Complex Roots of Quadratics: Free Response

With $a = 2$, $b = -10$ and $c = 17$, compute the discriminant first, because its sign decides what kind of answer you are heading for.

$\Delta = (-10)^2 - 4(2)(17) = 100 - 136 = -36$

It is negative, so no real number solves this equation and the formula is about to ask for the square root of a negative number. Read that root in $i$-form the moment it appears, $\sqrt{-36} = 6i$, and then carry on exactly as usual with $2a = 4$:

$x = \frac{-(-10) \pm \sqrt{-36}}{2(2)} = \frac{10 \pm 6i}{4}$

The last step is the one that goes wrong most often. The denominator sits under the entire numerator, so both $10$ and $6i$ are divided by $4$, not just the first of them:

$x = \frac{10}{4} \pm \frac{6}{4}i = \frac{5}{2} \pm \frac{3}{2}i$

So the two roots are $\frac{5}{2} + \frac{3}{2}i$ and $\frac{5}{2} - \frac{3}{2}i$, with real part $\frac{5}{2}$ in both and imaginary parts $\frac{3}{2}$ and $-\frac{3}{2}$. Two cheap checks are available and worth the seconds. The roots should add to $-\frac{b}{a} = 5$, and they do, because the imaginary parts cancel. Their product should be $\frac{c}{a} = \frac{17}{2}$, and it is:

$\left(\frac{5}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{25}{4} + \frac{9}{4} = \frac{34}{4} = \frac{17}{2}$

Take the discriminant first again, with $a = 1$, $b = 2$ and $c = 8$:

$\Delta = 2^2 - 4(1)(8) = 4 - 32 = -28$

Now $\sqrt{-28} = i\sqrt{28} = i\sqrt{4 \cdot 7} = 2i\sqrt{7}$, and with $2a = 2$ both terms of the numerator are halved:

$x = \frac{-2 \pm 2i\sqrt{7}}{2} = -1 \pm i\sqrt{7}$

The sum is $-2 = -\frac{b}{a}$ and the product is $1 + 7 = 8 = \frac{c}{a}$, so both roots check out.

Now the comparison. In both equations the imaginary part is

$q = \frac{\sqrt{|\Delta|}}{2a},$

and $2a$ is an integer, so everything turns on whether $\sqrt{|\Delta|}$ is rational. Here $|\Delta|$ is a positive integer, and the square root of a positive integer is either an integer or irrational, never a fraction in between. So, for an equation of this kind with $\Delta < 0$, $q$ is rational exactly when $|\Delta|$ is a perfect square. Both directions of that hold: if $|\Delta| = m^2$ then $\sqrt{|\Delta|} = m$ and $q$ is rational, and if $q$ is rational then $\sqrt{|\Delta|} = 2aq$ is rational and therefore an integer, which makes $|\Delta|$ a perfect square.

That is exactly what separated the two equations:

$|-36| = 6^2, \qquad |-28| \ne m^2 \text{ for any integer } m$

Keep the negative discriminant attached to that statement, because it is what puts the radical in the imaginary part in the first place. When $\Delta > 0$ the same radical lands in the real part instead, and the roots of $x^2 - 3$ have imaginary part $0$ though $12$ is no perfect square. The mechanism is not new: the imaginary part inherits whatever the radical leaves behind, exactly as an irrational real root did when the discriminant was positive.

The formula produces the two roots from one expression, and the only difference between them is the sign in front of the second piece. So whatever the first piece is, both roots contain it unchanged, and whatever the second piece is, one root adds it and the other subtracts it.

The first piece is $-\frac{b}{2a}$. It is built from the coefficients alone, with no radical in it, so it is a real number for every real $a \ne 0$ and $b$, no matter what the discriminant does. It is the axis of symmetry, exactly as it was when the roots were real.

