P1.2.2 - Surds and rationalising denominators

P1.2.2 - Surds and rationalising denominators

Surds let you keep exact answers instead of replacing them with rounded decimals. In this lesson you will simplify, multiply, and collect square-root expressions, then use rationalising to rewrite fractions so the denominator is rational. The key idea is simple: choose a form of 1 that changes the denominator without changing the value of the fraction.

Exact surd form

A square root is exact notation. For example, \sqrt{2} is not a rounded number; it is the exact positive number whose square is 2.

Surd

A surd is an irrational root left in exact form, such as \sqrt{2}, \sqrt{3}, or 5\sqrt{7}. In this lesson, all surds are square-root surds.

When x >= 0, the notation \sqrt{x} means the non-negative square root of x. This is why

(x)2=x.(\sqrt{x})^2=x.

For non-negative x and y, the multiplication rule is

Square-root product rule

xy=xy\sqrt{xy}=\sqrt{x}\sqrt{y}

This rule is useful in both directions. To simplify \sqrt{72}, look for a square factor:

72=36×2=362=62.\sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\sqrt{2}=6\sqrt{2}.

A surd is in a clean simplified form when the number under the square root has no square factor greater than 1. So \sqrt{72} is not simplified, but 6\sqrt{2} is.

Worked example: simplify 3\sqrt{18}-\sqrt{50}.

First simplify each surd separately:

318=39×2=3(32)=92,3\sqrt{18}=3\sqrt{9\times 2}=3(3\sqrt{2})=9\sqrt{2},

and

50=25×2=52.\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}.

Now the terms are like surds, so they can be collected:

318−50=92−52=42.3\sqrt{18}-\sqrt{50}=9\sqrt{2}-5\sqrt{2}=4\sqrt{2}.

The common trap is to use the multiplication rule for addition. You may write \sqrt{18}=\sqrt{9\times 2}, but you may not write \sqrt{9+2}=\sqrt{9}+\sqrt{2}.

Multiplying and expanding surds

Surds behave like algebraic terms when you multiply brackets: expand carefully, then simplify each product.

For example,

(2+3)(5−3)=10−23+53−(3)2.(2+\sqrt{3})(5-\sqrt{3}) =10-2\sqrt{3}+5\sqrt{3}-(\sqrt{3})^2.

Since (\sqrt{3})^2=3, this becomes

10+33−3=7+33.10+3\sqrt{3}-3=7+3\sqrt{3}.

The most important product in this spec point is the difference of two squares:

(x+y)(x−y).(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y}).

Expanding shows why the middle terms vanish:

(x+y)(x−y)=x−xy+xy−y=x−y.(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y}) =x-\sqrt{x}\sqrt{y}+\sqrt{x}\sqrt{y}-y=x-y.

So, for non-negative x and y,

Difference of two square-root terms

(x+y)(x−y)=x−y(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y

Worked example: simplify (3+\sqrt{5})(2-\sqrt{5}).

Expand every term:

(3+5)(2−5)=6−35+25−(5)2.(3+\sqrt{5})(2-\sqrt{5}) =6-3\sqrt{5}+2\sqrt{5}-(\sqrt{5})^2.

Now simplify the square-root square and collect:

6−35+25−5=1−5.6-3\sqrt{5}+2\sqrt{5}-5=1-\sqrt{5}.

This is exact. There is no reason to replace \sqrt{5} with a decimal unless a later question explicitly asks for a decimal approximation.

Rationalising a single-surd denominator

To rationalise a denominator means to rewrite a fraction so that the denominator is rational. The value of the fraction must not change, so you multiply by a form of 1.

For a single square-root denominator, multiply numerator and denominator by the square root that will make the denominator a square.

Worked example: rationalise

63.\frac{6}{\sqrt{3}}.

Multiply by \sqrt{3}/\sqrt{3}:

63×33=633=23.\frac{6}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} =\frac{6\sqrt{3}}{3} =2\sqrt{3}.

