Question

Find the value of ‘$k$’ for which the following quadratic equation has real roots:

$k{x}^2+6x+1=0$

Answer 1

Step 1:

Comparison of the coefficients:

Compare the coefficients of the given equation with the standard quadratic equation, $a{x}^2+bx+c=0$.

$$\begin{gathered} a=k \\ b=6 \\ c=1 \end{gathered}$$

Step 2:

Find the discriminant:

The discriminant $D$ of the given equation can be calculated as,

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& {\left(6\right)}^2-\left(4\times k\times 1\right)\\ & =& 36-4k\end{array}$

Since, the given equation has real roots, so the discriminant must be greater than or equal to 0.

$\begin{array}{rcl}D& \ge & 0\\ 36-4k& \ge & 0\end{array}$

Solve the inequality for $k$.

$\begin{array}{rcl}36-4k& \ge & 0\\ & \Rightarrow & 4k\le 36\\ & \Rightarrow & k\le \frac{36}{4}\\ & \Rightarrow & k\le 9\end{array}$

Hence, the value of $k$ is $k\le 9$.