Evaluate, Estimate and Simplify Square Roots

The definition picks one number on purpose, because two numbers can share a square. Both $7^2 = 49$ and $(-7)^2 = 49$ are true, so the equation $x^2 = 49$ has two solutions, $7$ and $-7$. In general, the equation $x^2 = k$ with $k > 0$ has two solutions: the positive root $\sqrt{k}$ and the negative root $-\sqrt{k}$. The symbol $\sqrt{49}$ names only the nonnegative root, so $\sqrt{49} = 7$. For the square with area $49$, you also reject $-7$ because a side length must be positive.

This shows that a statement and its converse, the statement with its two parts swapped, can differ. For $k \ge 0$, the statement "if $x = \sqrt{k}$, then $x^2 = k$" is true by the definition. Its converse, "if $x^2 = k$, then $x = \sqrt{k}$," is false. The value $x = -7$ makes $x^2 = 49$ true, but $-7 \ne \sqrt{49}$.

A perfect square is the square of a whole number, such as $0$, $1$, $4$, $9$ or $16$. To evaluate the square root of a perfect square, use the definition: find the nonnegative number whose square equals the radicand. For example, $\sqrt{0} = 0$, because $0 \ge 0$ and $0^2 = 0$. The table lists the positive perfect squares up to $144$. Knowing them by heart speeds up all three skills in this lesson.

Now take a radicand that is not a perfect square, such as $30$. No whole number has a square of $30$, and the decimal form of $\sqrt{30}$ never ends and never repeats. So any decimal you write for it is an estimate, a close value that you mark with the symbol $\approx$.

You can trap the value with the order property of square roots: for nonnegative numbers $a$ and $b$, if $a < b$, then $\sqrt{a} < \sqrt{b}$. Here is the reason. If $\sqrt{a}$ were greater than or equal to $\sqrt{b}$, then squaring these nonnegative numbers would give $a \ge b$. That contradicts the given fact $a < b$. You use the property in this direction: compare the radicands first, then conclude how the roots compare.

Some radicands contain a perfect square as a factor. For example, $72 = 36 \cdot 2$. To simplify a radical, you factor out perfect squares, which changes its form but not its value. A radical with a whole-number radicand is in simplest form when its radicand has no perfect-square factor greater than $1$.

The tool for simplifying is the product property of square roots. You can prove it from the definition.

Given: $a \ge 0$ and $b \ge 0$.

To prove: $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$.

Proof: The number $\sqrt{a} \cdot \sqrt{b}$ is nonnegative, because it is a product of two nonnegative numbers. The commutative and associative properties of multiplication, then the definition, give $(\sqrt{a} \cdot \sqrt{b})^2 = (\sqrt{a})^2 (\sqrt{b})^2 = ab$. So $\sqrt{a} \cdot \sqrt{b}$ is the nonnegative number whose square is $ab$. By the definition, that number is $\sqrt{ab}$.

The property is an equation, so it reads both ways. To simplify, you read it from left to right, turning one radical into two:

When the radicand is a fraction, use the quotient property of square roots: for $a \ge 0$ and $b > 0$, $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$. The proof follows the same pattern. The number $\dfrac{\sqrt{a}}{\sqrt{b}}$ is nonnegative and its square is $\dfrac{a}{b}$, so the definition makes it equal to $\sqrt{\dfrac{a}{b}}$. The condition $b > 0$ keeps the divisor nonzero and the radicand nonnegative. The next example uses both properties, each read from left to right.

Takeaway: Every skill in this lesson rests on one definition: for a nonnegative radicand, the square root symbol names the nonnegative number whose square is the radicand. You evaluate by finding that number, you estimate by trapping it between known squares, and you simplify by factoring out perfect squares. Next, you will solve equations using square roots, where both solutions of $x^2 = k$ with $k > 0$ matter, and you will multiply and divide radical expressions.