Square roots

Squaring a number means multiplying it by itself. A square root reverses this - it asks "what non-negative number, when squared, gives $a$?"

A perfect square is a number whose square root is a whole number. Recognising these from memory makes working with square roots much faster.

Evaluate $\sqrt{49}$, $\sqrt{0}$, and $\sqrt{-9}$.

Solution

Evaluating $\sqrt{49}$

We need a non-negative number that, when squared, gives $49$. Since $7^2 = 49$ and $7 \geq 0$:

Evaluating $\sqrt{0}$

We need a non-negative number that, when squared, gives $0$. Since $0^2 = 0$ and $0 \geq 0$:

Evaluating $\sqrt{-9}$

There is no real number whose square is $-9$. Any real number squared gives a non-negative result, so $\sqrt{-9}$ is not defined in the real numbers.

Every positive number has two square roots - one positive and one negative - because both a number and its negative square to the same result. The symbol $\sqrt{a}$ picks out just one of them.

The key distinction is between evaluating $\sqrt{a}$ (one answer, always non-negative) and solving $x^2 = a$ (two answers, $\pm\sqrt{a}$).

Solve $x^2 = 36$. Then evaluate $\sqrt{36}$.

Solution

Solving $x^2 = 36$

We need all values of $x$ whose square is $36$. Both $6$ and $-6$ satisfy this:

So the equation has two solutions:

Written compactly: $x = \pm 6$.

Evaluating $\sqrt{36}$

The square root symbol always returns the principal (non-negative) root. Since $6 \geq 0$ and $6^2 = 36$:

The answer is $6$, not $\pm 6$. The $\pm$ only appears when solving an equation like $x^2 = 36$.

Most positive integers are not perfect squares - their square roots are irrational numbers (decimals that go on forever without repeating). We cannot write these roots exactly, but we can estimate where they fall.

This gives a quick way to place any square root between two consecutive integers - useful for sanity-checking calculations and building numerical intuition.

Estimate $\sqrt{50}$. Between which two consecutive integers does it fall, and which is it closer to?

Solution

First, we identify the perfect squares on either side of $50$:

Taking square roots:

So $\sqrt{50}$ lies between $7$ and $8$.

To determine which integer it is closer to, we compare distances: $50$ is only $1$ away from $49$, but $14$ away from $64$. Since $50$ is much closer to $49$, $\sqrt{50}$ is very close to $7$.

So $\sqrt{50}$ falls between $7$ and $8$, and is much closer to $7$ (approximately $7.07$).