Question

Without solving, examine the nature of the roots of the equation:

$x^2+px-q^2=0$

Answer 1

Step 1: Nature of the roots of an equation.

The given equation is as follows:

$x^2+px-q^2=0$.

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

$a=1,b=p\text{and}c=-q^2$

Now solving for discriminant as follows:

$\begin{array}{rcl}D& =& {\left(b\right)}^2-4ac\\ & =& {\left(p\right)}^2-4\left(1\right)\left(-q^2\right)\\ & =& p^2+4q^2\ge 0\left(\because p^2\text{and}{q}^2\ge 0\mathrm{always}\right)\end{array}$

Step 2: Conclusion.

From, the above relation it can be concluded that the discriminant of an equation is always non-negative.

When $p$ and $q$ are zero then $D=0$, the roots are real and equal.

If either $p$ and $q$is non-zero, we have real and distinct roots.

Hence, in either of the above cases, the roots are always real for all real values of $p$ and $q$.