The second piece is where the sign of $\Delta$ lands. When $\Delta < 0$, write $\sqrt{\Delta} = i\sqrt{|\Delta|}$, and the offset becomes

$\frac{\sqrt{\Delta}}{2a} = \frac{\sqrt{|\Delta|}}{2a}\,i,$

which is a real number times $i$. Naming the two pieces $p = -\frac{b}{2a}$ and $q = \frac{\sqrt{|\Delta|}}{2a}$, both of them real, the formula reads

$x = p + qi \quad\text{and}\quad x = p - qi.$

The shared real part is forced by the shape of the formula, and the opposite imaginary parts are forced by the $\pm$. So the answer to the second question is that the centre does not change character at all: it is real before and after. The offset does. When $\Delta > 0$ it is a real distance and the $\pm$ moves the two roots left and right along the number line; when $\Delta < 0$ the same $\pm$ moves them the same distance up and down instead, off the line and on opposite sides of it. The roots still straddle the axis. They have simply stopped straddling it along the line.

Conjugation is a function, so it may be applied to both sides of a true equation and the two sides stay equal. Apply it to $az^2 + bz + c = 0$ and work on each side.

The right side is $0$, a real number, so it is its own conjugate. That is the third fact, used here on the right side and due to return at the end for the coefficients:

$\overline{0} = 0$

The left side is a sum of three terms, and conjugation respects sums, so it passes through both plus signs. That is the first fact:

$\overline{az^2 + bz + c} = \overline{az^2} + \overline{bz} + \overline{c}$

Each of the first two terms is a product, and conjugation respects products, so it passes through those as well. That is the second fact, and it is applied three times here: once to separate $a$ from $z^2$, once to separate $b$ from $z$, and once more on $z^2 = z \cdot z$, which gives $\overline{z^2} = \overline{z}\,\overline{z} = \overline{z}^{\,2}$:

$\overline{az^2} + \overline{bz} + \overline{c} = \overline{a}\,\overline{z}^{\,2} + \overline{b}\,\overline{z} + \overline{c}$

Stop and look at that line, because it is true for every quadratic whatsoever, with any complex coefficients at all. No hypothesis has been spent yet.

The hypothesis is spent on the very next line, and only there. Since $a$, $b$ and $c$ are real, each is its own conjugate, so $\overline{a} = a$, $\overline{b} = b$ and $\overline{c} = c$, and the three barred coefficients turn back into the original three:

$0 = a\overline{z}^{\,2} + b\overline{z} + c$

That expression is the original quadratic evaluated at $\overline{z}$, and it equals $0$, so $\overline{z}$ is a root. Notice what the theorem does and does not say. If $z$ happens to be real then $\overline{z} = z$ and the conclusion is no news. Its whole force is on non-real roots, which it forbids from ever turning up alone.

Hunting for coefficients would be slow. Build the quadratic from the roots you want instead, since a quadratic with roots $r$ and $s$ and leading coefficient $1$ is $(x - r)(x - s)$. The root inventory to aim for is the one the theorem forbids for real coefficients: one real root and one non-real root. Take $r = 2$ and $s = 3i$ and multiply out:

$(x - 2)(x - 3i) = x^2 - 3ix - 2x + 6i = x^2 - (2 + 3i)x + 6i$

Both of the chosen numbers are roots, which the factored form already guarantees, but substitute to be sure. At $x = 2$:

$4 - (2 + 3i)(2) + 6i = 4 - 4 - 6i + 6i = 0$

At $x = 3i$, using $(3i)^2 = -9$ and $(2 + 3i)(3i) = 6i + 9i^2 = -9 + 6i$:

$-9 - (-9 + 6i) + 6i = -9 + 9 - 6i + 6i = 0$

Now test the conjugate of the non-real root. At $x = -3i$, using $(-3i)^2 = -9$ and $(2 + 3i)(-3i) = -6i - 9i^2 = 9 - 6i$:

$-9 - (9 - 6i) + 6i = -18 + 12i$

That is not $0$, so $-3i$ is not a root. The quadratic has exactly two roots, one real and one not, and the non-real one stands alone.

