Complex Roots of a Quadratic

A curve that misses the axis

The curve $y=x^2-1$y = x² − 1 crosses the x-axis at x = −1 and x = 1, so $x^2-1=0$x² − 1 = 0 has those two roots. Lift the curve by 2 and it becomes $y=x^2+1$y = x² + 1. Now its lowest point is (0, 1), the curve never comes down to the axis, and $x^2+1=0$x² + 1 = 0 has no real root.

The discriminant says the same. For $x^2+1$x² + 1, a = 1, b = 0 and c = 1, so $b^2-4ac=0-4=-4$b² − 4ac = 0 − 4 = −4. A negative discriminant means no real roots, because the quadratic formula needs its square root.

The curve $y=x^2-1$y = x² − 1, with its lowest point as the handle. Slide it up and the two crossings close in, meet at 0 when the curve is $y=x^2$y = x², and leave the axis above that. When the curve is $y=x^2+1$y = x² + 1, its roots are $\pm i$±i.

Solve it anyway

$x^2+1=0$x² + 1 = 0 gives $x^2=-1$x² = −1. The number i is defined by $i^2=-1$i² = −1, so x = i is a root. So is x = −i, because ${\left(-i\right)}^2=i^2=-1$(−i)² = i² = −1. The equation has two roots, i and −i.

In the same way $x^2+9=0$x² + 9 = 0 gives $x^2=-9$x² = −9 and $x=\pm 3i$x = ±3i, since ${\left(3i\right)}^2=9i^2=-9$(3i)² = 9i² = −9. In general $x^2+k^2=0$x² + k² = 0 has the roots ki and −ki.

The formula with a negative discriminant

Solve $x^2-4x+13=0$x² − 4x + 13 = 0. The discriminant is $b^2-4ac=16-52=-36$b² − 4ac = 16 − 52 = −36, and $\sqrt{-36}=6i$√(−36) = 6i. The quadratic formula gives $x=\frac{4\pm 6i}{2}=2\pm 3i$x = (4 ± 6i)/2 = 2 ± 3i.

Check 2 + 3i by substitution. ${\left(2+3i\right)}^2=4+12i+9i^2=-5+12i$(2 + 3i)² = 4 + 12i + 9i² = −5 + 12i, and −4(2 + 3i) = −8 − 12i. Adding these and 13: (−5 − 8 + 13) + (12 − 12)i = 0. The same working with −3i in place of 3i shows that 2 − 3i is a root too.

The sign of the discriminant sorts the cases: positive gives two real roots, zero gives one repeated root, and negative gives two non-real roots.

The curve $y=x^2-4x+13$y = x² − 4x + 13. Its lowest point is (2, 9), 9 above the x-axis, so the curve never meets the axis and the equation has no real root. Its roots are $2\pm 3i$2 ± 3i.

Completing the square

Completing the square shows the same pair. $x^2-4x+13={\left(x-2\right)}^2+9$x² − 4x + 13 = (x − 2)² + 9, so the equation is ${\left(x-2\right)}^2=-9$(x − 2)² = −9. Then $x-2=\pm 3i$x − 2 = ±3i and $x=2\pm 3i$x = 2 ± 3i.

The two descriptions match. The real part 2 is the x-coordinate of the lowest point, and $9=3^2$9 = 3² is its height above the axis. The formula works for the same reason: it comes from completing the square, which never needs the square root to be real.

Why the roots are a conjugate pair

The formula gives $x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$x = −b/2a ± √(b² − 4ac)/2a. When the discriminant is negative, $\sqrt{b^2-4ac}=i\sqrt{4ac-b^2}$√(b² − 4ac) = i√(4ac − b²), with a real square root. So the two roots are p + qi and p − qi, where $p=-\frac{b}{2a}$p = −b/2a and $q=\frac{\sqrt{4ac-b^2}}{2a}$q = √(4ac − b²)/2a: the same real part and opposite imaginary parts. They are conjugates.

