New Zealand
Level 8 - NCEA Level 3

# Multiply algebraic surds

Lesson

The square root function $f\left(x\right)=\sqrt{x}$f(x)=x is defined for all non-negative real values of $x$x. This means we can compute expressions like $\sqrt{9}=3$9=3 and $\sqrt{17}=4.1231\dots$17=4.1231, but expressions like $\sqrt{-16}$16 and $\sqrt{-29}$29 have no equivalent values in the real numbers.

It is important to remember how we have defined $\sqrt{x}$x when we are manipulating algebraic expressions involving surds. We will be able to make use of some convenient rules once we are sure that the variables in our expressions are non-negative.

#### Exploration

Let's consider some real number $a$a that is greater than zero. Since $a$a is positive, then we know that $\sqrt{a}$a is also a positive real number.

For values of $x$x greater than or equal to zero, the squaring function $x^2$x2 and the square root function $\sqrt{x}$x have the effect of undoing each other. So we will find that $\left(\sqrt{a}\right)^2=a$(a)2=a and $\sqrt{a^2}=a$a2=a.

We can use this result to derive another useful rule. Consider now two real numbers $a$a and $b$b that are both greater than zero. The product $ab$ab can be rewritten as follows.

 $ab$ab $=$= $\left(\sqrt{a}\right)^2\left(\sqrt{b}\right)^2$(√a)2(√b)2 (Since $\left(\sqrt{a}\right)^2=a$(√a)2=a and $\left(\sqrt{b}\right)^2=b$(√b)2=b) $=$= $\sqrt{a}\sqrt{a}\sqrt{b}\sqrt{b}$√a√a√b√b (Writing the powers as repeated multiplication) $=$= $\sqrt{a}\sqrt{b}\sqrt{a}\sqrt{b}$√a√b√a√b (Rearranging the terms in the product) $=$= $\left(\sqrt{a}\sqrt{b}\right)\times\left(\sqrt{a}\sqrt{b}\right)$(√a√b)×(√a√b) (Grouping into two identical factors) $=$= $\left(\sqrt{a}\sqrt{b}\right)^2$(√a√b)2 (Writing the repeated multiplication using powers)

Now that we know $ab=\left(\sqrt{a}\sqrt{b}\right)^2$ab=(ab)2, we can take the square root of both sides of this equation.

 $\sqrt{ab}$√ab $=$= $\sqrt{\left(\sqrt{a}\sqrt{b}\right)^2}$√(√a√b)2 $=$= $\sqrt{a}\sqrt{b}$√a√b

In the final step, the inner square function was "undone" by the outer square root function. The result is the rule $\sqrt{ab}=\sqrt{a}\sqrt{b}$ab=ab.

Remember!

If $a$a and $b$b are non-negative real numbers, then

$\sqrt{a^2}=a$a2=a

$\left(\sqrt{a}\right)^2=a$(a)2=a

$\sqrt{ab}=\sqrt{a}\sqrt{b}$ab=ab

Notice that if $a$a is less than zero, then $\left(\sqrt{a}\right)^2$(a)2 is not defined, and $\sqrt{a^2}=\left|a\right|$a2=|a|, which has the opposite sign to $a$a. That is, $\left|a\right|$|a| is positive and $a$a is negative, so that $\left|a\right|\ne a$|a|a and $\sqrt{a^2}\ne a$a2a.

#### Worked example

Assuming $u$u and $v$v are non-negative real numbers, write the expression $\sqrt{u^3}\sqrt{v}$u3v using a single surd and simplify where possible.

Think: We want the final form to contain a single surd of the form $\sqrt{a}$a, and we can use the rules outlined above to get there.

Do:

 $\sqrt{u^3}\sqrt{v}$√u3√v $=$= $\sqrt{u^2u}\sqrt{v}$√u2u√v (Since $u^3=u^2u$u3=u2u) $=$= $\sqrt{u^2}\sqrt{u}\sqrt{v}$√u2√u√v (Using the fact that $\sqrt{ab}=\sqrt{a}\sqrt{b}$√ab=√a√b) $=$= $u\sqrt{u}\sqrt{v}$u√u√v (Since $u=\sqrt{u^2}$u=√u2) $=$= $u\sqrt{uv}$u√uv (Once again using the fact that $\sqrt{ab}=\sqrt{a}\sqrt{b}$√ab=√a√b)

### Cube roots and upward

Each of the results obtained above can be applied to expressions involving cube roots, fourth roots, and so on.

#### Worked example

Write the expression $\sqrt[3]{4}\sqrt[3]{2m}\sqrt[3]{n}$3432m3n using a single surd and simplify where possible.

Think: To write the expression so that it contains only one surd of the form $\sqrt[3]{a}$3a we can first combine two of the cube roots together.

Do:

 $\sqrt[3]{4}\sqrt[3]{2m}\sqrt[3]{n}$3√43√2m3√n $=$= $\sqrt[3]{8m}\sqrt[3]{n}$3√8m3√n (Using the fact $\sqrt[3]{a}\sqrt[3]{b}=\sqrt[3]{ab}$3√a3√b=3√ab) $=$= $\sqrt[3]{8mn}$3√8mn (Using the fact $\sqrt[3]{a}\sqrt[3]{b}=\sqrt[3]{ab}$3√a3√b=3√ab again) $=$= $\sqrt[3]{2^3mn}$3√23mn (Since $2^3=8$23=8) $=$= $\sqrt[3]{2^3}\sqrt[3]{mn}$3√233√mn (Once again using the fact that $\sqrt[3]{a}\sqrt[3]{b}=\sqrt[3]{ab}$3√a3√b=3√ab) $=$= $2\sqrt[3]{mn}$23√mn (Simplifying $\sqrt[3]{2^3}$3√23)

Reflect: Notice that it took two steps to get from $\sqrt[3]{4}\sqrt[3]{2m}\sqrt[3]{n}$3432m3n to $\sqrt[3]{8mn}$38mn, but we used the same rule for both steps. We can generalise this rule to account for any type of root function and any number of terms in the product.

Simplifying a product of surds

If $a_1$a1, $a_2$a2, $\dots$, $a_m$am are non-negative real numbers, and $\sqrt[n]{a}$na is the $n$nth root of $a$a, then

$\sqrt[n]{a_1}\times\sqrt[n]{a_2}\times\dots\times\sqrt[n]{a_m}=\sqrt[n]{a_1\times a_2\times\dots\times a_m}$na1×na2××nam=na1×a2××am

#### Practice questions

##### Question 1

Assuming $p$p and $q$q are non-negative, write the expression $3\sqrt{p}\sqrt{q}$3pq using a single surd.

##### Question 2

Assuming $u$u and $v$v are non-negative values, simplify the expression $7\sqrt{u}\times8\sqrt{uv}$7u×8uv.

##### Question 3

Write the expression $\sqrt[3]{u}\sqrt[3]{u^2}\sqrt[3]{v}$3u3u23v using a single surd and simplify where possible.

### Outcomes

#### M8-9

Manipulate complex numbers and present them graphically

#### 91577

Apply the algebra of complex numbers in solving problems