Solve with Square Roots, Multiply and Divide Radicals | Geometry | LevelKite

In geometry, you often know a square's area and need its side length. Finding that length means undoing a square, and square roots do that job. In this lesson, you will solve equations using square roots, then multiply and divide radical expressions.

A square root of a number $k$ is a number whose square is $k$ . For example, $7$ and $-7$ are both square roots of $49$ . The symbol $\sqrt{\ }$ is the radical symbol, the number under it is the radicand, and an expression that contains the symbol is a radical expression. In this course, every radicand is nonnegative, which means zero or positive.

Key concept

What the square root symbol means

For any number $k \ge 0$ , the symbol $\sqrt{k}$ names exactly one number: the nonnegative number whose square is $k$ . So $\sqrt{49} = 7$ , not $-7$ . Every rule in this lesson comes from this definition.

\sqrt{k} \ge 0 \quad \text{and} \quad (\sqrt{k})^2 = k \qquad \text{for } k \ge 0

A symbol and an equation can ask different questions. The symbol $\sqrt{49}$ names one number. The equation $x^2 = 49$ asks for every solution, which is a value of $x$ that makes the equation true. The table compares them and shows what happens when the right side of the equation is zero or negative.

What you see	What it asks for	Result
$\sqrt{49}$	The nonnegative number whose square is $49$	$7$
$x^2 = 49$	Every number whose square is $49$	Two solutions: $x = 7$ or $x = -7$
$x^2 = 0$	Every number whose square is $0$	One solution: $x = 0$
$x^2 = -49$	Every real number whose square is $-49$	No real solution, because a real number's square is never negative

The $x^2 = 49$ row is one case of a theorem, a statement proved from definitions and earlier facts.

Square Root Property: If $x^2 = k$ and $k > 0$ , then $x = \sqrt{k}$ or $x = -\sqrt{k}$ . These are the positive root and the negative root.

Why it holds. You are given $x^2 = k$ with $k > 0$ , and you want to show that $x$ must be one of those two values. Subtract $k$ from both sides (Subtraction Property of Equality) to get $x^2 - k = 0$ . By the definition of square root, $k = (\sqrt{k})^2$ , so the left side is a difference of squares: $(x - \sqrt{k})(x + \sqrt{k}) = 0$ . The Zero Product Property says that if a product is $0$ , at least one factor is $0$ . So $x = \sqrt{k}$ or $x = -\sqrt{k}$ .

Which direction you use. Solving uses this theorem forward, from the equation to the two values. Its converse swaps the if and then parts: if $k > 0$ and $x$ is $\sqrt{k}$ or $-\sqrt{k}$ , then $x^2 = k$ . The converse is also true, because squaring either value gives $k$ . That is what your check tests.

Worked example

Example 1: Isolate the square first

Solve $3x^2 - 5 = 43$ .

Add 5 to both sides, using the Addition Property of Equality, so the squared term stands alone.
$3x^2 - 5 + 5 = 43 + 5, \text{ so } 3x^2 = 48$
Divide both sides by 3, using the Division Property of Equality. This is valid because the divisor 3 is not zero.
$\frac{3x^2}{3} = \frac{48}{3}, \text{ so } x^2 = 16$
Apply the Square Root Property in its forward direction. Its condition holds because 16 is greater than 0, so the equation has two solutions, the positive root and the negative root.
$x = \sqrt{16} \text{ or } x = -\sqrt{16}$
Evaluate the radical using the definition of square root: the square root of 16 is 4, because 4 is nonnegative and 4 squared is 16.
$x = 4 \text{ or } x = -4$
Check each solution by substituting it into the original equation. A negative number squared is positive, so both values make the equation true.
$3(4)^2 - 5 = 48 - 5 = 43, \quad 3(-4)^2 - 5 = 48 - 5 = 43$

Answer

The equation $3x^2 - 5 = 43$ has two solutions, $x = 4$ and $x = -4$ .

The Square Root Property also applies when a whole expression is squared. If an expression squared equals $k$ and $k > 0$ , then the expression equals $\sqrt{k}$ or $-\sqrt{k}$ .

