Question

In the following, determine the set of values of $p$ for which the quadratic equation has real roots:

$p{x}^2+4x+1=0$

Answer 1

Step 1: Obtain coefficients of the given equation.

The given equation is as follows:

$p{x}^2+4x+1=0$

Compare the given quadratic equation with $a{x}^2+bx+c=0$ to get:

$a=p,b=4,c=1$

Solve for discriminant as:

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& {\left(4\right)}^2-4\times p\times 1\\ & =& 16-4p\end{array}$

Step 2: Use the condition of real roots.

For the real root, the value of the discriminant will be:

$\begin{array}{rcl}& \Rightarrow & b^2-4ac\ge 0\\ & \Rightarrow & 16-4p\ge 0\\ & \Rightarrow & 4p\le 16\\ & \Rightarrow & p\le 4\end{array}$

For the values $p\le 4$ , the given quadratic equation has real roots.