$\begin{array}{rcl} D & = & {(- 2 k)}^{2} - 4 \times 1 (7 k - 12) \\ = & 4 k^{2} - 28 k + 48 \end{array}$
Step 3: Finding the value of $k$
Given equation will have real and equal roots if $D = 0$ . So,
$\begin{array}{rcl} 4 k^{2} - 28 k + 48 & = & 0 \\ \Rightarrow & 4 (k^{2} - 7 k + 12) = 0 \\ \Rightarrow & k^{2} - 7 k + 12 = 0 \end{array}$
Factorizing the equation,
$\begin{array}{rcl} k^{2} - (4 + 3) k + 12 & = & 0 \\ \Rightarrow & k^{2} - 4 k - 3 k + 12 = 0 \\ \Rightarrow & k (k - 4) - 3 (k - 4) = 0 \\ \Rightarrow & (k - 4) (k - 3) = 0 \end{array}$
Equating each factor to zero,
$\Rightarrow k - 4 = 0$ or $k - 3 = 0$
$∴ k = 4$ or $k = 3$
Therefore, the value of $k$ for which equation $x^{2} - 2 k x + 7 k - 12 = 0$ has real and equal roots is either $3$ or $4$ .

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