The denominator is now rational. The numerator may still contain a surd, which is fine.

Worked example: rationalise and simplify

4+222.\frac{4+\sqrt{2}}{2\sqrt{2}}.

Multiply numerator and denominator by \sqrt{2}:

4+222×22=42+24.\frac{4+\sqrt{2}}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} =\frac{4\sqrt{2}+2}{4}.

Now simplify the fraction:

42+24=22+12.\frac{4\sqrt{2}+2}{4}=\frac{2\sqrt{2}+1}{2}.

Notice that the whole numerator is multiplied by \sqrt{2}, not just one term. That is what preserves the value of the original fraction.

Rationalising a two-term denominator

If the denominator has two terms, such as 3+\sqrt{5} or 4-\sqrt{3}, multiplying by a single square root will not remove every surd term. Instead, use the conjugate.

Conjugate

The conjugate of a+b\sqrt{c} is a-b\sqrt{c}. The conjugate of a-b\sqrt{c} is a+b\sqrt{c}.

The conjugate works because the product becomes a difference of squares:

(a+bc)(a−bc)=a2−b2c.(a+b\sqrt{c})(a-b\sqrt{c})=a^2-b^2c.

That denominator is rational.

Worked example: rationalise

43+5.\frac{4}{3+\sqrt{5}}.

The conjugate of 3+\sqrt{5} is 3-\sqrt{5}, so

43+5×3−53−5=4(3−5)32−(5)2.\frac{4}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}} =\frac{4(3-\sqrt{5})}{3^2-(\sqrt{5})^2}.

The denominator is

9−5=4,9-5=4,

so

4(3−5)4=3−5.\frac{4(3-\sqrt{5})}{4}=3-\sqrt{5}.

Worked example: rationalise and simplify

1+23−2.\frac{1+\sqrt{2}}{3-\sqrt{2}}.

Use the conjugate 3+\sqrt{2}:

1+23−2×3+23+2=(1+2)(3+2)9−2.\frac{1+\sqrt{2}}{3-\sqrt{2}}\times \frac{3+\sqrt{2}}{3+\sqrt{2}} =\frac{(1+\sqrt{2})(3+\sqrt{2})}{9-2}.

Expand the numerator:

(1+2)(3+2)=3+2+32+2=5+42.(1+\sqrt{2})(3+\sqrt{2})=3+\sqrt{2}+3\sqrt{2}+2=5+4\sqrt{2}.

Therefore

1+23−2=5+427.\frac{1+\sqrt{2}}{3-\sqrt{2}}=\frac{5+4\sqrt{2}}{7}.

The sign is the detail to watch. If the denominator is a+\sqrt{b}, use a-\sqrt{b}. If the denominator is a-\sqrt{b}, use a+\sqrt{b}.

Choosing a clean final form

In exam-style algebra, rationalising is often only one step in a longer exact simplification. After rationalising, check whether the expression can still be collected, factorised, or written in a requested form.

Worked example: show that

25−1−5=1−52.\frac{2}{\sqrt{5}-1}-\sqrt{5}=\frac{1-\sqrt{5}}{2}.

First rationalise the fraction:

25−1×5+15+1=2(5+1)5−1=5+12.\frac{2}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1} =\frac{2(\sqrt{5}+1)}{5-1} =\frac{\sqrt{5}+1}{2}.

Now subtract \sqrt{5}. Use a common denominator:

5+12−5=5+1−252=1−52.\frac{\sqrt{5}+1}{2}-\sqrt{5} =\frac{\sqrt{5}+1-2\sqrt{5}}{2} =\frac{1-\sqrt{5}}{2}.

This style of question rewards the rationalising method and the final exact form. Do not switch to decimals: a decimal answer would hide the exact structure.

Choosing a clean final form Continued

The final habit is to read the requested form. If the question asks for p+q\sqrt{r}, the numbers p and q may be fractions or negative. They only need to be rational.