3.4 Quadratic Equations

Solving by Factoring

A quadratic equation has the standard form ax² + bx + c = 0 where a ≠ 0. If the quadratic factors, the fastest route is the zero-product property: if a product equals zero, at least one factor must be zero. So you factor, set each factor equal to zero, and solve.

Worked example: Solve x² − 5x + 6 = 0.

• Factor: two numbers multiplying to 6 and adding to −5 are −2 and −3, so (x − 2)(x − 3) = 0.
• Set each factor to zero: x − 2 = 0 → x = 2; x − 3 = 0 → x = 3.
• Solutions: x = 2 and x = 3.
• Check x = 2: 4 − 10 + 6 = 0. ✓

Always move every term to one side so the equation equals zero BEFORE factoring — the zero-product property only works against zero.

The Square-Root Method

When there is no middle (bx) term, isolate the squared expression and take the square root of both sides, remembering the ± (plus-or-minus).

Worked example: Solve x² = 49 → x = ±√49 = ±7.

Worked example: Solve (x − 3)² = 16. Take roots: x − 3 = ±4, so x = 3 ± 4, giving x = 7 or x = −1.

Completing the Square

Completing the square turns any quadratic into a perfect-square binomial. Move the constant to the right, take half of the coefficient of x, square it, and add it to BOTH sides.

Worked example: Solve x² + 6x − 7 = 0.

• Move the constant: x² + 6x = 7.
• Half of 6 is 3; 3² = 9. Add 9 to both sides: x² + 6x + 9 = 16.
• Write the left as a perfect square: (x + 3)² = 16.
• Take square roots: x + 3 = ±4, so x = −3 ± 4.
• Solutions: x = 1 or x = −7.

This technique also derives the quadratic formula and is essential later for circles and conic sections.

The Quadratic Formula

The quadratic formula solves EVERY quadratic, even those that do not factor:

x = (−b ± √(b² − 4ac)) / 2a

Identify a, b, and c from ax² + bx + c = 0, substitute carefully (mind the signs), and simplify.

Quadratic Formula: Worked Example

Worked example: Solve 2x² + 3x − 5 = 0 by the quadratic formula.

• Identify: a = 2, b = 3, c = −5.
• Discriminant: b² − 4ac = 3² − 4(2)(−5) = 9 + 40 = 49.
• Apply the formula: x = (−3 ± √49) / (2·2) = (−3 ± 7) / 4.
• Two solutions: x = (−3 + 7)/4 = 4/4 = 1, and x = (−3 − 7)/4 = −10/4 = −5/2.

Same equation by factoring: 2x² + 3x − 5 = (2x + 5)(x − 1) = 0, giving x = −5/2 and x = 1 — the SAME answers, confirming both methods agree.

The Discriminant

The discriminant is the expression b² − 4ac under the radical. It tells you the number and type of solutions before solving:

Worked example: For x² − 4x + 4 = 0, the discriminant is (−4)² − 4(1)(4) = 16 − 16 = 0, so there is exactly one repeated solution (x = 2).

Common Mistakes

• Forgetting the ±: the square-root method and the formula both yield two values.
• Sign errors with b² − 4ac when b or c is negative — note (−4)² = +16 and −4ac can flip sign.
• Not setting the equation equal to zero before factoring.
• Dividing only PART of the numerator by 2a; the entire −b ± √(...) is over 2a.
• Taking half of b incorrectly when completing the square.

Choosing the Best Method

All four methods give the same answers, but some are faster for a given equation. Use this guide to decide quickly on the test.

Worked example (square-root method on a shifted square): Solve 3x² = 48. Divide by 3 to isolate the square: x² = 16. Take roots: x = ±4. Solutions x = 4 or x = −4.

A Word Problem Application

Quadratics model areas and projectile motion. After setting up the equation, move everything to one side and solve.

Worked example: A rectangular garden is 3 feet longer than it is wide and has an area of 40 square feet. Find the width w.

• Set up: width is w, length is w + 3, so w(w + 3) = 40.
• Expand: w² + 3w = 40, then w² + 3w − 40 = 0.
• Factor: (w + 8)(w − 5) = 0, so w = −8 or w = 5.
• A width cannot be negative, so w = 5 feet (and the length is 8 feet).

This discarding of a negative root is a frequent ALEKS step: always check that each solution makes sense in the real-world context before reporting it.