Transforming and simplifying expressions with square roots

Mrs Lan has three square plots of land side by side in a row, with areas of $12$ m², $27$ m² and $48$ m². She wants a fence along the front edge of all three plots. How long is the fence? Can we write its exact length in a neat form?

Three square plots side by side

The three sides are $\sqrt{12}$ , $\sqrt{27}$ and $\sqrt{48}$ m. There is no rule for adding square roots, but we can transform them so that they share the same "root part".

Taking factors outside and inside the root sign

Explore

We split each number under the root sign into a perfect square times another number, then use $\sqrt{ab}=\sqrt a\cdot\sqrt b$ :

$\sqrt{12}=\sqrt{4\cdot3}=2\sqrt3,\qquad \sqrt{27}=\sqrt{9\cdot3}=3\sqrt3,\qquad \sqrt{48}=\sqrt{16\cdot3}=4\sqrt3.$

The three roots are now multiples of the same number $\sqrt3$ . A square root such as $\sqrt3$ that cannot be written as a whole number or a fraction is called a surd. Surds that are multiples of the same root are like surds, and we add them just as we add like terms.

The first formula holds because $\sqrt{A^2B}=\sqrt{A^2}\cdot\sqrt B=|A|\sqrt B$ . The second is the first one read backwards; when $A<0$ the number $A\sqrt B$ is negative, so the minus sign must stay outside.

When there are letters under the root sign, remember that $\sqrt{A^2}=|A|$ , so we must look at the sign of $A$ .

Moving a factor inside the root sign lets us compare two numbers without a calculator.

Removing denominators and rationalising

Numbers like $\dfrac{6}{\sqrt3}$ are hard to estimate because the denominator is irrational. We transform them so that the denominator has no root.

Theorem The transformations

Removing the denominator under the root sign: $\sqrt{\dfrac AB}=\dfrac{\sqrt{AB}}{|B|}$ for $AB\ge0$ , $B\neq0$ .
Rationalising the denominator: $\dfrac{A}{\sqrt B}=\dfrac{A\sqrt B}{B}$ for $B>0$ .
For $A\ge0$ , $A\neq B^2$ : $\dfrac{C}{\sqrt A+B}=\dfrac{C\left(\sqrt A-B\right)}{A-B^2}$ and $\dfrac{C}{\sqrt A-B}=\dfrac{C\left(\sqrt A+B\right)}{A-B^2}$ .
For $A$ , $B\ge0$ , $A\neq B$ : $\dfrac{C}{\sqrt A+\sqrt B}=\dfrac{C\left(\sqrt A-\sqrt B\right)}{A-B}$ and $\dfrac{C}{\sqrt A-\sqrt B}=\dfrac{C\left(\sqrt A+\sqrt B\right)}{A-B}$ .

The common idea: multiply the numerator and the denominator by an expression that turns the denominator into the identity $(m+n)(m-n)=m^2-n^2$ , where squaring removes the root. The expressions $\sqrt A+\sqrt B$ and $\sqrt A-\sqrt B$ are called conjugates.

Example Rationalising denominators

Rationalise the denominator: a) $\dfrac{6}{\sqrt3}$ ; b) $\dfrac{4}{\sqrt5-1}$ ; c) $\dfrac{1}{\sqrt7+\sqrt5}$ .

Solution. In a), we multiply the numerator and the denominator by $\sqrt3$ ; in b) and c), we multiply them by the conjugate of the denominator. We have

$\begin{aligned} \text{a) }& \frac{6}{\sqrt3}=\frac{6\sqrt3}{3}=2\sqrt3;\\ \text{b) }& \frac{4}{\sqrt5-1}=\frac{4\left(\sqrt5+1\right)}{5-1}=\sqrt5+1;\\ \text{c) }& \frac{1}{\sqrt7+\sqrt5}=\frac{\sqrt7-\sqrt5}{7-5}=\frac{\sqrt7-\sqrt5}{2}. \end{aligned}$

So a) $2\sqrt3$ ; b) $\sqrt5+1$ ; c) $\dfrac{\sqrt7-\sqrt5}{2}$ .

Simplifying expressions with square roots

A simplification question in the grade 10 entrance exam usually follows these steps: find the condition for the expression to be defined, factorise the numerators and denominators, write over a common denominator, and collect terms.

