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$3² = 9. A negative number times a negative number is also positive: ${\left(-3\right)}^2=\left(-3\right)\times \left(-3\right)=9$(−3)² = (−3) × (−3) = 9. And $0^2=0$0² = 0.

So for every real number x, $x^2$x² is 0 or more. The graph of $y=x^2$y = x² shows this: its lowest point is (0, 0), and on either side it rises. No part of it lies below the x-axis.

Where $y=x^2$y = x² meets a level line

The solutions of $x^2=k$x² = k are the values of x where the curve $y=x^2$y = x² meets the level line y = k. For k = 9 the line crosses the curve twice, at x = −3 and x = 3, so $x^2=9$x² = 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. $x^2=-1$x² = −1 has no solution among the real numbers, and the same is true for $x^2=k$x² = k whenever k is negative.

The curve $y=x^2$y = x² 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: $\sqrt{9}=3$√9 = 3. 0 has one square root, 0.

A negative number has no real square root. $\sqrt{-4}$√(−4) is not 2, because $2^2=4$2² = 4, and it is not −2, because ${\left(-2\right)}^2=4$(−2)² = 4 as well. No real number squares to −4.

Equations whose curve misses the axis

$x^2+1=0$x² + 1 = 0 is the same equation as $x^2=-1$x² = −1, so it has no real solution either. On a graph, $y=x^2+1$y = x² + 1 is the curve $y=x^2$y = x² 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 $x^2+2x+5=0$x² + 2x + 5 = 0, completing the square gives ${\left(x+1\right)}^2+4=0$(x + 1)² + 4 = 0, so ${\left(x+1\right)}^2=-4$(x + 1)² = −4, and no real number squared gives −4. The discriminant says the same: $b^2-4ac=4-20=-16$b² − 4ac = 4 − 20 = −16, which is negative, and the quadratic formula would need $\sqrt{-16}$√(−16).

The gold curve is $y=x^2+1$y = x² + 1, lowest at (0, 1). The plain curve is $y=x^2+2x+5={\left(x+1\right)}^2+4$y = x² + 2x + 5 = (x + 1)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 $x^2=-1$x² = −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. ${\left(-3\right)}^2=9$(−3)² = 9, not −9: minus times minus is plus, so no negative number squares to a negative.

Reading $-3^2$−3² as ${\left(-3\right)}^2$(−3)². Without brackets the square is done first, so $-3^2=-\left(3^2\right)=-9$−3² = −(3²) = −9. That does not make −3 a square root of −9, because the square of −3 is ${\left(-3\right)}^2=9$(−3)² = 9.

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