📖 Quadratics Have Two Solutions
A quadratic equation has a variable squared — like x². Unlike regular equations, quadratics almost always have two answers. Missing one of them is the most common GED mistake on this topic.
The key to solving quadratics is factoring — which you already know. Once an equation is factored, solving it is fast.
The connection: Factoring creates the multiplication structure we need to solve quadratics. If you can factor, you can solve.
⚡ The Zero Product Rule
There is one simple rule that makes solving factored equations possible:
The Zero Product Rule
If A × B = 0, then A = 0 or B = 0
If two things multiply to zero, at least one of them must be zero.
There is no other way to multiply and get zero.
This means: when you have a factored equation set equal to zero, you just set each factor equal to zero and solve separately.
Notice the sign flip: x − 3 = 0 gives x = +3. x + 2 = 0 gives x = −2. The solution is always the opposite sign of what's written inside the parentheses.
🗂️ Three Situations You'll See
Situation 1
Already factored
(x − 4)(x + 1) = 0
→ Just solve each factor
Situation 2
Factor then solve
x² + 5x + 6 = 0
→ Factor first, then solve
Situation 3
Set = 0 first
x² + 5x = 14
→ Move all terms, then factor
The equation must equal zero before you can use the Zero Product Rule. That's why Situation 3 requires a first step — moving everything to one side.
🔢 Worked Examples
Example 1 — Already Factored
Solve: (x − 4)(x + 3) = 0
The equation is already factored and equals zero — use the Zero Product Rule directly.
Factor 1: x − 4 = 0 → x = 4
Factor 2: x + 3 = 0 → x = −3
Two solutions: x = 4 and x = −3
Check x = 4: (4−4)(4+3) = (0)(7) = 0 ✓
Check x = −3: (−3−4)(−3+3) = (−7)(0) = 0 ✓
Example 2 — Factor Then Solve
Solve: x² + 7x + 12 = 0
Step 1: Factor. Need product +12 and sum +7 → 3 and 4.
x² + 7x + 12 = (x + 3)(x + 4)
Step 2: Set each factor to zero.
x + 3 = 0 → x = −3 | x + 4 = 0 → x = −4
Two solutions: x = −3 and x = −4
Example 3 — Negative Constant
Solve: x² + x − 6 = 0
Step 1: Factor. Negative constant → one positive, one negative. Need product −6, sum +1.
(+3)(−2) = −6 ✓ and 3 + (−2) = +1 ✓ → factors: (x + 3)(x − 2)
Step 2: Solve. x + 3 = 0 → x = −3 | x − 2 = 0 → x = 2
Two solutions: x = −3 and x = 2
Example 4 — Set Equal to Zero First
Solve: x² + 5x = 14
Step 1: Subtract 14 from both sides. The equation must equal zero.
x² + 5x − 14 = 0
Step 2: Factor. Need product −14, sum +5 → (+7)(−2) = −14 and 7+(−2) = 5 ✓
(x + 7)(x − 2) = 0
Step 3: Solve. x + 7 = 0 → x = −7 | x − 2 = 0 → x = 2
Two solutions: x = −7 and x = 2
Example 5 — Answer Choice Strategy
Which value is a solution to
x² − 3x − 10 = 0?
A) x = 2
B) x = 3
C) x = −2
D) x = 5 ✓
Strategy: test each choice by substituting into the equation.
A) x=2: (4) − 6 − 10 = −12 ✗
B) x=3: (9) − 9 − 10 = −10 ✗
C) x=−2: (4) + 6 − 10 = 0 — wait, that works too! But check D first.
D) x=5: (25) − 15 − 10 = 0 ✓
Both C and D equal zero — both are solutions. x² − 3x − 10 = (x−5)(x+2) = 0 → x=5 and x=−2.
🚫 Common Mistakes
Mistake 1 — Most Common
Writing only one solution
Quadratics almost always have TWO answers. Always set both factors to zero.
Mistake 2
Forgetting to set equal to zero first
x² + 5x = 14 cannot be factored yet. Subtract 14 first → x² + 5x − 14 = 0.
Mistake 3
Getting the sign wrong on solutions
(x + 3) = 0 gives x = −3, not +3. The sign flips when you solve.
Mistake 4
Moving a term to the wrong side
x² + 6x = 7 → subtract 7 → x² + 6x − 7 = 0. Adding 7 instead gives the wrong equation.
✏️ Practice Questions