Choose a solving method from the initial form of a quadratic equation, and write complex solutions as a ± bi

Learning goal: Solve quadratic equations in one variable by the method that fits the initial form of the equation: inspection for x² = 49, taking square roots, completing the square, applying the quadratic formula, or utilizing the Zero-Product Property after factoring; and, for quadratic equations with real coefficients that have complex solutions (Ohio's N.CN.7; Ohio's wording of A.REI.4b itself ends with the list of methods), recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

Matching the method to the initial form. Every quadratic equation can be solved by the quadratic formula, but the formula is often the slowest route, and choosing a method from the way the equation is written is a skill in its own right. Ask, in order: Is it of the form x² = k or (x − p)² = q? Take square roots. Is one side zero and the other side factorable? Factor and use the Zero-Product Property. Is the leading coefficient 1 and the x-coefficient even? Completing the square is quick. Otherwise, apply the quadratic formula. Whatever the method, every solution is substituted back into the original equation before it is reported.

Inspection and taking square roots. The equation x² = 49 has two solutions, 7 and −7, because both 7² and (−7)² equal 49; a student who writes only x = 7 has found half the solution set. This can be read off by inspection, but the reasoning is "taking square roots of both sides", which always produces ±. The same move handles (x − 3)² = 20: x − 3 = ±√20 = ±2√5, so x = 3 ± 2√5, approximately 7.4721 and −1.4721. Check the exact form: (3 + 2√5 − 3)² = (2√5)² = 20. The equation (x + 1)² = 12 gives x = −1 ± 2√3. When the equation is already a squared expression equal to a number, expanding it and using the formula is wasted effort.

The Zero-Product Property and factoring. The Zero-Product Property says that if a product of real (or complex) numbers is zero, then at least one of the factors is zero: if ab = 0, then a = 0 or b = 0. The condition is that the product equals zero, not any other number. Solve x² − 2x − 8 = 0: factor as (x − 4)(x + 2) = 0, so x − 4 = 0 or x + 2 = 0, giving x = 4 or x = −2. Check: 16 − 8 − 8 = 0 and 4 + 4 − 8 = 0. Now the misconception: faced with (x − 2)(x + 5) = 8, a student sets x − 2 = 8 and x + 5 = 8. That is wrong because a product can equal 8 in infinitely many ways, 1 × 8, 2 × 4, 16 × 1/2, and nothing forces either factor to be 8. The correct route is to expand, x² + 3x − 10 = 8, move everything to one side, x² + 3x − 18 = 0, and factor again, (x + 6)(x − 3) = 0, so x = −6 or x = 3. Check in the original: (−8)(−1) = 8 and (1)(8) = 8. A second misconception is dividing by the variable: for 5x² = 20x, dividing both sides by x gives x = 4 and silently discards x = 0, which is a solution because 0 = 0. Instead factor: 5x(x − 4) = 0, so x = 0 or x = 4.

Factoring with a leading coefficient. Solve 6x² + x − 12 = 0. Look for two binomials whose outer and inner products combine to x: (3x − 4)(2x + 3) expands to 6x² + 9x − 8x − 12 = 6x² + x − 12, so the factoring is right. Then 3x − 4 = 0 or 2x + 3 = 0, giving x = 4/3 or x = −3/2. Check x = 4/3: 6(16/9) + 4/3 − 12 = 32/3 + 4/3 − 36/3 = 0. Check x = −3/2: 6(9/4) − 3/2 − 12 = 27/2 − 3/2 − 24/2 = 0. Verifying the factoring by expanding before using it is the step students skip and the step that catches sign errors.

Applying the quadratic formula. When factoring over the integers fails, use x = (−b ± √(b² − 4ac))/(2a). Solve 3x² − 7x + 1 = 0: a = 3, b = −7, c = 1, discriminant 49 − 12 = 37, so x = (7 ± √37)/6, approximately 2.1805 and 0.1529. Check the approximation 2.1805: 3(4.7546) − 15.2635 + 1 ≈ 14.2638 − 15.2635 + 1 = 0.0003, zero to the precision of the rounding. Report the exact form; use the decimal to check and to locate the solution on a graph. The discriminant b² − 4ac decides the count and type of solutions: positive gives two distinct real solutions, zero gives one repeated real solution, negative gives two complex solutions that are conjugates of each other.

Complex solutions written a ± bi. The equation x² + 4x + 13 = 0 has discriminant 16 − 52 = −36. There is no real number whose square is −36, but the imaginary unit i, defined by i² = −1, gives √(−36) = 6i. So x = (−4 ± 6i)/2 = −2 ± 3i. The two solutions, −2 + 3i and −2 − 3i, are complex conjugates. Completing the square reaches the same place: (x + 2)² = −9, so x + 2 = ±3i. Check x = −2 + 3i by substitution, using i² = −1: (−2 + 3i)² = 4 − 12i + 9i² = 4 − 12i − 9 = −5 − 12i; then 4(−2 + 3i) = −8 + 12i; adding 13 gives −5 − 12i − 8 + 12i + 13 = 0. The imaginary parts cancel and the real parts sum to zero, so the check passes. When the leading coefficient is not 1, the same arithmetic applies: 2x² − 2x + 5 = 0 has discriminant 4 − 40 = −36, so x = (2 ± 6i)/4 = 1/2 ± 3i/2. In a ± bi form, a = 1/2 and b = 3/2.

A fully worked graphical reading. The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4, the two real solutions of x² − 2x − 8 = 0, and its lowest point is (1, −9). The graph of y = x² − 6x + 10 has lowest point (3, 1) and never touches the x-axis, which is the picture of a negative discriminant: 36 − 40 = −4, and the solutions 3 ± i are complex, so no real x makes the expression zero. A graph can show how many real solutions there are and roughly where; the algebra gives the exact values, and only the algebra can produce complex solutions.

Worked selection of method, four equations: (1) x² = 81: take square roots, x = ±9; check 81 = 81 both ways. (2) (x + 1)² = 12: square roots, x + 1 = ±√12 = ±2√3, so x = −1 ± 2√3 ≈ 2.4641 and −4.4641; check (−1 + 2√3 + 1)² = (2√3)² = 12. (3) x² + 5x − 14 = 0: one side is zero and the expression factors, (x + 7)(x − 2) = 0, so x = −7 or x = 2 by the Zero-Product Property; check 49 − 35 − 14 = 0 and 4 + 10 − 14 = 0. (4) 2x² + 3x − 4 = 0: the discriminant 9 + 32 = 41 is not a perfect square, so factoring over the integers is impossible; apply the formula, x = (−3 ± √41)/4 ≈ 0.8508 and −2.3508; check 2(0.8508)² + 3(0.8508) − 4 ≈ 1.4477 + 2.5524 − 4 = 0.0001.

Worked complex solutions by completing the square: Solve x² − 6x + 13 = 0. Then x² − 6x = −13, add 9: (x − 3)² = −4, so x − 3 = ±2i and x = 3 ± 2i. Check x = 3 + 2i: (3 + 2i)² = 9 + 12i + 4i² = 9 + 12i − 4 = 5 + 12i; −6(3 + 2i) = −18 − 12i; then 5 + 12i − 18 − 12i + 13 = 0. The conjugate 3 − 2i checks the same way with every i replaced by −i.

Worked complex solutions by the formula with a fractional real part: Solve 2x² − 2x + 5 = 0. a = 2, b = −2, c = 5, discriminant 4 − 40 = −36, √(−36) = 6i. x = (2 ± 6i)/4 = 1/2 ± 3i/2. Check x = 1/2 + 3i/2: x² = 1/4 + 2(1/2)(3i/2) + 9i²/4 = 1/4 + 3i/2 − 9/4 = −2 + 3i/2; 2x² = −4 + 3i; −2x = −1 − 3i; sum with 5: −4 + 3i − 1 − 3i + 5 = 0.

Worked misconception, product equal to a nonzero number: Solve (x − 2)(x + 5) = 8. Setting each factor to 8 is invalid because the Zero-Product Property applies only when the product is zero. Expand: x² + 3x − 10 = 8, so x² + 3x − 18 = 0 and (x + 6)(x − 3) = 0, giving x = −6 or x = 3. Check: (−6 − 2)(−6 + 5) = (−8)(−1) = 8 and (3 − 2)(3 + 5) = (1)(8) = 8. The wrong method would have given x = 10 and x = 3; at x = 10 the product is (8)(15) = 120, not 8.

Method map: Make a table with the columns initial form, method, example equation, solutions, and check. Fill six rows using x² = 49, (x − 3)² = 20, x² − 2x − 8 = 0, 6x² + x − 12 = 0, 3x² − 7x + 1 = 0 and x² + 4x + 13 = 0, and in the last column show the substitution of one solution with the arithmetic written out (including i² = −1 where a solution is complex). Beneath the table, write one sentence explaining what the sign of the discriminant tells you about the number and type of solutions.

Independent performance: Write eight quadratic equations of your own, at least one for each initial form in the method map, including two whose solutions are complex and one that is a product of two factors set equal to a nonzero number. Solve each by the method appropriate to its initial form and state in one line why that method was the appropriate choice. Give every solution in exact form, give a four-decimal approximation for every irrational real solution, write every complex solution as a ± bi, and substitute each solution into the original equation to show it works. Then solve one of your real-solution equations a second way (for example by the formula after having factored) and show the two methods agree. Success criteria: no solution is dropped when square roots are taken, the Zero-Product Property is applied only to a product that equals zero, no equation is divided by the variable, and every complex solution is checked with i² = −1 shown.