4.04 Operations with radical expressions
Adding and subtracting radicals
In previous lessons, we have seen that, in general, $\sqrt{a}+\sqrt{b}$√a+√b $\ne$≠ $\sqrt{a+b}$√a+b, and similarly $\sqrt{a}-\sqrt{b}$√a−√b $\ne$≠ $\sqrt{a-b}$√a−b. We needed to use the idea of like terms to add and subtract expressions involving radicals. We can summarized this as:
$c\sqrt{a}+d\sqrt{a}$c√a+d√a = $\left(c+d\right)\sqrt{a}$(c+d)√a
$c\sqrt{a}-d\sqrt{a}$c√a−d√a = $\left(c-d\right)\sqrt{a}$(c−d)√a
$a$a, $b$b, $c$c and $d$d can be numerical or algebraic.
Using simplified radicals
Sometimes we are asked to add and subtract radicals that have different radicands (arguments). In this case, we can try to simplify one of the radicals so that we have the same radicands.
Worked example
Question 1
Simplify $\sqrt{12x}-\sqrt{3x}$√12x−√3x
Think: $\sqrt{12x}$√12x and $\sqrt{3x}$√3x do not have the same radicand, so can't be subtracted as they are. However, we can simplify $\sqrt{12x}$√12x using our technique for simplifying radicals.
Do: Start by simplifying $\sqrt{12x}$√12x and then combine like terms.
| $\sqrt{12x}-\sqrt{3x}$√12x−√3x | $=$= | $\sqrt{4\times3x}-\sqrt{3x}$√4×3x−√3x |
| $=$= | $\sqrt{4}\sqrt{3x}-\sqrt{3x}$√4√3x−√3x | |
| $=$= | $2\sqrt{3x}-\sqrt{3x}$2√3x−√3x | |
| $=$= | $\sqrt{3x}$√3x |
When adding and subtracting radicals, they must have the same radicand before we can simplify
Be sure to simplify all radicals first. This may involve variables!
Practice questions
Question 2
Simplify completely: $\sqrt{243}+\sqrt{3}$√243+√3
Question 3
Simplify the expression $\sqrt{ax^5}+x^2\sqrt{ax}$√ax5+x2√ax, where $x$x represents a positive number.
Multiplying and dividing radicals
So far we know that radicals only like to combine and split when multiplication and division are involved and not addition and subtraction. This gives us a lot of freedom when multiplying and dividing more complicated radicals.
In general, we found that:
$\sqrt{a}\times\sqrt{b}=\sqrt{a\times b}$√a×√b=√a×b and $\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$√a×b=√a×√b
as well as that:
$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√a√b=√ab and $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$√ab=√a√b
Working with algebraic factors
When algebraic factors are involved in products it is important to remember to simplify them as well. The properties below will help us.
If $a$a and $b$b are non-negative real numbers, then
$\sqrt{a^2}=a$√a2=a
$\sqrt{ab}=\sqrt{a}\sqrt{b}$√ab=√a√b
It is important to note that these properties apply to higher powers as well, so we can state that:
For variables $a$a and $b$b,
$\sqrt[3]{a^3}=a$3√a3=a
$\sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}$3√ab=3√a3√b
Worked examples
Question 4
Assuming $u$u and $v$v are non-negative real numbers, write the expression $\sqrt{u^3}\sqrt{v}$√u3√v using a single radical and simplify where possible.
Think: We want the final form to contain a single radical of the form $a\sqrt{b}$a√b, 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) |
Question 5
Write the expression $\sqrt[3]{4}\sqrt[3]{2m}\sqrt[3]{n}$3√43√2m3√n using a single radical and simplify where possible.
Think: To write the expression so that it contains only one radical of the form $\sqrt[3]{a}$3√a 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}$3√43√2m3√n to $\sqrt[3]{8mn}$3√8mn, but we used the same rule for both steps. We can generalize this rule to account for any type of root function and any number of terms in the product
Distributing binomials products involving radical expressions
We have used the distributive property to distribute expressions of the form $\left(a+b\right)\left(c+d\right)$(a+b)(c+d) previously. We can use that same property with any of the terms involve radicals.
