Interpret solutions as intersection points and determine how many solutions exist. | Systems of equations in two variables | SAT Math

Use the discriminant to turn an intersection-count question into a quick solve for the parameter.

Core Idea

The number of solutions equals the number of intersection points. For a line meeting a parabola, the discriminant of the resulting quadratic tells you: positive → 2, zero → 1, negative → 0.

Understanding

Count the intersections, not the algebra alone.

Substitute to get a quadratic in one variable.
Put it in $𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0$ form.
Use the discriminant $𝑏 2 - 4 𝑎 𝑐$ : positive = 2 solutions, zero = 1, negative = 0.

If a parameter is involved, set the discriminant to the condition the question asks for and solve for that value.

Step by Step

Substitute to eliminate one variable and get a quadratic equation.
Write it in standard form $𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0$ .
Compute the discriminant: $Δ = 𝑏 2 - 4 𝑎 𝑐$ .
Interpret: $Δ > 0$ → two solutions, $Δ = 0$ → exactly one, $Δ < 0$ → none.
If a parameter is involved, set the discriminant to the required condition and solve for it.

Misconceptions

Confusing discriminant = 0 (one solution) with discriminant < 0 (no solutions).
Forgetting that a line and parabola can have at most two intersection points — never three.
Computing the discriminant before fully simplifying to standard form, leading to wrong coefficients.

Question

Worked Example

$𝑦 = 2 𝑥 + 𝑐$
$𝑦 = 𝑥 2 + 4 𝑥 + 7$

For what value of $𝑐$ does the system of equations above have exactly one solution?

Select an answer to see the explanation