Systems with a nonlinear equation
Advanced Math is about 35% of SAT Math (13–15 of 44 questions). A system with one linear and one quadratic equation appears in most tests, as a multiple-choice or typed-answer question that asks for a solution, the number of solutions, or a constant that makes the line touch the parabola.
Solutions are intersection points
Each solution of a system is a point that lies on both graphs. A line and a parabola can meet in 0, 1 or 2 points, so such a system has 0, 1 or 2 solutions. The picture shows y = x² − 4x + 3 and y = x − 1: they cross at (1, 0) and (4, 3).
Substitution: make one equation
Because both equations are equal to y, set the right sides equal to each other. x² − 4x + 3 = x − 1. Move everything to one side: x² − 5x + 4 = 0. Solve: (x − 1)(x − 4) = 0, so x = 1 or x = 4. Then find each y with the simpler (linear) equation: y = x − 1 gives y = 0 and y = 3. Write the solutions as pairs. The question may ask only for the x-values, only for the y-values, or for a sum, so reread the last line.
How many solutions? Use the discriminant
After substitution you have a quadratic ax² + bx + c = 0. Its discriminant D = b² − 4ac tells the number of intersections: D > 0 means two points, D = 0 means one point (the line is tangent to the parabola), D < 0 means no points. When a constant k is in the line, put k into c and solve D = 0 for k.
| D | Intersections | Line and parabola |
|---|---|---|
| D > 0 | 2 | line cuts the parabola |
| D = 0 | 1 | line touches (tangent) |
| D < 0 | 0 | line misses the parabola |
Reading a graph
When the question shows a graph, a solution is an exact crossing point whose coordinates you can read. Check the point in both equations if the picture is unclear. A point that lies on only one curve is not a solution of the system. Count crossings for the number of solutions.
A line and a circle
The same idea works with a circle. Substitute y from the line into x² + y² = r². For y = x − 1 and x² + y² = 25: x² + (x − 1)² = 25, so 2x² − 2x − 24 = 0, then x² − x − 12 = 0, and x = 4 or x = −3. The points are (4, 3) and (−3, −4). Take care when you square (x − 1): it is x² − 2x + 1.
- Set the right sides equal: x² − 4x + 3 = x − 1.
- x² − 5x + 4 = 0, so (x − 1)(x − 4) = 0.
- x = 1 or x = 4. The sum is 5.
- Shortcut: for ax² + bx + c = 0, the sum of the roots is −b/a = 5.
- Set equal: x² − 6x + 10 = 2x + k.
- x² − 8x + (10 − k) = 0.
- One point means D = 0: (−8)² − 4(1)(10 − k) = 0.
- 64 − 40 + 4k = 0, so 4k = −24 and k = −6.
- Check: x² − 8x + 16 = (x − 4)², so the touching point is x = 4, y = 2.
- (−2, 0)
- (0, 2)
- (0, 4)
- (2, 0)
- A solution must lie on both graphs.
- (−2, 0): −(4) + 4 = 0 ✓ and −2 + 2 = 0 ✓.
- (0, 4) lies only on the parabola, and (0, 2) lies only on the line. (2, 0) is on the parabola but 2 + 2 = 4 ≠ 0.
- x² − 6 = 3x + 4, so x² − 3x − 10 = 0.
- (x − 5)(x + 2) = 0, so x = 5 or x = −2. The positive one is a = 5.
- Find b with the line: b = 3(5) + 4 = 19.
- Check in the parabola: 25 − 6 = 19 ✓
- Giving only the x-valuesCheck the last line of the question. If it asks for y, put each x back into the linear equation.
- Putting x back into the quadraticUse the linear equation to find y: it is simpler, and only one value comes out for each x.
- Sign mistakes when moving termsMove every term to one side and recompute b² − 4ac with its signs. A common slip is 10 − k written as k − 10.
- Counting a point on one graph as a solutionA solution must satisfy both equations. Test it in both.
- Forgetting the case D = 0“Exactly one solution” means the line is tangent to the parabola: set the discriminant equal to 0.
- Squaring a bracket wrongly(x − 1)² = x² − 2x + 1, not x² − 1.
Type both equations on separate lines. Desmos draws the line and the parabola. Click on each place where they cross to see the coordinates; a grey dot with a label appears. If a constant k is in the line, type it as k: Desmos offers a slider, and you can move it until the line just touches the parabola, which gives the value of k that makes exactly one solution. Type x² + y² = 25 for a circle.
- Type y = x^2 − 6x + 10
- Type y = 2x + k and accept the slider
- Move the slider until the line touches the parabola at one point
- Read k = −6
Use Desmos for questions about intersections, counting solutions and finding k, and to check an algebra answer. Hand substitution is as fast when the numbers are small.
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