y = x^2 + 3x + 5y = x - 2Consider the system of equations above. In the xy-plane, what...

1. INFER the connection between intersection points and equal y-values

• Given information:
  • First equation: \(\mathrm{y = x^2 + 3x + 5}\) (parabola)
  • Second equation: \(\mathrm{y = x - 2}\) (line)
  • Need: number of intersection points
• Key insight: Intersection points occur where both equations have the same x and y coordinates, so we need to find where the y-values are equal.

2. TRANSLATE this insight into an equation

• Set the right sides equal to each other:
  \(\mathrm{x^2 + 3x + 5 = x - 2}\)

3. SIMPLIFY to standard quadratic form

• Move all terms to one side:
  \(\mathrm{x^2 + 3x + 5 = x - 2}\)
  \(\mathrm{x^2 + 3x - x + 5 + 2 = 0}\)
  \(\mathrm{x^2 + 2x + 7 = 0}\)

4. INFER that we need the discriminant to count real solutions

• For any quadratic \(\mathrm{ax^2 + bx + c = 0}\), the discriminant \(\mathrm{Δ = b^2 - 4ac}\) tells us:
  • If \(\mathrm{Δ \gt 0}\): two real solutions (two intersection points)
  • If \(\mathrm{Δ = 0}\): one real solution (one intersection point)
  • If \(\mathrm{Δ \lt 0}\): no real solutions (zero intersection points)

5. SIMPLIFY the discriminant calculation

• From \(\mathrm{x^2 + 2x + 7 = 0}\), we have \(\mathrm{a = 1, b = 2, c = 7}\)
• \(\mathrm{Δ = (2)^2 - 4(1)(7)}\)
  \(= 4 - 28\)
  \(= -24\)

6. INFER the final answer from the discriminant

• Since \(\mathrm{Δ = -24 \lt 0}\), there are no real solutions to the equation
• This means no real x-values where the graphs intersect
• Therefore, zero intersection points

Answer: D) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when rearranging to standard form, particularly with the signs.

For example, they might incorrectly rearrange \(\mathrm{x^2 + 3x + 5 = x - 2}\) as:
\(\mathrm{x^2 + 3x + 5 - x - 2 = 0}\) → \(\mathrm{x^2 + 2x + 3 = 0}\)

This gives discriminant \(\mathrm{Δ = 4 - 12 = -8}\), which is still negative, but if they made a different sign error, they could get a positive discriminant and conclude there are two intersection points. This may lead them to select Choice B (Exactly two).

Second Most Common Error:

Conceptual confusion about discriminant interpretation: Students correctly calculate \(\mathrm{Δ = -24}\) but don't understand what a negative discriminant means.

They might think "negative discriminant = negative number of solutions" doesn't make sense, so they guess or assume it means one solution. This may lead them to select Choice A (Exactly one).

The Bottom Line:

This problem tests whether students can connect the geometric concept of intersection points with the algebraic process of solving quadratic equations, requiring both accurate algebraic manipulation and understanding of the discriminant's meaning.