Distributive Property:
$\left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$(a+b)(c+d)=ac+ad+bc+bd
Special patterns:
$\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2
$\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2
Worked examples
Question 6
Distribute and simplify $\left(\sqrt{3}-\sqrt{10}\right)\left(\sqrt{3}+\sqrt{12}\right)$(√3−√10)(√3+√12)
Think: The distributive property says to multiply each term in the first bracket by each term in the second bracket. We need to simplify square root expressions as much as possible.
Do:
| $\left(\sqrt{3}-\sqrt{10}\right)\left(\sqrt{3}+\sqrt{12}\right)$(√3−√10)(√3+√12) | $=$= | $\sqrt{3}\sqrt{3}+\sqrt{3}\sqrt{12}-\sqrt{10}\sqrt{3}-\sqrt{10}\sqrt{12}$√3√3+√3√12−√10√3−√10√12 |
| $=$= | $\left(\sqrt{3}\right)^2+\sqrt{3\times12}-\sqrt{10\times3}-\sqrt{10\times12}$(√3)2+√3×12−√10×3−√10×12 | |
| $=$= | $3+\sqrt{36}-\sqrt{30}-\sqrt{120}$3+√36−√30−√120 | |
| $=$= | $3+6-\sqrt{30}-\sqrt{4\times30}$3+6−√30−√4×30 | |
| $=$= | $9-\sqrt{30}-2\sqrt{30}$9−√30−2√30 | |
| $=$= | $9-3\sqrt{30}$9−3√30 |
Question 7
Distribute and simplify $\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}+\sqrt{7}\right)$(2√5−√7)(2√5+√7)
Think: Notice that this fits one of our special patterns to give us a difference of squares, $\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2. Let's see what happens.
Do:
| $\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}+\sqrt{7}\right)$(2√5−√7)(2√5+√7) | $=$= | $\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2$(2√5)2−(√7)2 |
| $=$= | $2^2\left(\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2$22(√5)2−(√7)2 | |
| $=$= | $4\times5-7$4×5−7 | |
| $=$= | $20-7$20−7 | |
| $=$= | $13$13 |
Reflect: The product of two irrational numbers has resulted in a rational number. This could be a useful concept.
Practice questions
Question 8
Distribute and simplify the given expression: $\sqrt{13}\left(\sqrt{11}+2\right)$√13(√11+2)
Question 9
Assuming $j$j and $k$k are non-negative, write the expression $\sqrt{4j}\sqrt{j}\sqrt{4k}$√4j√j√4k using a single radical and simplify where possible.
Conjugates and rationalizing denominators
A binomial is an expression of the form $A+B$A+B, containing two terms. Changing the sign of the second term gives us the binomial $A-B$A−B, which we call a conjugate for the original binomial $A+B$A+B.
If we then try to find a conjugate for the binomial $A-B$A−B by changing the sign of the second term, we obtain the original binomial $A+B$A+B. That is, any binomial is a conjugate of its own conjugate. We often refer to two such binomials as a conjugate pair.
Notice that the product of a conjugate pair has a familiar form $\left(A+B\right)\left(A-B\right)$(A+B)(A−B) which is the factored form of the difference of two squares $A^2-B^2$A2−B2. This observation motivates us to look at binomials containing radicals - note that the expression $A^2-B^2$A2−B2 will be rational even if the terms $A$A or $B$B are square roots.
Conjugates of binomials with radicals
Consider a binomial such as $1+\sqrt{2}$1+√2. We can find a conjugate for this expression in the same way - by switching the sign of the second term. Doing so, we find that $1-\sqrt{2}$1−√2 is a conjugate for $1+\sqrt{2}$1+√2.
The process is the same even if the expression is more complicated, such as $\sqrt{x}-4\sqrt{3}$√x−4√3. A conjugate for this expression would be $\sqrt{x}+4\sqrt{3}$√x+4√3.
For any binomial expression $A+B$A+B, we can find a conjugate $A-B$A−B by changing the sign of the second term.
A binomial and its conjugate are sometimes called a conjugate pair.
