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Every quadratic has roots

Numbers · level 7, lesson 5 of 6

A question to start

Draw the graph of \(y = x^2 + 2x + 5\). It is a U-shaped parabola whose lowest point is at \((-1, 4)\), so the whole curve floats above the \(x\)-axis and never touches it. The equation \(x^2 + 2x + 5 = 0\) has no real solutions.

Yet with complex numbers it has exactly two:

\[x = -1 + 2i \qquad\text{and}\qquad x = -1 - 2i.\]

Look at the real part, \(-1\). That is exactly the \(x\)-coordinate of the lowest point of the graph. Coincidence? And why do the two answers come as a mirror pair, \(+2i\) and \(-2i\)? This lesson answers both questions.

The idea

Completing the square works in \(\mathbb{C}\)

Solve \(x^2 + 2x + 5 = 0\) by completing the square: rewriting the \(x^2\) and \(x\) terms as a perfect square plus a constant. Since \((x + 1)^2 = x^2 + 2x + 1\),

\[x^2 + 2x + 5 = (x + 1)^2 + 4.\]

So the equation says \((x + 1)^2 = -4\). In the real numbers we stop here: a square cannot be \(-4\). In the complex numbers, \(-4\) has two square roots, \(2i\) and \(-2i\) (lesson 2). So

\[x + 1 = \pm 2i \qquad\Longrightarrow\qquad x = -1 \pm 2i.\]

Check \(x = -1 + 2i\): \(x^2 = 1 - 4i + 4i^2 = -3 - 4i\), and \(2x = -2 + 4i\). So \(x^2 + 2x + 5 = (-3 - 4i) + (-2 + 4i) + 5 = 0\). ✓

The quadratic formula, with nothing left out

The same steps on the general equation \(ax^2 + bx + c = 0\) (with \(a \ne 0\)) give the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

The derivation (divide by \(a\), complete the square, take square roots) uses only the rules of arithmetic, which all still hold in \(\mathbb{C}\). The only step that could fail is "take the square root of \(b^2 - 4ac\)". In \(\mathbb{R}\) it fails when \(b^2 - 4ac < 0\). In \(\mathbb{C}\) it never fails, because every real number, positive or negative, has square roots there. So:

Every quadratic equation with real coefficients has solutions in \(\mathbb{C}\).

(Coefficients are the numbers \(a\), \(b\), \(c\) in front of the powers of \(x\).) In fact the same is true for quadratics whose coefficients are themselves complex; see the last section.

The discriminant, read again

The number under the root, \(D = b^2 - 4ac\), is the discriminant: it "discriminates" between the kinds of answer. For real \(a\), \(b\), \(c\):

Discriminant Solutions Graph of \(y = ax^2 + bx + c\)
\(D > 0\) two different real numbers crosses the \(x\)-axis twice
\(D = 0\) one real number, repeated: \(x = -\dfrac{b}{2a}\) touches the \(x\)-axis at its vertex
\(D < 0\) two non-real complex numbers, \(-\dfrac{b}{2a} \pm \dfrac{\sqrt{-D}}{2a}\,i\) misses the \(x\)-axis

In the old real-number view, the bottom row said "no solutions". Now it says "two solutions, both off the real line". The total is always two, if a repeated solution is counted twice. A quadratic always has two roots; the discriminant only tells you where they are.

Why the real part is the vertex

When \(D < 0\), both roots have real part \(-\dfrac{b}{2a}\). That is the \(x\)-coordinate of the vertex of the parabola, which sits on its line of symmetry. So the two complex roots sit directly "above and below" the line of symmetry, in the complex plane.

(−1, 4) −1 y = x² + 2x + 5 x −1 + 2i −1 − 2i real imag.

Why non-real roots come in conjugate pairs

The two roots \(-1 + 2i\) and \(-1 - 2i\) are conjugates: reflections of each other in the real axis. This is no accident, and the reason works for polynomials of any degree, not just quadratics.

Let \(p(x) = a_n x^n + \cdots + a_1 x + a_0\) be a polynomial whose coefficients \(a_0, \ldots, a_n\) are all real. Suppose \(p(z) = 0\) for some complex \(z\). Take the conjugate of both sides. In lesson 3 we proved that the conjugate of a sum is the sum of the conjugates and the conjugate of a product is the product of the conjugates, so \(\overline{z^k} = \bar z^{\,k}\); and a real coefficient is its own conjugate. Therefore

\[0 = \bar 0 = \overline{p(z)} = \overline{a_n z^n + \cdots + a_0} = a_n \bar z^{\,n} + \cdots + a_0 = p(\bar z).\]

So \(\bar z\) is a root too. ∎

If a polynomial has real coefficients, its non-real roots come in conjugate pairs.

