P1 1.2 - Simplifying surds

Surd

A surd is a root that is irrational. For example, $\sqrt2$ and $\sqrt{12}$ are surds; $\sqrt{16}=4$ is not. An exact form retains the value without rounding.

Extracting a square factor

For $k\ge0$ and $m\ge0$,

If $k$ can be negative, the factor outside is $|k|$, because a square root is nonnegative.

Not in the formula booklet: learn it.

Worked example 1: one square factor

Simplify $\sqrt{18}$.

Plan. Separate the square factor $9$ from the remaining factor $2$.

Answer. $3\sqrt2$.

Check. The proposed result is positive and $(3\sqrt2)^2=9\times2=18$, so it is the positive square root of $18$.

Worked example 2: a larger square factor

Simplify $\sqrt{72}$.

Plan. Use $36\times2$, rather than stopping after extracting $4$.

Answer. $6\sqrt2$.

Check. $(6\sqrt2)^2=72$, and $6\sqrt2>0$. Also, $8^2<72<9^2$, so the value must lie between $8$ and $9$, as $6\sqrt2$ does.

Watch out

$\sqrt{9+16}=5$, whereas $\sqrt9+\sqrt{16}=7$. The rule for $\sqrt{ab}$ does not turn $\sqrt{a+b}$ into $\sqrt a+\sqrt b$.

Worked example 3: addition after simplification

Simplify $\sqrt{12}+\sqrt{27}$.

Plan. Rewrite each term as a multiple of $\sqrt3$, then collect coefficients.

Answer. $5\sqrt3$.

Check. $\sqrt{12}\approx3.464$ and $\sqrt{27}\approx5.196$; their sum and $5\sqrt3$ both give approximately $8.660$.

Worked example 4: multiplying the same surds

Simplify $\sqrt{12}\times\sqrt{27}$.

Plan. Multiply inside one root, then recognise a square number.

Answer. $18$, which is rational even though both factors are irrational.

Check. Using the simplified factors instead gives $(2\sqrt3)(3\sqrt3)=6\times3=18$.

Watch out

$(\sqrt2+3)^2$ has a middle term:

Squaring each term alone misses the two cross-products.

Rationalising a denominator

To rationalise a denominator is to rewrite a fraction so that its denominator is rational. Multiplying the top and bottom by the same nonzero expression multiplies the fraction by $1$.

Worked example 5: a single root below the line

Rationalise the denominator of $6/\sqrt3$.

Plan. Multiply both numerator and denominator by $\sqrt3$.

Answer. $2\sqrt3$.

Check. Multiplying the answer by the original denominator gives $(2\sqrt3)\sqrt3=6$, the original numerator.

Worked example 6: a coefficient in the denominator

Rationalise the denominator of $6/(2\sqrt3)$.

Plan. Keep the factor $2$ when multiplying the denominator by $\sqrt3$.

Answer. $\sqrt3$.

Check. $\sqrt3(2\sqrt3)=6$. Also, doubling a denominator must halve the value found in the previous example.

Watch out

$1/\sqrt5\ne1/5$. Multiplying only the denominator changes the value. The valid result is $1/\sqrt5=\sqrt5/5$.

Conjugate

For a two-term surd expression, its conjugate has the same two terms with the sign between them reversed. For example, $3+\sqrt2$ and $3-\sqrt2$ are conjugates.

Worked example 7: a sum in the denominator

Rationalise the denominator of $1/(\sqrt5+2)$.

Plan. Use $\sqrt5-2$, so the denominator becomes a difference of squares.

Answer. $\sqrt5-2$.

Check. The answer is positive and approximately $0.236$; its product with $\sqrt5+2$ is exactly $1$.

Worked example 8: a difference in the denominator

Rationalise the denominator of $1/(\sqrt5-2)$.

Plan. Multiply both parts by $\sqrt5+2$.

Answer. $\sqrt5+2$, exactly as claimed at the start.

Check. $(\sqrt5+2)(\sqrt5-2)=1$. The original denominator is small and positive, so its reciprocal should be greater than $1$, as this answer is.

Watch out

For $2-\sqrt5$, the conjugate gives $(2-\sqrt5)(2+\sqrt5)=4-5=-1$. The denominator is negative; it cannot be replaced by $1$.