We can rewrite the binomial $A+B$A+B in the equivalent form $B+A$B+A by changing the order of the terms. By doing so we can see that $B-A$B−A is also a conjugate for this expression, as well as $A-B$A−B.
That is, a binomial has two possible conjugates (since there are two orders in which the binomial can be written).
Rationalizing the denominator
As mathematicians, we want to avoid having expressions involving radicals in the denominator of fractions as they can be more difficult to work with.
The process of rewriting expressions like $\frac{2}{\sqrt{3}}$2√3 or $\frac{3+\sqrt{5}}{1-\sqrt{2}}$3+√51−√2 so they don't have radicals in the denominator is called rationalizing the denominator, and it does not change the value of the fraction.
Rationalizing the denominator using perfect squares
We know that when we square a radical, the answer is always going to be rational and without radicals! This is the key we need to rationalize fractions such as $\frac{8}{3\sqrt{7}}$83√7.
What do you think will happen if we multiply the denominator here by $\sqrt{7}$√7? Well, the square root sign will disappear right? However, we need the fraction to still have the same value, so let's try multiplying the fraction by $\frac{\sqrt{7}}{\sqrt{7}}=1$√7√7=1, which will not change the fraction at all!
$\frac{8}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{8\times\sqrt{7}}{3\times\sqrt{7}\times\sqrt{7}}$83√7×√7√7=8×√73×√7×√7
which simplifies down to
$\frac{8\sqrt{7}}{3\left(\sqrt{7}\right)^2}=\frac{8\sqrt{7}}{3\times7}$8√73(√7)2=8√73×7
This answer can again be finally simplified down to $\frac{8\sqrt{7}}{21}$8√721.
The fraction now looks completely different! But try putting it in your calculator, do you get the same value as $\frac{8}{3\sqrt{7}}$83√7?
So now we know one way of rationalizing the denominator is to multiply top and bottom by the radical in the denominator.
Rationalizing the denominator using differences of two squares
The above technique is great, but it only works for monomial denominators.
Let's take a look at an example: $\frac{5}{\sqrt{6}-1}$5√6−1.
We saw in question 7, that when we distributed to get a difference of two squares, that an irrational product became rational. We can use this idea here.
Let's see what happens when we multiply $\left(\sqrt{6}-1\right)$(√6−1) by $\left(\sqrt{6}+1\right)$(√6+1).
| $\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)$(√6−1)(√6+1) | $=$= | $\left(\sqrt{6}\right)^2-1^2$(√6)2−12 |
| $=$= | $6-1$6−1 | |
| $=$= | $5$5 |
We actually use the conjugate of the denominator to rationalize it.
In general, if we have $\frac{n}{a+b}$na+b we need to multiply both the numerator and denominator by the conjugate of the denominator. That is:
$\frac{n}{a+b}=\frac{n}{a+b}\times\frac{a-b}{a-b}$na+b=na+b×a−ba−b
Worked example
Question 10
Rationalize the denominator of $\frac{5}{\sqrt{6}-1}$5√6−1, simplify fully.
Think: First we need to determine the conjugate of $\sqrt{6}-1$√6−1 and then multiply both the numerator and denominator by it. Finally, we simplify fully.
Do:
| $\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}+1}{\sqrt{6}+1}$5√6−1×√6+1√6+1 | $=$= | $\frac{5\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}$5(√6+1)(√6−1)(√6+1) |
| $=$= | $\frac{5\sqrt{6}+5}{5}$5√6+55 | |
| $=$= | $\frac{5\left(\sqrt{6}+1\right)}{5}$5(√6+1)5 | |
| $=$= | $\sqrt{6}+1$√6+1 |
Reflect: We can apply the same principal to expressions involving variables.
Practice questions
Question 11
Determine a conjugate for $-5\sqrt{3}+4\sqrt{v}$−5√3+4√v.
Question 12
Rationalize the denominator for the given expression:
$\frac{1}{\sqrt{11}}$1√11
Question 13
Simplify, expressing your answer with a rational denominator:
$\frac{9}{\sqrt{5}-7}$9√5−7