6.2 Solving Quadratics by Factoring, Square Roots & Completing the Square

6.2 Solving Quadratics by Factoring, Square Roots & Completing the Square

Skill 8 of Competency 2 asks you to solve quadratics using a variety of methods, and it names four: factoring, the quadratic formula, completing the square, and graphing. Items often specify a method, so you cannot rely on one technique for everything.

Choosing a method

+---------------------------------------------------------------------------+
|  Form you see                        Best method                          |
+---------------------------------------------------------------------------+
|  x^2 = 49,  3(x-2)^2 = 27            Square root method                    |
|  x^2 + 7x + 12 = 0  (factors easily) Factoring                            |
|  x^2 + 6x - 4 = 0   (b is even)      Completing the square                |
|  Anything, especially ugly ones      Quadratic formula                    |
|  "Approximate the solutions"         Graphing / estimation                |
+---------------------------------------------------------------------------+

The zero product property and factoring

If AB = 0, then A = 0 or B = 0. This holds only for a product equal to zero, which is why every factoring solution begins by moving all terms to one side.

Solve x² + 5x = 24 Set to zero: x² + 5x − 24 = 0 Factor: two numbers multiplying to −24 and adding to 5 are 8 and −3 → (x + 8)(x − 3) = 0 Zero product: x + 8 = 0 or x − 3 = 0 → x = −8 or x = 3

[!WARNING] A very common student error — and a Competency 5 diagnostic item — is applying the zero product property to a nonzero product. From (x + 8)(x − 3) = 10, you cannot write x + 8 = 10 or x − 3 = 10. There is no "ten product property." Expand and reset to zero first.

With a leading coefficient, factor by grouping:

Solve 3x² − 11x − 4 = 0 a · c = −12; two numbers multiplying to −12 and adding to −11 are −12 and 1 3x² − 12x + x − 4 = 0 → 3x(x − 4) + 1(x − 4) = 0 → (x − 4)(3x + 1) = 0 x = 4 or x = −1/3

Always factor out a GCF first: 2x² − 18 = 0 becomes 2(x² − 9) = 0 → 2(x + 3)(x − 3) = 0, giving x = ±3.

The square root method

Use it whenever the variable appears inside exactly one squared expression. Isolate the square, then take the square root of both sides — including the ± sign.

Solve 3(x − 4)² − 15 = 33 Isolate: 3(x − 4)² = 48 → (x − 4)² = 16 Root: x − 4 = ±4 x = 8 or x = 0

Solve 2x² − 90 = 0 x² = 45 → x = ±√45 = ±3√5

Dropping the negative root is the standard error. The equation x² = 16 has two solutions, 4 and −4; only the principal square root √16 is uniquely 4. Contexts sometimes discard the negative root — a length cannot be negative — but the equation still has both.

Completing the square

This method converts ax² + bx + c into a(x − h)² + k, which both solves the equation and exposes the vertex. That dual payoff is why the blueprint lists it separately from the quadratic formula.

The core move: for x² + bx, add (b/2)² to complete the perfect square trinomial, since x² + bx + (b/2)² = (x + b/2)².

Leading coefficient 1:

Solve x² + 10x − 24 = 0 Move the constant: x² + 10x = 24 Half of 10 is 5, and 5² = 25. Add 25 to both sides: x² + 10x + 25 = 49 Factor the left: (x + 5)² = 49 Root: x + 5 = ±7 → x = 2 or x = −12

Leading coefficient not 1: factor a out of the x-terms first.

Solve 2x² − 12x + 5 = 0 Move the constant: 2x² − 12x = −5 Factor 2 from the left: 2(x² − 6x) = −5 Half of −6 is −3, and (−3)² = 9. Adding 9 inside the parentheses actually adds 2 · 9 = 18 to the left side, so add 18 to the right: 2(x² − 6x + 9) = −5 + 18 → 2(x − 3)² = 13 (x − 3)² = 13/2 → x − 3 = ±√(13/2) → x = 3 ± √(6.5) ≈ 3 ± 2.55

The bolded caution is the step candidates miss: what you add inside the parentheses is scaled by the factored-out coefficient before it reaches the other side.

Converting to vertex form

The same procedure rewrites a function without solving anything.

f(x) = x² − 8x + 3 f(x) = (x² − 8x + 16) + 3 − 16 [add and subtract 16 to preserve value] f(x) = (x − 4)² − 13, so the vertex is (4, −13)

Cross-check with the axis formula: x = −(−8)/2 = 4, and f(4) = 16 − 32 + 3 = −13. ✓

When the leading coefficient is not 1, the same scaling caution applies:

g(x) = 3x² + 12x + 1 = 3(x² + 4x) + 1 = 3(x² + 4x + 4) + 1 − 12 = 3(x + 2)² − 11

Here 4 was added inside, which added 3 · 4 = 12, so 12 was subtracted outside to keep the function unchanged. The vertex is (−2, −11).