TheMastermind wrote:

This question deserves a better explanation. Anyone who could offer some help?

Hi

TheMastermind ,

Here I go.

For this question, you need to understand two things:

1. If the equation is intersecting x axis, then y must be zero. Or I can say \(x^2\) + 2qx + r = 0

2. Roots of a quadratic equation(\(ax^2 + bx + c = 0\)) are given by the formula:

x = \([ -b + \sqrt{b^2 - 4ac}]/2a\)

and

x = \([ -b - \sqrt{b^2 - 4ac}]/2a\)

Now, in order to have real roots, the values inside the square root MUST be positive.

or I can say \(b^2 - 4ac > = 0\)

Thus, when you make the similar equation with the question in hand,you will say you need \(q^2 - r\).

To get the real roots you will say \(q^2 >= r\).

This is what option A is doing.

if we have \(q^2 > r\), we will have 2 roots.

if we have \(q^2 = r\), we will have 1 root.

Hence, A is sufficient. It tells us that we have two roots or two intersecting points.

Does that make sense?