Simplifying Square Roots and Surds

Hello. In the previous lesson, you learned that fractional exponents and roots are two notations for the same idea:

$$x^{1/2}=\sqrt{x}$$

This lesson turns that connection into an exam-useful skill: simplifying square roots, combining surds correctly, and removing a square root from a denominator when needed. These questions reward method more than cleverness. A clean factorisation and one reliable checking routine will prevent most errors.

A surd is a root expression whose value is irrational, such as $\sqrt{2}$, $\sqrt{3}$, or $5\sqrt{7}$. In a final simplified answer, the number inside a square root should have no square factor greater than $1$.

---

Square roots: extract pairs

The square-root symbol means the principal, non-negative square root. Thus,

$$\sqrt{49}=7$$

not both $7$ and $-7$. Both numbers square to $49$, but $\sqrt{49}$ specifically denotes the positive one.

The central rule is the product property:

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

provided $a$ and $b$ are non-negative real numbers.

This is exactly compatible with the exponent rule from the previous lesson:

$$\sqrt{ab}=(ab{)}^{1/2}=a^{1/2}{b}^{1/2}$$

The practical consequence is simple: if a number inside a root contains a perfect-square factor, take that factor outside.

For example,

$$\sqrt{72}$$

Since $72=36\cdot 2$,

$$\sqrt{72}=\sqrt{36\cdot 2}$$ $$=\sqrt{36}\sqrt{2}$$ $$=6\sqrt{2}$$

The number $2$ has no perfect-square factor other than $1$, so $6\sqrt{2}$ is simplified.

Choose the largest square factor

For speed, recognise these common perfect squares:

Number	Perfect squares worth testing
Divisible by $4$	$4,16,36,64,\dots$
Divisible by $9$	$9,36,81,\dots$
Divisible by $25$	$25,100,\dots$
Around $50$ to $150$	$36,49,64,81,100,121,144$

For instance:

$$\sqrt{108}$$

The largest convenient square factor is $36$:

$$\sqrt{108}=\sqrt{36\cdot 3}=6\sqrt{3}$$

Using a smaller factor, such as $4\cdot 27$, would still work, but it creates an unnecessary extra step:

$$\sqrt{108}=\sqrt{4\cdot 27}=2\sqrt{27}=2\sqrt{9\cdot 3}=6\sqrt{3}$$

Both are correct. The first is more efficient under time pressure.

When no large square factor is immediately visible, use prime factorisation. Every pair of equal prime factors produces one factor outside the root:

$$\sqrt{360}$$ $$=\sqrt{2^3\cdot 3^2\cdot 5}$$ $$=\sqrt{2^2\cdot 3^2\cdot 10}$$ $$=2\cdot 3\sqrt{10}$$ $$=6\sqrt{10}$$

Think of it as “pairs come out; unpaired factors remain inside.”

---

What can and cannot be done with roots

Square roots behave well with multiplication and division:

$$\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$$ $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}(b>0)$$

But they do not distribute over addition or subtraction:

$$\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$$

For a quick proof, take $a=9$ and $b=16$:

$$\sqrt{9+16}=\sqrt{25}=5$$

whereas

$$\sqrt{9}+\sqrt{16}=3+4=7$$

Since $5\ne 7$, the supposed rule is false.

This is a classic entrance-exam trap. Do not split a root across a plus or minus sign.

Add and subtract only like surds

Surds are like terms only when their radical parts match exactly. First simplify every term; only then combine coefficients.

Consider:

$$3\sqrt{8}+2\sqrt{18}-\sqrt{50}$$

Simplify each root:

$$3\sqrt{8}=3(2\sqrt{2})=6\sqrt{2}$$ $$2\sqrt{18}=2(3\sqrt{2})=6\sqrt{2}$$ $$\sqrt{50}=5\sqrt{2}$$

Now all three terms are like surds:

$$6\sqrt{2}+6\sqrt{2}-5\sqrt{2}$$ $$=7\sqrt{2}$$

In contrast,

$$2\sqrt{3}+5\sqrt{2}$$

cannot be combined. The radical parts differ, just as $2x+5y$ cannot be combined.