Example Simplifying an expression with

\sqrt x

Simplify $A=\dfrac{x-\sqrt x}{\sqrt x-1}+\dfrac{x-4}{\sqrt x+2}$ for $x\ge0$ , $x\neq1$ .

Solution. Since $x\ge0$ , $x=\left(\sqrt x\right)^2$ . Hence $x-\sqrt x=\sqrt x\left(\sqrt x-1\right)$ and $x-4=\left(\sqrt x-2\right)\left(\sqrt x+2\right)$ . We have

$A=\frac{\sqrt x\left(\sqrt x-1\right)}{\sqrt x-1}+\frac{\left(\sqrt x-2\right)\left(\sqrt x+2\right)}{\sqrt x+2}=\sqrt x+\sqrt x-2=2\sqrt x-2.$

So $A=2\sqrt x-2$ for $x\ge0$ , $x\neq1$ .

Watch out Common mistakes

Adding unlike surds. Wrong: $\sqrt2+\sqrt3=\sqrt5$ . Why it is wrong: a root does not split over addition. Right: $\sqrt2+\sqrt3$ is already simplest; we can only add $a\sqrt3+b\sqrt3=(a+b)\sqrt3$ .
Moving a negative number inside the root. Wrong: $-2\sqrt3=\sqrt{12}$ . Why it is wrong: the left side is negative and the right side is positive. Right: $-2\sqrt3=-\sqrt{12}$ .
Multiplying only the denominator by the conjugate. Wrong: $\dfrac{4}{\sqrt5-1}=\dfrac{4}{4}=1$ . Why it is wrong: if the denominator is multiplied by $\sqrt5+1$ , so must the numerator be. Right: $\dfrac{4\left(\sqrt5+1\right)}{4}=\sqrt5+1$ .
Forgetting the conditions when simplifying. Wrong: simplifying $\dfrac{x-\sqrt x}{\sqrt x-1}$ without writing $x\neq1$ . Why it is wrong: at $x=1$ the denominator is $0$ . Right: state $x\ge0$ , $x\neq1$ .

Key terms

English	Tiếng Việt
take a factor outside the square root	đưa thừa số ra ngoài dấu căn
move a factor inside the square root	đưa thừa số vào trong dấu căn
like surds	căn thức đồng dạng
remove the denominator under the root sign	khử mẫu của biểu thức lấy căn
rationalise the denominator	trục căn thức ở mẫu
conjugate	biểu thức liên hợp
simplify	rút gọn
factorise	phân tích thành nhân tử

Summary

$\sqrt{A^2B}=|A|\sqrt B$ ( $B\ge0$ ): split off a square and take it outside.
$A\sqrt B=\sqrt{A^2B}$ when $A\ge0$ ; $A\sqrt B=-\sqrt{A^2B}$ when $A<0$ . Useful for comparing.
Like surds: $a\sqrt c+b\sqrt c=(a+b)\sqrt c$ .
$\dfrac{A}{\sqrt B}=\dfrac{A\sqrt B}{B}$ ; with a denominator $\sqrt A\pm B$ or $\sqrt A\pm\sqrt B$ , multiply the numerator and the denominator by the conjugate.
Simplifying: conditions → factorise → common denominator → collect terms → check with one value.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Take the factor outside the root sign: a) $\sqrt{75}$ ; b) $\sqrt{200}$ ; c) $\sqrt{28a^2}$ for $a<0$ .

What you need and how to start

What you need: $\sqrt{A^2B}=|A|\sqrt B$ .

Steps:

Find the largest perfect square that divides the number under the root.
Take its root outside; with letters, look at the sign.

Show solution

a) $\sqrt{75}=\sqrt{25\cdot3}=5\sqrt3$ . b) $\sqrt{200}=\sqrt{100\cdot2}=10\sqrt2$ . c) $\sqrt{28a^2}=\sqrt{(2a)^2\cdot7}=2|a|\sqrt7=-2a\sqrt7$ (since $a<0$ ).

Answer. a) $5\sqrt3$ ; b) $10\sqrt2$ ; c) $-2a\sqrt7$ .

Exercise 2

Evaluate $\sqrt{20}-\sqrt{45}+3\sqrt5$ .

What you need and how to start

What you need: take factors outside the root sign to get like surds.

