Roots of Negative Numbers

A square is never negative

Squaring multiplies a number by itself. A positive number times a positive number is positive: $3^{2} = 9$. A negative number times a negative number is also positive: $\left(-3\right)^{2} = \left(-3\right) \times \left(-3\right) = 9$. And $0^{2} = 0$.

So for every real number x, $x^{2}$ is 0 or more. The graph of $y = x^{2}$ shows this: its lowest point is (0, 0), and on either side it rises. No part of it lies below the x-axis.

Where $y = x^{2}$ meets a level line

The solutions of $x^{2} = k$ are the values of x where the curve $y = x^{2}$ meets the level line y = k. For k = 9 the line crosses the curve twice, at x = −3 and x = 3, so $x^{2} = 9$ has two solutions. For k = 0 the line touches the curve once, at x = 0.

For k = −1 the line y = −1 lies below the lowest point of the curve, so they never meet. $x^{2} = -1$ has no solution among the real numbers, and the same is true for $x^{2} = k$ whenever k is negative.

The curve $y = x^{2}$ meets the line y = 9 at the dots, x = −3 and x = 3. The line y = −1 runs below the curve’s lowest point, (0, 0), and never meets it.

Square roots

A square root of a number is a number whose square is that number. 9 has two, 3 and −3, and the sign $\surd$ means the positive one: $\sqrt{9} = 3$. 0 has one square root, 0.

A negative number has no real square root. $\sqrt{-4}$ is not 2, because $2^{2} = 4$, and it is not −2, because $\left(-2\right)^{2} = 4$ as well. No real number squares to −4.

Equations whose curve misses the axis

$x^{2} + 1 = 0$ is the same equation as $x^{2} = -1$, so it has no real solution either. On a graph, $y = x^{2} + 1$ is the curve $y = x^{2}$ lifted up by 1. Its lowest point is (0, 1), so it never reaches the x-axis, where y = 0.

Any quadratic whose curve misses the x-axis is in the same position. For $x^{2} + 2x + 5 = 0$, completing the square gives $\left(x + 1\right)^{2} + 4 = 0$, so $\left(x + 1\right)^{2} = -4$, and no real number squared gives −4. The discriminant says the same: $b^{2} - 4ac = 4 - 20 = -16$, which is negative, and the quadratic formula would need $\sqrt{-16}$.

The gold curve is $y = x^{2} + 1$, lowest at (0, 1). The plain curve is $y = x^{2} + 2x + 5 = \left(x + 1\right)^{2} + 4$, lowest at (−1, 4). Both stay above the x-axis, so neither equation has a real solution.

A number that is not on the line

Every real number lies on the number line, and none of them squares to −1. An answer to $x^{2} = -1$ would have to be a new kind of number, off the number line.

Negative numbers and fractions were once new in the same way: 3 − 5 has no answer among the counting numbers, and 1 ÷ 2 has none among the whole numbers. Each time a new number was defined with the property that was missing, and the old rules of arithmetic kept working. The number defined to square to −1 is called i.

The usual mistakes

Taking a negative number as the answer. $\left(-3\right)^{2} = 9$, not −9: minus times minus is plus, so no negative number squares to a negative.

Reading $-3^{2}$ as $\left(-3\right)^{2}$. Without brackets the square is done first, so $-3^{2} = -\left(3^{2}\right) = -9$. That does not make −3 a square root of −9, because the square of −3 is $\left(-3\right)^{2} = 9$.

Thinking a perfect square helps. $\sqrt{9} = 3$, but $\sqrt{-9}$ is not 3 or −3. The obstacle is the minus sign: every real square is 0 or more.