A frequent wrong move is:

$$\sqrt{12}+\sqrt{3}=\sqrt{15}$$

This is false. Simplify the first term instead:

$$\sqrt{12}+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}$$

---

Multiplying and dividing surds

With multiplication, combine the numbers under the roots first, then simplify:

$$\sqrt{6}\cdot \sqrt{15}$$ $$=\sqrt{90}$$ $$=\sqrt{9\cdot 10}$$ $$=3\sqrt{10}$$

If coefficients are present, multiply them normally:

$$(2\sqrt{3})(5\sqrt{6})$$ $$=10\sqrt{18}$$ $$=10(3\sqrt{2})$$ $$=30\sqrt{2}$$

For division, simplify coefficients and radicals carefully:

$$\frac{4\sqrt{18}}{2\sqrt{2}}$$

First simplify the numerical coefficient:

$$=2\cdot \frac{\sqrt{18}}{\sqrt{2}}$$

Then use the quotient property:

$$=2\sqrt{\frac{18}{2}}$$ $$=2\sqrt{9}$$ $$=6$$

---

Roots of fractions and rationalising the denominator

A square root of a fraction may be separated as follows:

$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}(a\ge 0,b>0)$$

For example,

$$\sqrt{\frac{72}{25}}$$ $$=\frac{\sqrt{72}}{\sqrt{25}}$$ $$=\frac{6\sqrt{2}}{5}$$

Sometimes a radical appears in the denominator, as in:

$$\frac{\sqrt{8}}{\sqrt{10}}$$

In standard simplified form, we usually remove that radical from the denominator. This is called rationalising the denominator.

Multiply the fraction by $1$, written in a useful form:

$$\frac{\sqrt{10}}{\sqrt{10}}=1$$

So,

$$\frac{\sqrt{8}}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}$$ $$=\frac{\sqrt{80}}{10}$$ $$=\frac{4\sqrt{5}}{10}$$ $$=\frac{2\sqrt{5}}{5}$$

The reason this works is not a special trick: multiplying numerator and denominator by the same non-zero quantity preserves the value of a fraction. The particular multiplier is chosen so that the denominator becomes a perfect square:

$$\sqrt{n}\cdot \sqrt{n}=\sqrt{n^2}=n$$

For a denominator containing just one square root, use this pattern:

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

For example,

$$\frac{7}{\sqrt{5}}$$ $$=\frac{7}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}$$ $$=\frac{7\sqrt{5}}{5}$$

For now, focus on denominators containing a single root. Denominators such as $\sqrt{a}+\sqrt{b}$ need a further idea called a conjugate; that will fit naturally once algebraic identities have been covered.

---

Variables inside square roots: one important caution

When variables are assumed positive, simplify them just like numerical factors:

$$\sqrt{45x^2}$$ $$=\sqrt{9\cdot 5\cdot x^2}$$ $$=3x\sqrt{5}$$

However, without a condition such as $x\ge 0$, the completely correct statement is:

$$\sqrt{x^2}=|x|$$

For example, if $x=-4$, then:

$$\sqrt{x^2}=\sqrt{16}=4$$

but $x=-4$. The square root cannot equal a negative number.

In most basic entrance-exam simplification questions, the variable is stated or understood to be positive. Still, check any conditions given in the question before writing $\sqrt{x^2}=x$.

---

A three-pass routine for exam questions

Use this routine every time you simplify a surd expression.

1. Simplify each radicand. Extract every perfect-square factor.
2. Perform the operation. Combine only like surds for addition or subtraction; combine radicands only for multiplication or division.
3. Standardise the final form. Combine coefficients, leave a square-free radicand, and rationalise a single-root denominator if one remains.

For example:

$$\frac{\sqrt{27}+2\sqrt{12}}{\sqrt{3}}$$

First pass: simplify the numerator.

$$\sqrt{27}=3\sqrt{3}$$ $$2\sqrt{12}=2(2\sqrt{3})=4\sqrt{3}$$

Thus,

$$\frac{3\sqrt{3}+4\sqrt{3}}{\sqrt{3}}$$

Second pass: combine like surds.

$$=\frac{7\sqrt{3}}{\sqrt{3}}$$

Third pass: simplify the quotient.

$$=7$$

Notice that trying to divide before simplifying would not be wrong, but it would be slower and make an error more likely.

---

Key takeaways

A simplified square-root expression has no perfect-square factor left inside a radicand:

$$\sqrt{72}=6\sqrt{2}$$

The valid structural rules are:

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

But do not apply either rule across addition or subtraction:

$$\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$$

Simplify before adding or subtracting, and combine only like surds. When a single square root remains in a denominator, multiply numerator and denominator by that root to rationalise it.

Next, you will use these foundations to expand algebraic expressions with standard identities. The same structural discipline matters there: brackets, signs, and the exact operation between terms determine which rule is valid.