Two consequences:

  • If you know one non-real root, you know another one for free.
  • A real polynomial of odd degree (a cubic, a quintic, ...) always has at least one real root. Its roots, counted properly, number an odd amount (level 8 shows a degree-\(n\) polynomial has \(n\) of them), and the non-real ones pair off, so at least one is left over and must be its own conjugate, that is, real.

The proof needed the coefficients to be real. With complex coefficients the pairing can fail: \((x - i)(x - 2) = x^2 - (2 + i)x + 2i\) has roots \(i\) and \(2\), which are not a conjugate pair.

Sum and product of the roots

If the roots of \(ax^2 + bx + c = 0\) are \(r\) and \(s\), then \(ax^2 + bx + c = a(x - r)(x - s) = a\big(x^2 - (r + s)x + rs\big)\). Comparing coefficients,

\[r + s = -\frac ba \qquad rs = \frac ca\]

These hold in \(\mathbb{C}\) as well. For a conjugate pair \(r = p + qi\), \(s = p - qi\): the sum is \(2p\) and the product is \(p^2 + q^2 = |r|^2\), both real, as they must be. This lets you build a quadratic from a root: for the root \(3 - 2i\) (and hence also \(3 + 2i\)), the sum is 6 and the product is \(9 + 4 = 13\), so the equation is \(x^2 - 6x + 13 = 0\).

The discriminant in the complex world

What if \(a\), \(b\), \(c\) are complex? The quadratic formula still works; the only new task is finding a square root of a complex number \(D\), which you did in lesson 3 by writing \((x + yi)^2 = D\) and comparing parts. So every quadratic, with any complex coefficients, has two roots in \(\mathbb{C}\) (counted with repetition). The discriminant now has only two cases: \(D = 0\) (one repeated root) and \(D \ne 0\) (two different roots). "Positive" and "negative" no longer make sense, because complex numbers have no order (lesson 1). Nothing outside \(\mathbb{C}\) is ever needed; level 8 shows this is true for polynomials of every degree.

Worked examples

1. Solve \(2x^2 - 6x + 5 = 0\).

  1. \(a = 2\), \(b = -6\), \(c = 5\). \(D = 36 - 40 = -4\), so the roots are non-real.
  2. \(\sqrt{-4} = 2i\). So \(x = \dfrac{6 \pm 2i}{4} = \dfrac32 \pm \dfrac12 i\).
  3. Check with sum and product: sum \(= 3 = -\dfrac{-6}{2}\) ✓; product \(= \dfrac94 + \dfrac14 = \dfrac{10}{4} = \dfrac52 = \dfrac ca\) ✓.

2. Classify without solving. What kind of roots do these have?

  1. \(x^2 - 5x + 3 = 0\): \(D = 25 - 12 = 13 > 0\), two different real roots.
  2. \(4x^2 + 4x + 1 = 0\): \(D = 16 - 16 = 0\), one repeated real root, \(x = -\tfrac12\).
  3. \(x^2 + x + 1 = 0\): \(D = 1 - 4 = -3 < 0\), two non-real conjugate roots, \(-\tfrac12 \pm \tfrac{\sqrt3}{2}i\).

3. Build an equation. Find a quadratic with real coefficients that has \(2 + 5i\) as a root.

  1. The other root must be the conjugate, \(2 - 5i\).
  2. Sum: \(4\). Product: \(4 + 25 = 29\).
  3. Equation: \(x^2 - 4x + 29 = 0\). (Any multiple, such as \(3x^2 - 12x + 87 = 0\), also works.)

4. A cubic with one known root. Solve \(x^3 - 3x^2 + 7x - 5 = 0\), given that \(x = 1\) is a root.

  1. Since \(x = 1\) is a root, \(x - 1\) is a factor. Divide: \(x^3 - 3x^2 + 7x - 5 = (x - 1)(x^2 - 2x + 5)\). Check by expanding: \(x^3 - 2x^2 + 5x - x^2 + 2x - 5 = x^3 - 3x^2 + 7x - 5\). ✓
  2. Solve \(x^2 - 2x + 5 = 0\): \(D = 4 - 20 = -16\), so \(x = \dfrac{2 \pm 4i}{2} = 1 \pm 2i\).
  3. The roots are \(1\), \(1 + 2i\), \(1 - 2i\): one real root and a conjugate pair, as the odd-degree argument promised.