There is no contradiction with part A, because part A was never available here. Its hypothesis is that $a$, $b$ and $c$ are real, and this quadratic has $b = -(2 + 3i)$ and $c = 6i$. The proof used that hypothesis at exactly one line, replacing $\overline{b}$ by $b$ and $\overline{c}$ by $c$. Withdraw it and that line fails, and with it the conclusion. This is why the theorem is worth stating with its hypothesis attached every time: what fails without it is not a technicality but the whole result.

An if and only if is two claims, and they have to be tested one at a time.

The forward direction, that real coefficients force the non-real roots into conjugate pairs, is precisely what part A proves. Nothing more is needed for it.

The converse says that conjugate pairing forces the coefficients to be real, and it is false. The reason is that multiplying an equation through by any nonzero constant leaves its roots exactly where they were, and that constant does not have to be real. Start from a quadratic whose roots really are a conjugate pair:

$x^2 - 6x + 34 = 0 \quad\text{has}\quad x = \frac{6 \pm \sqrt{36 - 136}}{2} = \frac{6 \pm 10i}{2} = 3 \pm 5i$

Now multiply through by $i$:

$ix^2 - 6ix + 34i = 0$

A number satisfies this equation exactly when it satisfies the previous one, since $i \ne 0$, so the roots are still $3 + 5i$ and $3 - 5i$, still a conjugate pair. Yet every one of its three coefficients is non-real. The converse therefore fails.

It fails a second way, more cheaply. The quadratic $i(x - 1)(x - 2) = ix^2 - 3ix + 2i$ has the two real roots $1$ and $2$, so it has no non-real roots at all and the converse's hypothesis holds vacuously, while its conclusion is still false.

What repairs it is a condition on the leading coefficient: with $a$ real and nonzero, a quadratic whose non-real roots are accompanied by their conjugates has real coefficients throughout. Do not assume such a quadratic factors and then read $b$ and $c$ off the factorization; build it instead, since the hypothesis permits only two root inventories and each can be multiplied out directly.

If both roots are real, say $r_1$ and $r_2$, then

$a(x - r_1)(x - r_2) = ax^2 - a(r_1 + r_2)x + ar_1r_2,$

and every coefficient is a product of real numbers. If instead the roots are a conjugate pair $p \pm qi$ with $q \ne 0$, their sum $2p$ and their product $p^2 + q^2$ are real, so

$a\bigl(x - (p + qi)\bigr)\bigl(x - (p - qi)\bigr) = ax^2 - 2apx + a\left(p^2 + q^2\right)$

is real throughout as well.

One case deserves naming, because it looks like a third inventory that would sink the repair: a repeated non-real root, as in $x^2 - 2ix - 1 = (x - i)^2$, whose leading coefficient $1$ is perfectly real. It is not a counterexample, because the hypothesis rules it out rather than covering it. The root $i$ is non-real and its conjugate $-i$ is not a root, since $(-i - i)^2 = -4 \ne 0$, so this quadratic's non-real roots do not come in conjugate pairs. The two counterexamples earlier are excluded from the other side, by their leading coefficient of $i$.

The coefficients are real and one root is not, so the conjugate root theorem applies and hands over the second root at no cost: it is $1 - 4i$. With both roots in hand the two totals finish the job.

The sum of the roots is

$(1 + 4i) + (1 - 4i) = 2,$

and the sum equals $-\frac{b}{a}$, so $-\frac{b}{3} = 2$ and $b = -6$.

The product is a conjugate product, so the $i$ terms cancel and what survives is the squared modulus of either root:

$(1 + 4i)(1 - 4i) = 1^2 + 4^2 = 17$

The product equals $\frac{c}{a}$, so $\frac{c}{3} = 17$ and $c = 51$. The quadratic is $3x^2 - 6x + 51$.

Verify by solving it. The discriminant is $(-6)^2 - 4(3)(51) = 36 - 612 = -576$, and $\sqrt{-576} = 24i$, so

$x = \frac{6 \pm 24i}{6} = 1 \pm 4i,$

returning the given root and its conjugate. One non-real root and a leading coefficient determined all three coefficients.

