Nonlinear equations and systems
Solve quadratics four ways, handle radical, rational and absolute value equations, and find where a line meets a curve. Plus the checks that catch fake solutions.
Solutions are x-intercepts
Solving x² + bx + c = 0 means finding the x-values where y = x² + bx + c equals zero. On a graph those are the points where the parabola crosses the x-axis. A parabola can cross twice, touch once, or miss completely.
Factor, then use the zero product rule
If two things multiply to zero, at least one of them is zero. So once a quadratic is factored and set equal to zero, each bracket gives a solution.
- Move everything to one side: 2x² − 8x = 0
- Factor out the common factor: 2x(x − 4) = 0
- Set each factor to zero: 2x = 0 or x − 4 = 0, so x = 0 or x = 4
Keep every solutionDividing both sides of 2x² = 8x by x would lose the solution x = 0. Factor instead of dividing by a variable.
Square roots and completing the square
When the equation is a square equal to a number, take the square root of both sides and remember the ±. Completing the square turns any quadratic into that shape.
- Move the constant: x² + 6x = 7
- Half of 6 is 3, and 3² = 9. Add 9 to both sides: x² + 6x + 9 = 16
- The left side is a perfect square: (x + 3)² = 16
- Square root with ±: x + 3 = 4 or x + 3 = −4, so x = 1 or x = −7
The quadratic formula and the discriminant
For ax² + bx + c = 0, the quadratic formula always works:
The part under the root, b² − 4ac, is the discriminant. The SAT often asks only how many solutions there are, and the discriminant answers that in one line.
- a = 2, b = −3, c = −4
- Discriminant: (−3)² − 4(2)(−4) = 9 + 32 = 41, which is positive, so two solutions
- x = (3 ± √41) ÷ 4
Radical and rational equations
To remove a square root, square both sides. To remove fractions, multiply by the denominators. Both moves can add solutions that do not work in the original, called extraneous solutions. The fix is simple: check every answer.
- Square both sides: x + 7 = x² + 2x + 1
- Get zero on one side: x² + x − 6 = 0, so (x + 3)(x − 2) = 0
- Candidates: x = −3 or x = 2
- Check x = −3: √4 = 2 but −3 + 1 = −2. It fails. Check x = 2: √9 = 3 and 2 + 1 = 3. The only solution is x = 2.
Absolute value equations
Absolute value is distance from zero, so it is never negative. |x − 4| = 6 splits into x − 4 = 6 or x − 4 = −6, giving x = 10 or x = −2. If an absolute value is set equal to a negative number, there is no solution.
When a line meets a curve
A system like y = x² − 2 and y = mx + b has its solutions where the graphs cross. To solve it, set the two right sides equal and solve the quadratic you get. Move the line to see all three cases.
- Set them equal: x² − 2 = x + 4
- Zero on one side: x² − x − 6 = 0, so (x − 3)(x + 2) = 0
- x = 3 or x = −2. Use the line to find y: (3, 7) and (−2, 2)
Solve it in Desmos
Every equation in this lesson can be solved by graphing. Graph each side as its own equation and tap where they meet. It also shows extraneous solutions clearly, because a fake solution has no crossing point.
Spot the trap
Each solution has one faulty step. All three lose or invent a solution, which is exactly how the wrong answer choices are built.