Geometry adds a second test. A solution of the equation must also fit the figure, and a length must be positive. So an equation can have two solutions even when only one of them gives a possible figure. Solve first, then test each solution against the figure's conditions.

Worked example

Example 2: A squared expression as a side length

Given: a square has side length $x + 2$ centimeters and area $50$ square centimeters. Find: the value of $x$ and the side length.

Apply the area formula for a square: area equals side length squared. Substitute the given side length and area.
$(x + 2)^2 = 50$
Treat x + 2 as the number being squared. Because 50 is greater than 0, the Square Root Property gives the positive root and the negative root.
$x + 2 = \sqrt{50} \text{ or } x + 2 = -\sqrt{50}$
Simplify the radical by factoring out the perfect square 25, using the Product Property for nonnegative radicands from your work on simplifying radicals.
$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}, \text{ so } x + 2 = 5\sqrt{2} \text{ or } x + 2 = -5\sqrt{2}$
Subtract 2 from both sides of each equation, using the Subtraction Property of Equality.
$x = 5\sqrt{2} - 2 \text{ or } x = -5\sqrt{2} - 2$
Test the geometric condition: the side length x + 2 must be positive. The first case gives a positive length, so keep it. The second case gives a negative length, so reject it.
$x + 2 = 5\sqrt{2} > 0 \text{ (keep)}, \quad x + 2 = -5\sqrt{2} < 0 \text{ (reject)}$
Check the kept value in the equation from the first step. Square each factor (Power of a Product Property) and use the definition of square root: the square root of 2, squared, is 2.
$(5\sqrt{2} - 2 + 2)^2 = (5\sqrt{2})^2 = 5^2 (\sqrt{2})^2 = 25 \cdot 2 = 50$

Answer

The value of $x$ that fits the square is $x = 5\sqrt{2} - 2$ , so the side length is $5\sqrt{2}$ centimeters.

Now you will multiply and divide radicals. Start with numbers you know: $\sqrt{4} \cdot \sqrt{9} = 2 \cdot 3 = 6$ , and $\sqrt{4 \cdot 9} = \sqrt{36} = 6$ . One case does not prove a rule, so here is the reason.

You are given $a \ge 0$ and $b \ge 0$ , and you want to show $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ . The left side is nonnegative, because both factors are nonnegative by the definition of square root. Its square is $(\sqrt{a} \cdot \sqrt{b})^2 = (\sqrt{a})^2(\sqrt{b})^2 = ab$ , by the Power of a Product Property and the same definition. So the left side is the nonnegative number whose square is $ab$ , and by definition that number is $\sqrt{ab}$ .

The same reasoning gives the quotient rule when $a \ge 0$ and $b > 0$ . You must exclude $b = 0$ because $\sqrt{0} = 0$ and division by zero is undefined. These two results are the Product Property and the Quotient Property of radicals.

\text{Product Property: } \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \quad \text{for } a \ge 0,\ b \ge 0

\text{Quotient Property: } \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \quad \text{for } a \ge 0,\ b > 0

Each property is an equation, so you can read it in either direction. To multiply or divide radicals, read it left to right and combine the radicands. When you simplified radicals, you read the Product Property right to left and split the radicand.

A coefficient is a number that multiplies a radical, like the $2$ in $2\sqrt{6}$ . To multiply radical expressions, multiply the coefficients, multiply the radicands, and then simplify by factoring out perfect squares. In symbols, $p\sqrt{a} \cdot q\sqrt{b} = pq\sqrt{ab}$ for $a \ge 0$ and $b \ge 0$ .

Worked example

Example 3: Multiply and simplify

Multiply and simplify $2\sqrt{6} \cdot 5\sqrt{10}$ .