That needs the coefficients to be real. $x^2-\left(3i\right)x-2=0$x² − (3i)x − 2 = 0 has a non-real coefficient, 3i, and its roots are i and 2i: $i^2-3i\times i-2=-1+3-2=0$i² − 3i × i − 2 = −1 + 3 − 2 = 0 and ${\left(2i\right)}^2-3i\times 2i-2=-4+6-2=0$(2i)² − 3i × 2i − 2 = −4 + 6 − 2 = 0. They are not a conjugate pair, and −i is not a root.

Check with the sum and the product

For $a{x}^2+bx+c=0$ax² + bx + c = 0 the roots add to $-\frac{b}{a}$−b/a and multiply to $\frac{c}{a}$c/a. For $x^2-4x+13=0$x² − 4x + 13 = 0, $\left(2+3i\right)+\left(2-3i\right)=4=-\frac{b}{a}$(2 + 3i) + (2 − 3i) = 4 = −b/a, and $\left(2+3i\right)\left(2-3i\right)=4+9=13=\frac{c}{a}$(2 + 3i)(2 − 3i) = 4 + 9 = 13 = c/a. So $x^2-4x+13=\left(x-\left(2+3i\right)\right)\left(x-\left(2-3i\right)\right)$x² − 4x + 13 = (x − (2 + 3i))(x − (2 − 3i)).

Two more

$x^2+2x+5=0$x² + 2x + 5 = 0: the discriminant is 4 − 20 = −16 and $\sqrt{-16}=4i$√(−16) = 4i, so $x=\frac{-2\pm 4i}{2}=-1\pm 2i$x = (−2 ± 4i)/2 = −1 ± 2i. Check: ${\left(-1+2i\right)}^2=1-4i-4=-3-4i$(−1 + 2i)² = 1 − 4i − 4 = −3 − 4i, and −3 − 4i + 2(−1 + 2i) + 5 = 0.

$2x^2-2x+5=0$2x² − 2x + 5 = 0: the discriminant is 4 − 40 = −36, so $x=\frac{2\pm 6i}{4}=\frac{1}{2}\pm \left(\frac{3}{2}\right)i$x = (2 ± 6i)/4 = ½ ± (3/2)i. Both parts are divided by 2a = 4. Check: ${\left(\frac{1}{2}+\left(\frac{3}{2}\right)i\right)}^2=\frac{1}{4}+\left(\frac{3}{2}\right)i-\frac{9}{4}=-2+\left(\frac{3}{2}\right)i$(½ + (3/2)i)² = ¼ + (3/2)i − 9/4 = −2 + (3/2)i, so $2x^2=-4+3i$2x² = −4 + 3i, and $-4+3i-2\left(\frac{1}{2}+\left(\frac{3}{2}\right)i\right)+5=-4+3i-1-3i+5=0$−4 + 3i − 2(½ + (3/2)i) + 5 = −4 + 3i − 1 − 3i + 5 = 0.

The usual mistakes

Giving real roots. $x^2+9=0$x² + 9 = 0 does not have roots $\pm 3$±3: ${\left(\pm 3\right)}^2=9$(±3)² = 9, so $\pm 3$±3 solve $x^2-9=0$x² − 9 = 0, the curve that does cross the axis.

Stopping at "no solution". There is no real solution, but the formula still gives two complex roots: $x^2+9=0$x² + 9 = 0 has roots $\pm 3i$±3i.

Dividing only part of the top by 2a. $\frac{4\pm 6i}{2}$(4 ± 6i)/2 is $2\pm 3i$2 ± 3i, not $2\pm 6i$2 ± 6i.

Losing the sign of −b. For $x^2-4x+13=0$x² − 4x + 13 = 0, −b = 4, so the real part is 2, not −2.

A fountain under a glass roof

In the application below, a jet of water has height $6x-x^2$6x − x² meters, and the roof is 13 m up. Setting the height equal to 13 gives a quadratic with a negative discriminant. Its roots are not real, and that is the answer: no point of the jet reaches the roof.