Quadratic Equations
1. Core Idea
A quadratic has the form ax^2 + bx + c = 0 with a not equal to 0. It has at most two real solutions.
Three tools, in order of usefulness on the GRE:
- Factoring — fastest when it works, which on the GRE is most of the time.
- Vieta's formulas — when you only need the sum or product of the roots.
- The quadratic formula — the guaranteed method when factoring fails.
2. Must-Know Rules
| Concept | Formula |
|---|---|
| Standard form | ax^2 + bx + c = 0 |
| Quadratic formula | x = [-b +/- sqrt(b^2 - 4ac)] / (2a) |
| Discriminant | D = b^2 - 4ac |
| Sum of roots | -b/a |
| Product of roots | c/a |
| Zero-product property | AB = 0 means A = 0 or B = 0 |
The discriminant tells you the number of real roots:
| D | Roots |
|---|---|
| D > 0 | two distinct real roots |
| D = 0 | exactly one real root (a repeated root) |
| D < 0 | no real roots |
Vieta in practice. For 2x^2 - 10x + 8 = 0:
- Sum of roots = 10/2 = 5
- Product of roots = 8/2 = 4
(The roots are 1 and 4. Sum 5, product 4. Confirmed.)
Completing the square. x^2 + 6x + 5 = 0 -> (x + 3)^2 - 9 + 5 = 0 -> (x+3)^2 = 4 -> x + 3 = +/-2 -> x = -1 or -5.
Quadratic in disguise. x^4 - 5x^2 + 4 = 0 is quadratic in u = x^2: u^2 - 5u + 4 = 0 -> u = 1 or 4 -> x = +/-1 or +/-2. Four solutions.
3. Worked Examples
Example 1 — Factoring
Solve x^2 - x - 20 = 0.
Two numbers with product -20 and sum -1: -5 and +4.
(x - 5)(x + 4) = 0 -> x = 5 or x = -4.
Example 2 — Vieta shortcut
The roots of x^2 - 7x + k = 0 differ by 3. Find k.
Sum of roots = 7. Difference = 3.
So the roots are (7+3)/2 = 5 and (7-3)/2 = 2.
Product = k = 5 x 2 = 10.
Example 3 — Discriminant
For what values of m does x^2 + mx + 9 = 0 have exactly one real solution?
D = m^2 - 4(1)(9) = m^2 - 36 = 0
m^2 = 36 -> m = 6 or m = -6.
4. GRE Traps
- Missing the negative root. From x^2 = 49, x = 7 or -7. When the GRE gives x^2 = k, it is usually testing exactly this.
- Sum of roots = b/a. It is -b/a. The minus sign is forgotten constantly.
- Dividing by x. From x^2 = 3x, you lose the root x = 0.
- Assuming two roots always exist. If D < 0 there are none over the reals.
- Forgetting to set the equation to zero before factoring. x^2 = 5x + 24 must become x^2 - 5x - 24 = 0.
- Extraneous roots. After squaring both sides of an equation involving a square root, always substitute back — some roots will be invalid.
- Quadratic in disguise gives 2 solutions. x^4 = 16 has x = 2 and x = -2 over the reals; a degree-4 equation can have up to 4 real roots, so count carefully.
5. Speed Tricks
- Try factoring first. GRE quadratics are designed to factor about 80% of the time.
- Use Vieta when the question asks for the sum or product of roots — never solve for the roots individually.
- If the roots are r and s, the equation is x^2 - (r+s)x + rs = 0. Useful for building an equation from its roots.
- Recognise the perfect-square trinomials: x^2 + 2kx + k^2 and x^2 - 2kx + k^2.
- Back-solve from answer choices when a quadratic looks ugly — plug each choice in.
- Discriminant questions never need the roots. Just compute b^2 - 4ac.
6. Self-Check
Q1. Solve 2x^2 - 5x - 3 = 0.
Q2. What is the sum of the roots of 3x^2 + 12x - 7 = 0?
Q3. How many real solutions does x^2 + 2x + 5 = 0 have?
Answers
A1. ac = -6; two numbers with product -6 and sum -5: -6 and +1. Split: 2x^2 - 6x + x - 3 = 2x(x-3) + 1(x-3) = (2x+1)(x-3). x = 3 or x = -1/2.
A2. -b/a = -12/3 = -4.
A3. D = 4 - 20 = -16 < 0, so none.
7. One-Line Summary for the Board
Set it to zero, then factor. Sum of roots = -b/a, product = c/a. Never lose the negative root.