Regroup the factors, using the Commutative and Associative Properties of Multiplication, so the coefficients are together and the radicals are together.
$2\sqrt{6} \cdot 5\sqrt{10} = (2 \cdot 5)(\sqrt{6} \cdot \sqrt{10})$
Multiply the coefficients. Multiply the radicals with the Product Property, read left to right, which applies because 6 and 10 are nonnegative.
$(2 \cdot 5)(\sqrt{6} \cdot \sqrt{10}) = 10\sqrt{60}$
Simplify by factoring out the perfect square 4, because 60 = 4 times 15. This reads the Product Property right to left, with nonnegative radicands 4 and 15.
$10\sqrt{60} = 10\sqrt{4 \cdot 15} = 10 \cdot \sqrt{4} \cdot \sqrt{15} = 10 \cdot 2\sqrt{15}$
Multiply the coefficients. The radicand 15 has no perfect-square factor other than 1, so the radical is simplified.
$10 \cdot 2\sqrt{15} = 20\sqrt{15}$

Answer

The product $2\sqrt{6} \cdot 5\sqrt{10}$ simplifies to $20\sqrt{15}$ .

Dividing follows the same pattern: divide the coefficients and divide the radicands. In symbols, $\frac{p\sqrt{a}}{q\sqrt{b}} = \frac{p}{q}\sqrt{\frac{a}{b}}$ when $a \ge 0$ , $b > 0$ and $q \ne 0$ . These conditions keep the denominator nonzero. Below, the denominator $2\sqrt{3}$ is nonzero, $15 \ge 0$ and $3 > 0$ , so the Quotient Property applies, read left to right. The result is simplified because $5$ has no perfect-square factor other than $1$ .

\frac{6\sqrt{15}}{2\sqrt{3}} = \frac{6}{2} \cdot \frac{\sqrt{15}}{\sqrt{3}} = 3\sqrt{\frac{15}{3}} = 3\sqrt{5}

Worked example

Example 4: Mixed review with a rectangle

Given: a rectangle has width $x$ centimeters, length $2\sqrt{3}\,x$ centimeters and area $6\sqrt{12}$ square centimeters. Find: the value of $x$ .

This problem mixes both objectives, so ask at each step which property applies. Start with the area formula for a rectangle: length times width equals area.
$(2\sqrt{3}\,x)(x) = 6\sqrt{12}, \text{ so } 2\sqrt{3}\,x^2 = 6\sqrt{12}$
Divide both sides by the coefficient of the squared term (Division Property of Equality). This is valid because 2 times the square root of 3 is not zero.
$\frac{2\sqrt{3}\,x^2}{2\sqrt{3}} = \frac{6\sqrt{12}}{2\sqrt{3}}, \text{ so } x^2 = \frac{6\sqrt{12}}{2\sqrt{3}}$
Divide the coefficients, where the divisor 2 is nonzero. Apply the Quotient Property, which holds because 12 is nonnegative and 3 is positive. Then use the definition of square root: the square root of 4 is 2.
$x^2 = \frac{6}{2} \cdot \frac{\sqrt{12}}{\sqrt{3}} = 3\sqrt{\frac{12}{3}} = 3\sqrt{4} = 3 \cdot 2 = 6$
Because 6 is greater than 0, the Square Root Property gives two solutions of this equation.
$x = \sqrt{6} \text{ or } x = -\sqrt{6}$
Test the geometric condition: a width must be positive, so reject the negative root.
$\sqrt{6} > 0 \text{ (keep)}, \quad -\sqrt{6} < 0 \text{ (reject)}$
Check the kept value against the given area. Multiply with the Product Property, whose radicands here are nonnegative, and factor the perfect square 9 out of 18.
$\text{length} = 2\sqrt{3} \cdot \sqrt{6} = 2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}, \quad \text{area} = 6\sqrt{2} \cdot \sqrt{6} = 6\sqrt{12}$

Answer

The value is $x = \sqrt{6}$ , so the rectangle is $\sqrt{6}$ centimeters wide and $6\sqrt{2}$ centimeters long.

Everything here grows from one definition: for $k \ge 0$ , $\sqrt{k}$ names the one nonnegative number whose square is $k$ . So when $k > 0$ , the equation $x^2 = k$ has two solutions, the positive root and the negative root, and if $x$ is a length, only the positive root fits. The same definition lets you multiply radicals when $a \ge 0$ and $b \ge 0$ , and divide them when $a \ge 0$ and $b > 0$ , which keeps the denominator nonzero.