Run the same machinery and watch it stall. The theorem says that if $z$ is a root then so is $\overline{z}$. Here $z = 2$ is real, and a real number is its own conjugate:

$\overline{2} = 2$

So the theorem is perfectly true and completely uninformative. It returns the root already in hand and supplies no second root at all. Without a second root there is no sum and no product, so nothing determines $b$ and $c$.

That is not a gap in the argument; it is a fact about the quadratics themselves. Every quadratic $3(x - 2)(x - r)$ with $r$ real has real coefficients and has $2$ as a root, and different choices of $r$ give genuinely different quadratics:

$3(x - 2)(x - 5) = 3x^2 - 21x + 30, \qquad 3(x - 2)(x + 5) = 3x^2 + 9x - 30$

Both have $2$ as a root, both have real coefficients and leading coefficient $3$, and they are not the same polynomial, so the data cannot possibly determine one answer. Infinitely many more come from other choices of $r$.

The contrast with part A is the whole point, and it is not a contrast in difficulty. There the theorem manufactured a second root, and two roots plus a leading coefficient fix everything. Here it manufactures nothing, and one root plus a leading coefficient leaves a whole family standing. The theorem's force was never in the pairing rule as such; it is in what the pairing rule gives you when the root you are handed is not its own conjugate.

Write the root as $z = p + qi$ with $p$ and $q$ real and $q \ne 0$, since $z$ is not real. The coefficients are real, so the other root is $\overline{z} = p - qi$, and being monic the quadratic is $x^2 - (\text{sum})x + (\text{product})$.

The product of the pair is the squared modulus, and the modulus is exactly the datum you were given:

$c = z\overline{z} = p^2 + q^2 = |z|^2 = 7^2 = 49$

So the constant term is pinned down completely, without knowing either part of the root separately. The distance from the origin is precisely the quantity the product measures.

The sum is a different quantity, and the datum does not determine it, though it does not leave it entirely free either. The sum is $2p$, so

$b = -2p,$

and $p$ is not determined by $|z| = 7$. All that is forced is $p^2 + q^2 = 49$ with $q \ne 0$, which gives $p^2 < 49$, so $p$ may be anywhere strictly between $-7$ and $7$, and $b$ anywhere strictly between $-14$ and $14$. Two members of the family show the spread:

$x^2 + 49 \quad\text{with roots } \pm 7i, \qquad x^2 - 10x + 49 \quad\text{with roots } 5 \pm 2i\sqrt{6}$

Both are monic with real coefficients, and in both cases the root sits at distance $7$ from the origin, since $0^2 + 7^2 = 49$ and $5^2 + \left(2\sqrt{6}\right)^2 = 25 + 24 = 49$.

The difference between the two coefficients is therefore a difference between the two totals. The product depends only on how far the root is from the origin, which you were told; the sum depends on where it sits along that circle of possibilities, which you were not, and all the datum can do for the sum is confine it to a range.

The discriminant settles the first question on its own, with $a = 8$, $b = 4$ and $c = 1$:

$\Delta = 4^2 - 4(8)(1) = 16 - 32 = -16$

It is negative and the coefficients are real, so the roots are a non-real conjugate pair. Now the two totals, which are read straight off the coefficients and need no roots at all:

$r_1 + r_2 = -\frac{b}{a} = -\frac{4}{8} = -\frac{1}{2}, \qquad r_1 r_2 = \frac{c}{a} = \frac{1}{8}$

The third quantity is the product again, wearing a different hat. The roots are $p + qi$ and $p - qi$, so their product is a conjugate product:

$(p + qi)(p - qi) = p^2 - q^2i^2 = p^2 + q^2 = |p + qi|^2$

That is the squared modulus, which is the squared distance from the origin, and the two roots share it because $|p - qi| = |p + qi|$. So each root sits at squared distance $\frac{1}{8}$ from the origin, and no solving was needed to say so. Notice also that the product came out positive, which it had to: $p^2 + q^2$ is a sum of squares with $q \ne 0$.