Steps:

Write $\sqrt{20}=2\sqrt5$ , $\sqrt{45}=3\sqrt5$ .
Add and subtract the coefficients.

Show solution

We have $\sqrt{20}-\sqrt{45}+3\sqrt5=2\sqrt5-3\sqrt5+3\sqrt5=2\sqrt5$ .

Answer. $2\sqrt5$ .

Exercise 3

a) Compare $4\sqrt3$ and $3\sqrt5$ .

b) Put in increasing order: $2\sqrt7$ , $3\sqrt3$ , $4\sqrt2$ .

What you need and how to start

What you need: for $A\ge0$ , $A\sqrt B=\sqrt{A^2B}$ .

Steps:

Move the coefficients inside the root sign.
Compare the numbers under the roots.

Show solution

a) We have $4\sqrt3=\sqrt{48}$ and $3\sqrt5=\sqrt{45}$ . Since $48>45$ , $4\sqrt3>3\sqrt5$ .

b) We have $2\sqrt7=\sqrt{28}$ , $3\sqrt3=\sqrt{27}$ and $4\sqrt2=\sqrt{32}$ . Since $27<28<32$ , $3\sqrt3<2\sqrt7<4\sqrt2$ .

Answer. a) $4\sqrt3>3\sqrt5$ ; b) $3\sqrt3<2\sqrt7<4\sqrt2$ .

Exercise 4

Rationalise the denominator: a) $\dfrac{10}{\sqrt5}$ ; b) $\dfrac{2}{\sqrt3+1}$ ; c) $\dfrac{4}{\sqrt7-\sqrt3}$ .

What you need and how to start

What you need: $\dfrac A{\sqrt B}=\dfrac{A\sqrt B}B$ ; multiply by the conjugate of the denominator.

Steps:

Choose what to multiply the numerator and the denominator by.
Work out the denominator with the identity, then simplify.

Show solution

a) $\dfrac{10}{\sqrt5}=\dfrac{10\sqrt5}{5}=2\sqrt5$ . b) $\dfrac{2}{\sqrt3+1}=\dfrac{2\left(\sqrt3-1\right)}{3-1}=\sqrt3-1$ . c) $\dfrac{4}{\sqrt7-\sqrt3}=\dfrac{4\left(\sqrt7+\sqrt3\right)}{7-3}=\sqrt7+\sqrt3$ .

Answer. a) $2\sqrt5$ ; b) $\sqrt3-1$ ; c) $\sqrt7+\sqrt3$ .

Exercise 5

Remove the denominator under the root sign: a) $\sqrt{\dfrac23}$ ; b) $\sqrt{\dfrac58}$ .

What you need and how to start

What you need: $\sqrt{\dfrac AB}=\dfrac{\sqrt{AB}}{|B|}$ .

Steps:

Multiply the numerator and the denominator under the root so that the denominator is a perfect square.
Take the root of the denominator.

Show solution

a) $\sqrt{\dfrac23}=\sqrt{\dfrac{2\cdot3}{3^2}}=\dfrac{\sqrt6}{3}$ . b) $\sqrt{\dfrac58}=\sqrt{\dfrac{5\cdot2}{16}}=\dfrac{\sqrt{10}}{4}$ .

Answer. a) $\dfrac{\sqrt6}3$ ; b) $\dfrac{\sqrt{10}}4$ .

Exercise 6

Evaluate $\left(\sqrt{12}+\sqrt{27}\right)\cdot\sqrt3$ .

What you need and how to start

What you need: $\sqrt a\cdot\sqrt b=\sqrt{ab}$ (see square roots with multiplication and division).

Steps:

Expand the bracket.
Combine into one root and take the root.

Show solution

We have $\left(\sqrt{12}+\sqrt{27}\right)\cdot\sqrt3=\sqrt{36}+\sqrt{81}=6+9=15$ .

Answer. $15$ .

Exercise 7

Three square garden beds with areas of $18$ m², $32$ m² and $50$ m² are placed side by side in a row. Find the total length of their three sides along the row: write it in a neat form, then round to two decimal places.

What you need and how to start

What you need: a square of area $S$ has side $\sqrt S$ ; take factors outside the root sign.

Steps:

Write the three sides in the form $a\sqrt2$ .
Add, then use the calculator.

