In the xy-coordinate plane, how many points of intersection with the

The roots of the equation \(y = ax^2 + bx + c\) is given by

\(\frac{-b \pm \sqrt{b^2 - 4*a*c}}{2*a}\)

The graph represented by \(y = ax^2 + bx + c\) represents a parabola, and the points at which the graph intersects with the x-axis represent its roots.

Let's focus on the discriminant \(b^2 - 4ac\)

• \(b^2 - 4ac = 0\), only one root exists, i.e. the parabola intersects with the x-axis at only one point
• \(b^2 - 4ac > 0\), two roots exist, i.e. the parabola intersects with the x-axis at two points
• \(b^2 - 4ac < 0\), no real roots exists, i.e. the parabola does not intersect with the x-axis

The discriminant of the equation \(y= x^2 + 2qx + r\) is

\((2q)^2 - 4*1*r\)

\(4q^2 - 4r\)

Statement 1

\( (1) q^2 > r\)

If \(q^2\) is greater than \(r\), \(4*q^2\) is greater than \(4*r\). Hence, the value of the discriminant is positive. Therefore we can conclude that the graph of \(y= x^2 + 2qx + r\) intersects with the x-axis at two points.

The statement alone is sufficient. We can eliminate B, C, and E.

Statement 2

\((2) r^2 > q\)

The discriminant of the equation \(y= x^2 + 2qx + r\) is

\(4q^2 - 4r\)

If \(r = q^2\) and \(r^2 > q\) → The value of \(4q^2 - 4r\) equals 0. The graph of \(y= x^2 + 2qx + r\) will intersect with the x-axis at one point.

If \(q^2 > r\) and \(r^2 > q\) → Placement of the numbers on a number line is shown below -

---------- \(q\) ----- -1 ----- 0 ----- 1 -- \(r\) --------- \(q^2\) --------- \(r^2\) -------

The value of \(4q^2 - 4r\) is greater than 0. The graph of \(y= x^2 + 2qx + r\) will intersect with the x-axis at two points.

The statement alone is not sufficient. We can eliminate D.

Option A