7.3 Linear & Quadratic Equations and Inequalities
Key Takeaways
- Three linear forms: y = mx + b (slope-intercept), y - y1 = m(x - x1) (point-slope), Ax + By = C (standard).
- Parallel lines have equal slopes; perpendicular lines have slopes that multiply to -1.
- Quadratic formula: x = (-b +/- sqrt(b^2 - 4ac)) / 2a for ax^2 + bx + c = 0.
- Discriminant b^2 - 4ac: positive gives two real roots, zero gives one repeated root, negative gives no real roots.
- Solve quadratic inequalities by sign analysis: factor, find critical points, test intervals.
7.3 Linear & Quadratic Equations and Inequalities
Quick Answer: Linear: y = mx + b (slope-intercept), y - y1 = m(x - x1) (point-slope), Ax + By = C (standard). Parallel lines have equal slopes; perpendicular lines' slopes multiply to -1. Quadratic formula: x = (-b +/- sqrt(b^2 - 4ac)) / 2a; discriminant b^2 - 4ac decides root count.
Linear Equations - Three Forms
| Form | Equation | Use |
|---|---|---|
| Slope-intercept | y = mx + b | slope m, y-intercept b |
| Point-slope | y - y1 = m(x - x1) | given point + slope |
| Standard | Ax + By = C | intercepts at (C/A, 0) and (0, C/B) |
Slope
Given two points (x1, y1) and (x2, y2): m = (y2 - y1) / (x2 - x1).
Worked Example 6: Find the slope through (2, -3) and (5, 6).
- m = (6 - (-3)) / (5 - 2) = 9/3 = 3.
Parallel and Perpendicular Lines
- Parallel: same slope. y = 2x + 3 and y = 2x - 7 are parallel.
- Perpendicular: slopes multiply to -1 (i.e., m2 = -1/m1). y = 2x + 1 and y = -(1/2)x + 4 are perpendicular.
A horizontal line has slope 0; a vertical line has undefined slope. They are perpendicular to each other.
Solving Linear Equations
Worked Example 7: Solve 3(2x - 4) = 5x + 2.
- Distribute: 6x - 12 = 5x + 2.
- Subtract 5x: x - 12 = 2.
- Add 12: x = 14.
Linear Inequalities - Recap
Use the same algebra, but flip the inequality when multiplying or dividing by a negative. Solve 2x - 7 < 9: 2x < 16, x < 8.
Quadratic Equations
A quadratic has the form ax^2 + bx + c = 0, where a != 0. The GRE tests three solution methods.
Method 1: Factoring
If the quadratic factors nicely, this is fastest.
Worked Example 8: Solve x^2 + 5x + 6 = 0.
- Find two numbers multiplying to 6 and adding to 5: 2 and 3.
- Factor: (x + 2)(x + 3) = 0.
- Roots: x = -2 or x = -3.
Method 2: Quadratic Formula
For any quadratic ax^2 + bx + c = 0:
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
Worked Example 9: Solve 2x^2 - 7x + 3 = 0.
- a = 2, b = -7, c = 3.
- Discriminant: (-7)^2 - 4(2)(3) = 49 - 24 = 25.
- x = (7 +/- sqrt(25)) / 4 = (7 +/- 5) / 4.
- x = 12/4 = 3 or x = 2/4 = 1/2.
Method 3: Completing the Square (less common on GRE)
The GRE rarely requires completing the square, but it underlies the quadratic formula derivation. The GRE prefers factoring or the formula.
The Discriminant: Root Count at a Glance
The discriminant D = b^2 - 4ac tells you the nature of the roots before solving.
| D | Roots |
|---|---|
| D > 0 | two distinct real roots |
| D = 0 | one repeated real root |
| D < 0 | no real roots (two complex roots - not tested) |
Worked Example 10: How many real roots does 3x^2 + 2x + 5 = 0 have?
- D = 4 - 60 = -56 < 0, so no real roots.
Quadratic Inequalities
Solve quadratic inequalities by sign analysis.
Worked Example 11: Solve x^2 - x - 6 > 0.
- Factor: (x - 3)(x + 2) > 0.
- Critical points: x = 3 and x = -2 (where the expression equals 0).
- Test three intervals: (-infinity, -2), (-2, 3), (3, infinity).
- (-infinity, -2): try x = -3 -> (negative)(negative) = positive, satisfies.
- (-2, 3): try x = 0 -> (negative)(positive) = negative, does not satisfy.
- (3, infinity): try x = 4 -> (positive)(positive) = positive, satisfies.
- Solution: x < -2 OR x > 3.
Vertex and Axis of Symmetry
Every parabola y = ax^2 + bx + c has a vertical axis of symmetry at x = -b / (2a). The vertex sits on that axis. If a > 0, the parabola opens upward and the vertex is the minimum; if a < 0, it opens downward and the vertex is the maximum.
Worked Example 12: Find the vertex of y = x^2 - 6x + 11 and identify it as min or max.
- a = 1 > 0, so the parabola opens upward and the vertex is a minimum.
- x = -b / (2a) = -(-6) / (2 * 1) = 3.
- y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2.
- Vertex: (3, 2). The minimum value of y is 2.
The vertex form of a quadratic is y = a(x - h)^2 + k, where (h, k) is the vertex. Converting between standard and vertex form is rarely required on the GRE, but recognizing that the vertex is the minimum (a > 0) or maximum (a < 0) appears in QC questions comparing y-values.
Strategic GRE Tips
- Estimate roots first using the discriminant before solving.
- Try factoring first - the GRE often designs quadratics that factor cleanly.
- Plug answers in when stuck - substitute each choice to find which works.
- For QC questions, test x = 0, x = 1, x = -1, and a fraction like 1/2 when comparing variables.
- For inequalities with two variables, plug numbers carefully - multiplying by a variable could flip or not flip depending on the variable's sign.
- For maximum/minimum questions, locate the vertex at x = -b / (2a) and plug back in to find the y-value.
- For perpendicular-line slope questions, the product of slopes must be -1, so the second slope is the negative reciprocal of the first.
What is the slope of the line passing through (2, -3) and (5, 6)?
Using the quadratic formula, solve 2x^2 - 7x + 3 = 0. Which pair gives the solutions?
How many real roots does 3x^2 + 2x + 5 = 0 have?
Solve the quadratic inequality x^2 - x - 6 > 0. Which interval describes the solution?