Show solution

The three sides are $\sqrt{18}$ m, $\sqrt{32}$ m and $\sqrt{50}$ m. Hence the total length is $\sqrt{18}+\sqrt{32}+\sqrt{50}=3\sqrt2+4\sqrt2+5\sqrt2=12\sqrt2\approx16.97$ (m).

Answer. $12\sqrt2$ m, about $16.97$ m.

Exercise 8

Simplify the expression $B=\dfrac{a\sqrt b+b\sqrt a}{\sqrt a+\sqrt b}$ for $a>0$ , $b>0$ .

What you need and how to start

What you need: $a=\left(\sqrt a\right)^2$ for $a\ge0$ ; take out a common factor.

Steps:

Write the numerator as $\sqrt a\sqrt b\left(\sqrt a+\sqrt b\right)$ .
Cancel with the denominator.

Show solution

Since $a>0$ and $b>0$ , $a\sqrt b+b\sqrt a=\sqrt a\sqrt b\left(\sqrt a+\sqrt b\right)$ . Hence $B=\dfrac{\sqrt a\sqrt b\left(\sqrt a+\sqrt b\right)}{\sqrt a+\sqrt b}=\sqrt{ab}$ .

Answer. $B=\sqrt{ab}$ .

Exercise 9

Consider the expression $P=\dfrac{\sqrt x}{\sqrt x-3}-\dfrac{3}{\sqrt x+3}$ .

a) Find the condition for $P$ to be defined.

b) Simplify $P$ .

c) Find the value of $P$ when $x=36$ .

d) Find $x$ such that $P=2$ .

What you need and how to start

What you need: the common denominator is $\left(\sqrt x-3\right)\left(\sqrt x+3\right)=x-9$ .

Steps:

We need $x\ge0$ and $\sqrt x\neq3$ .
Write over the common denominator and collect the numerator.
Substitute $x=36$ ; solve the equation in d) and check the condition.

Show solution

a) The expression is defined when $x\ge0$ and $\sqrt x\neq3$ , that is, $x\ge0$ , $x\neq9$ .

b) For $x\ge0$ , $x\neq9$ , we have $P=\dfrac{\sqrt x\left(\sqrt x+3\right)-3\left(\sqrt x-3\right)}{x-9}=\dfrac{x+3\sqrt x-3\sqrt x+9}{x-9}=\dfrac{x+9}{x-9}$ .

c) Substituting $x=36$ (which satisfies the condition) into $P$ , we get $P=\dfrac{36+9}{36-9}=\dfrac{45}{27}=\dfrac53$ .

d) We have $P=2\iff\dfrac{x+9}{x-9}=2\iff x+9=2(x-9)\iff x=27$ (which satisfies the condition).

Answer. a) $x\ge0$ , $x\neq9$ ; b) $P=\dfrac{x+9}{x-9}$ ; c) $\dfrac53$ ; d) $x=27$ .

Exercise 10 Challenge

Evaluate $C=\sqrt{4-2\sqrt3}+\sqrt{4+2\sqrt3}$ .

What you need and how to start

What you need: $\left(\sqrt3\pm1\right)^2=4\pm2\sqrt3$ and $\sqrt{A^2}=|A|$ .

Steps:

Write $4-2\sqrt3=\left(\sqrt3-1\right)^2$ and $4+2\sqrt3=\left(\sqrt3+1\right)^2$ .
Remove the roots, looking at the sign of $\sqrt3-1$ .

Show solution

We have $4-2\sqrt3=\left(\sqrt3-1\right)^2$ and $4+2\sqrt3=\left(\sqrt3+1\right)^2$ . Hence $C=\left|\sqrt3-1\right|+\left|\sqrt3+1\right|$ . Since $\sqrt3>1$ , $C=\left(\sqrt3-1\right)+\left(\sqrt3+1\right)=2\sqrt3$ .

Answer. $C=2\sqrt3$ .

a) $\sqrt{50}=5\sqrt2$ .	True	False
b) $A=\sqrt{40}$ .	True	False
c) $A\cdot\sqrt2=8$ .	True	False
d) $\dfrac1A=\dfrac{\sqrt2}{8}$ .	True	False

Taking factors outside and inside the root sign

Explore

Removing denominators and rationalising

Simplifying expressions with square roots

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer