Solve quadratic equations with complex solutions, extend polynomial identities to the complex numbers, and show the Fundamental Theorem of Algebra holds for quadratics

Learning goal: Solve quadratic equations with real coefficients that have complex solutions; extend polynomial identities to the complex numbers, for example rewriting x² + 4 as (x + 2i)(x − 2i); know the Fundamental Theorem of Algebra and show that it is true for quadratic polynomials.

Scope note. N.CN.8 and N.CN.9 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

When a quadratic has no real solution. For ax² + bx + c = 0 with real coefficients and a ≠ 0, the quadratic formula gives x = (−b ± √(b² − 4ac))/(2a). The quantity D = b² − 4ac is the discriminant. If D > 0 there are two different real solutions; if D = 0 there is one real solution, counted twice; if D < 0 the square root is of a negative number, and before this unit that meant "no solution." With i available, √D = i√(−D), and the formula produces two complex solutions. Because the coefficients are real, those two solutions are conjugates of each other: (−b + i√(−D))/(2a) and (−b − i√(−D))/(2a) have the same real part −b/(2a) and opposite imaginary parts. So non-real solutions of a real-coefficient quadratic always come in conjugate pairs; a single non-real solution on its own is impossible and signals an arithmetic error.

A worked model inside the lesson. Solve x² − 4x + 13 = 0. Here a = 1, b = −4, c = 13, so D = 16 − 52 = −36. Then √D = √(−36) = 6i, and x = (4 ± 6i)/2 = 2 ± 3i. Check x = 2 + 3i by substitution, which is the step students most often skip: (2 + 3i)² = 4 + 12i + 9i² = −5 + 12i; −4(2 + 3i) = −8 − 12i; adding 13: (−5 + 12i) + (−8 − 12i) + 13 = (−5 − 8 + 13) + (12 − 12)i = 0 + 0i = 0. The check works because the imaginary parts cancel and the real parts sum to zero; both must happen. The solutions are 2 + 3i and 2 − 3i, a conjugate pair with real part 2 = −b/(2a).

Completing the square also works. Solve x² − 6x + 25 = 0 by completing the square: x² − 6x + 9 = −25 + 9, so (x − 3)² = −16, so x − 3 = ±4i and x = 3 ± 4i. The discriminant confirms: 36 − 100 = −64, √(−64) = 8i, and (6 ± 8i)/2 = 3 ± 4i. When the leading coefficient is not 1, keep it: 2x² − 2x + 5 = 0 has D = 4 − 40 = −36, so x = (2 ± 6i)/4 = 1/2 ± (3/2)i. A frequent error is dividing only the real part by 4; both parts of 2 ± 6i are divided.

Extending polynomial identities to the complex numbers. Over the real numbers, x² − 4 factors as (x − 2)(x + 2), but x² + 4 does not factor over the reals, because x² + 4 ≥ 4 > 0 for every real x, so it has no real zeros (a sum of squares x² + c² with a nonzero constant c has no real zeros; x² + y² as a polynomial in two variables is 0 at x = y = 0). Over the complex numbers it does: x² + 4 = (x + 2i)(x − 2i). Verify by expanding: (x + 2i)(x − 2i) = x² − 2ix + 2ix − 4i² = x² + 4, since −4i² = 4. This is the difference-of-squares identity x² − y² = (x − y)(x + y) with y = 2i, because (2i)² = −4 and x² − (−4) = x² + 4. In general, x² + y² = (x + yi)(x − yi) for any real x and y, an identity that has no counterpart in real arithmetic. The same idea factors x² + 9 = (x + 3i)(x − 3i), x² + 2 = (x + i√2)(x − i√2), and even numbers: 25 = 3² + 4² = (3 + 4i)(3 − 4i). A quadratic with a linear term factors over the complex numbers too, using its solutions: x² − 2x + 2 has D = 4 − 8 = −4, solutions 1 ± i, and factors as (x − 1 − i)(x − 1 + i). Check: ((x − 1) − i)((x − 1) + i) = (x − 1)² − i² = (x − 1)² + 1 = x² − 2x + 2. And higher-degree identities extend the same way: x⁴ − 16 = (x² − 4)(x² + 4) = (x − 2)(x + 2)(x − 2i)(x + 2i), four linear factors over the complex numbers where the real numbers allowed only two.

The Fundamental Theorem of Algebra. The theorem states: every polynomial of degree at least 1 with complex coefficients has at least one zero in the complex numbers. Real coefficients count, since real numbers are complex numbers. A consequence, obtained by dividing out the zero and repeating, is that a polynomial of degree n has exactly n complex zeros when each zero is counted with its multiplicity, and so factors completely into n linear factors over the complex numbers. The theorem is why the complex numbers are the natural home for solving polynomial equations: no further enlargement of the number system is ever needed for that purpose. The full proof is beyond this course; what the standard asks is to know the statement and to show it is true for quadratics.

Showing the theorem is true for quadratic polynomials. Take any ax² + bx + c with real coefficients and a ≠ 0. The quadratic formula produces a solution in every case, because the discriminant D is a real number and every real number has a complex square root: if D ≥ 0, √D is real; if D < 0, √D = i√(−D). So x = (−b + √D)/(2a) is always a complex number satisfying the equation, which is the "at least one zero" the theorem promises. Counting with multiplicity, there are exactly two: D > 0 gives two distinct real zeros, as with x² − 5x + 6 = (x − 2)(x − 3); D = 0 gives one zero of multiplicity 2, as with x² + 6x + 9 = (x + 3)², where −3 counts twice; D < 0 gives two distinct non-real zeros, as with x² + 4 = (x + 2i)(x − 2i). In all three cases the quadratic factors into two linear factors, exactly as the theorem's consequence predicts for degree 2. If the coefficients themselves are complex, the same formula still applies because every complex number has a square root, though finding it takes more work; this unit's examples keep the coefficients real.

A cube root example that ties the unit together. The equation x³ = 8 has the real solution 2, but the theorem says degree 3 means three zeros. Factor: x³ − 8 = (x − 2)(x² + 2x + 4). The quadratic factor has D = 4 − 16 = −12, so x = (−2 ± 2√3 i)/2 = −1 ± √3 i. The three cube roots of 8 are 2, −1 + √3 i and −1 − √3 i. The second is the number whose cube was shown to be 8 in the previous lesson by rotating through 120° three times; the third is its conjugate. Three zeros, as promised, and every non-real one paired with its conjugate.

A misconception about "no solution." The statement "x² + 1 = 0 has no solution" is true only with the qualifier "in the real numbers." Over the complex numbers it has exactly two solutions, i and −i. When a problem says "solve," the number system in use decides the answer, and in this unit that system is the complex numbers.

Worked quadratic with a conjugate pair: Solve x² + 2x + 5 = 0. D = 4 − 20 = −16, so √D = 4i and x = (−2 ± 4i)/2 = −1 ± 2i. Check x = −1 + 2i: (−1 + 2i)² = 1 − 4i + 4i² = −3 − 4i; 2(−1 + 2i) = −2 + 4i; sum with 5: (−3 − 2 + 5) + (−4 + 4)i = 0. The factored form over the complex numbers is (x + 1 − 2i)(x + 1 + 2i).

Worked leading coefficient not 1: Solve 4x² − 4x + 5 = 0. D = 16 − 80 = −64, so √D = 8i and x = (4 ± 8i)/8 = 1/2 ± i. Check x = 1/2 + i: (1/2 + i)² = 1/4 + i + i² = −3/4 + i; times 4 gives −3 + 4i; −4(1/2 + i) = −2 − 4i; sum with 5: (−3 − 2 + 5) + (4 − 4)i = 0.

Worked identity: Rewrite x² + 9 as a product of linear factors and verify. x² + 9 = x² − (3i)² = (x + 3i)(x − 3i). Expand: x² − 3ix + 3ix − 9i² = x² + 9, because −9i² = 9. So the sum of squares x² + 9, which has no real factorization, factors over the complex numbers with zeros ±3i.

Worked Fundamental Theorem check for all three discriminant cases: For x² − 5x + 6, D = 25 − 24 = 1 > 0, zeros 2 and 3, factors (x − 2)(x − 3): two real zeros. For x² + 6x + 9, D = 36 − 36 = 0, zero −3 of multiplicity 2, factors (x + 3)²: two zeros counting multiplicity. For x² − 4x + 13, D = −36 < 0, zeros 2 ± 3i, factors (x − 2 − 3i)(x − 2 + 3i): two non-real conjugate zeros. Each quadratic has exactly two complex zeros counted with multiplicity and factors into two linear factors, so the theorem holds in every case for degree 2.

Discriminant-case table: Make a table with columns equation, discriminant, number and type of zeros, zeros, and factored form over the complex numbers. Fill it for x² − 5x + 6 = 0, x² + 6x + 9 = 0, x² + 4 = 0, x² − 4x + 13 = 0, x² + 2x + 5 = 0 and 2x² − 2x + 5 = 0. Check every non-real zero by substitution, showing that both the real part and the imaginary part of the result are 0. Beneath the table, write two sentences: one stating the Fundamental Theorem of Algebra and one explaining how the table shows it is true for quadratics.

Independent performance: Write a quadratic equation with real coefficients whose solutions are 3 ± 5i, by expanding (x − 3 − 5i)(x − 3 + 5i), and then solve your own equation with the quadratic formula to confirm the solutions come back. Next, factor x⁴ − 81 completely over the complex numbers and state its four zeros. Finally, write a short explanation for a classmate of why the sum of two squares x² + y² factors over the complex numbers but not over the real numbers, including the expansion of (x + yi)(x − yi). Success criteria: your equation has integer coefficients and the formula returns exactly 3 ± 5i; all four zeros of x⁴ − 81 are checked by substitution or by multiplying the factors back; the explanation uses i² = −1 explicitly.