The discriminant is already known to be $-16$, so $\sqrt{-16} = 4i$ and, dividing the whole numerator by $2a = 16$:

$x = \frac{-4 \pm 4i}{16} = -\frac{1}{4} \pm \frac{1}{4}i$

A quadratic with leading coefficient $a$ and roots $z$ and $\overline{z}$ is $a(x - z)(x - \overline{z})$, and nothing in that changes when the roots are not real. With $z = -\frac{1}{4} + \frac{1}{4}i$:

$8x^2 + 4x + 1 = 8\left(x + \frac{1}{4} - \frac{1}{4}i\right)\left(x + \frac{1}{4} + \frac{1}{4}i\right)$

Expanding is the check. Group the two brackets around $x + \frac{1}{4}$ so the expansion is a difference of squares:

$8\left[\left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{4}i\right)^2\right] = 8\left[x^2 + \frac{1}{2}x + \frac{1}{16} + \frac{1}{16}\right] = 8x^2 + 4x + 1$

The two totals from part A agree with the roots you just found: they add to $-\frac{1}{2}$ because the imaginary parts cancel, and they multiply to $\frac{1}{16} + \frac{1}{16} = \frac{1}{8}$. Over the real numbers this quadratic was irreducible, since it has no real root and therefore no real linear factor. Over the complex numbers it splits, like every quadratic.

Start with the half that is right, because it is genuinely proved. If a real quadratic has non-real roots then they are a conjugate pair $p \pm qi$ with $q \ne 0$, so

$\frac{c}{a} = p^2 + q^2 > 0,$

and $a$ and $c$ must share a sign. That is a true implication, and it is what part A observed.

The classmate then read it in the other direction, and an implication does not travel backwards for free. A positive product says only that the two roots are not of opposite signs, and two real roots of the same sign multiply to a positive number just as happily as a conjugate pair does. The stated example settles it, and it is worth noticing that it has the same $a$ and the same $c$ as part A, so it satisfies the classmate's condition exactly:

$8x^2 - 6x + 1: \quad \Delta = (-6)^2 - 4(8)(1) = 36 - 32 = 4 > 0$

A positive discriminant means two distinct real roots, and they are

$x = \frac{6 \pm 2}{16} = \frac{1}{2} \quad\text{and}\quad \frac{1}{4},$

both real, both positive, and multiplying to $\frac{1}{8}$, the same product as before. So the condition on the signs of $a$ and $c$ cannot possibly decide the question: two quadratics that agree on $a$ and $c$ can disagree about where their roots live, because $b$ is free to change and the discriminant depends on it.

The accurate statement of what survives is the contrapositive. If $\frac{c}{a} \le 0$ then the roots are certainly real, since $\Delta = b^2 - 4ac \ge b^2 \ge 0$ in that case. So the sign of $\frac{c}{a}$ can rule non-real roots out but can never rule them in. Only the discriminant does that.

With $a = 2$, $b = 8$ and $c = k$, the discriminant is a function of $k$:

$\Delta = 8^2 - 4(2)(k) = 64 - 8k$

The coefficients are real, so the roots are non-real exactly when $\Delta < 0$. Solve that inequality:

$64 - 8k < 0 \quad\Longleftrightarrow\quad 64 < 8k \quad\Longleftrightarrow\quad k > 8$

Both directions hold, since every step is reversible, so $k > 8$ is not merely sufficient but exactly the condition.

At $k = 12$ the discriminant is $64 - 96 = -32$, and $\sqrt{-32} = i\sqrt{16 \cdot 2} = 4i\sqrt{2}$. Dividing the whole numerator by $2a = 4$:

$x = \frac{-8 \pm 4i\sqrt{2}}{4} = -2 \pm i\sqrt{2}$

Check against the totals: the sum should be $-\frac{b}{a} = -4$, and the two roots do add to $-4$; the product should be $\frac{c}{a} = 6$, and $(-2)^2 + \left(\sqrt{2}\right)^2 = 4 + 2 = 6$.