5. Complex coefficients. Solve \(z^2 - 3z + (3 - i) = 0\).

  1. \(D = 9 - 4(3 - i) = 9 - 12 + 4i = -3 + 4i\).
  2. A square root of \(-3 + 4i\): from lesson 3 (Try it question 7), \((1 + 2i)^2 = -3 + 4i\).
  3. \(z = \dfrac{3 \pm (1 + 2i)}{2}\), so \(z = 2 + i\) or \(z = 1 - i\).
  4. Check with sum and product: \((2 + i) + (1 - i) = 3\) ✓ and \((2 + i)(1 - i) = 2 - 2i + i - i^2 = 3 - i\) ✓. The roots are not conjugates, which is fine: the coefficients are not real.

Try it

  1. Solve \(x^2 - 6x + 13 = 0\).
  2. Solve \(x^2 - 4x + 8 = 0\) by completing the square.
  3. Solve \(3x^2 + x + 2 = 0\).
  4. Without solving, describe the roots of \(9x^2 - 12x + 4 = 0\), \(x^2 + x - 1 = 0\) and \(5x^2 + 2x + 1 = 0\).
  5. Find a quadratic with real coefficients and a root \(-3 + i\).
  6. Solve \(x^3 + 8 = 0\) completely. (Hint: \(x = -2\) is a root.)
  7. For which real numbers \(k\) does \(x^2 + kx + 9 = 0\) have non-real roots?
  8. A cubic with real coefficients has roots \(2\) and \(1 - i\). What is its third root? Write the cubic with leading coefficient 1.
Answers
  1. \(D = 36 - 52 = -16\), \(x = \dfrac{6 \pm 4i}{2} = 3 \pm 2i\).
  2. \(x^2 - 4x + 8 = (x - 2)^2 + 4\), so \((x - 2)^2 = -4\), \(x - 2 = \pm 2i\), \(x = 2 \pm 2i\).
  3. \(D = 1 - 24 = -23\), \(x = \dfrac{-1 \pm i\sqrt{23}}{6}\).
  4. \(D = 144 - 144 = 0\): one repeated real root, \(x = \tfrac{12}{18} = \tfrac23\). \(D = 1 + 4 = 5\): two different real roots. \(D = 4 - 20 = -16\): two non-real conjugate roots (\(-\tfrac15 \pm \tfrac25 i\)).
  5. The other root is \(-3 - i\). Sum \(-6\), product \(9 + 1 = 10\): \(x^2 + 6x + 10 = 0\).
  6. \(x^3 + 8 = (x + 2)(x^2 - 2x + 4)\) (check: \(x^3 - 2x^2 + 4x + 2x^2 - 4x + 8\) ✓). The quadratic has \(D = 4 - 16 = -12\), so \(x = \dfrac{2 \pm 2i\sqrt3}{2} = 1 \pm i\sqrt3\). Roots: \(-2\), \(1 + i\sqrt3\), \(1 - i\sqrt3\).
  7. Non-real roots need \(D = k^2 - 36 < 0\), that is \(k^2 < 36\): \(-6 < k < 6\).
  8. Real coefficients, so \(1 + i\) is also a root. The cubic is \((x - 2)(x - (1 - i))(x - (1 + i))\). The last two multiply to \(x^2 - 2x + 2\) (sum 2, product \(1 + 1 = 2\)). Then \((x - 2)(x^2 - 2x + 2) = x^3 - 4x^2 + 6x - 4\).

Common mistake

Writing the roots with the wrong denominator. In \(x = \dfrac{-b \pm \sqrt D}{2a}\) the \(2a\) divides both parts. For \(2x^2 - 6x + 5 = 0\) the roots are \(\dfrac{6 \pm 2i}{4} = \dfrac32 \pm \dfrac12 i\), not \(6 \pm \dfrac{2i}{4}\) and not \(\dfrac64 \pm 2i\).

Using the conjugate pair rule when the coefficients are not real. The pairing came from taking conjugates of real coefficients. For an equation like \(z^2 - 3z + (3 - i) = 0\) the roots (\(2 + i\) and \(1 - i\)) are not conjugates, and assuming they are gives wrong answers.

Where it leads