Algebra · Paper 1
Surds
Higher Level · Tap NEXT to begin
Section 1 of 9
What is a Surd?
Surds are square roots.
Simple Surd
$a\sqrt{b}$
$a$ = rational
$b$ = rational (i.e. $b$ ∈ $\tfrac{p}{q}$)
Compound Surd
Two parts:
$a \pm \sqrt{b}\qquad$ or $\qquad \sqrt{a} \pm \sqrt{b}$
Section 2 of 9
The Four Properties
Properties
(i)$\sqrt{ab} = \sqrt{a}\,\sqrt{b}$
(ii)$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$
(iii)$\sqrt{a^{2}} = \left(\sqrt{a}\right)^{2} = a$
(iv)$\sqrt{a \pm b}$ ≠ $\sqrt{a} \pm \sqrt{b}$
Warning — rule (iv). You cannot split a square root over a sum or difference. Check:
$\sqrt{4+9}$ ≠ $\sqrt{13}$
$\sqrt{4} + \sqrt{9}$ ≠ $\sqrt{13}$
$2 + 3$ ≠ $\sqrt{13}$
$5$ ≠ $\sqrt{13}$
You try
Spot the error. A student writes: $\sqrt{25 + 144} = \sqrt{25} + \sqrt{144} = 5 + 12 = 17$. What's wrong?
Which property did the student break? Compute $25 + 144$ first.
Property (iv): $\sqrt{a + b}$ ≠ $\sqrt{a} + \sqrt{b}$. The student split the root over a sum.
Correct: $\sqrt{25 + 144} = \sqrt{169} = 13$.
The answer is $13$, not $17$.
$\sqrt{169} = 13$ — student broke property (iv)
Section 3 of 9
A Note on $\sqrt{k}$ when $k<0$
$\sqrt{k}$ is impossible where $k < 0$.
Why? Any number squared is positive. Look at this:
$\sqrt{-36}$ ≠ $6 \quad$since$\quad (6)^{2} = 36$ ≠ $-36 \;\;\times$
$\sqrt{-36}$ ≠ $-6 \quad$since$\quad (-6)^{2} = 36$ ≠ $-36 \;\;\times$
Neither value works — and no real number squared will give a negative result.
$\therefore \;\; \sqrt{-36} = \text{impossible}$
You try
Is $\sqrt{-49}$ possible? Explain.
What happens when you square any real number?
Any real number squared is $\geq 0$. So nothing squared can give $-49$.
No — $\sqrt{-49}$ is impossible (in the reals).
No — impossible
Section 4 of 9
Simplifying Surds
Use property (i): pull out the largest square factor.
Simplify $\sqrt{8}$
$\sqrt{8} = \sqrt{4}\,\sqrt{2}$
$= 2\sqrt{2}$
Check: $\left(2\sqrt{2}\right)^{2} = 4(2) = 8 \;\checkmark$
Simplify $\sqrt{27}$
$\sqrt{27} = \sqrt{9}\,\sqrt{3}$
$= 3\sqrt{3}$
Simplify $\sqrt{72}$ — two ways
Slow way — extract a small square factor first:
$\sqrt{72} = \sqrt{4}\,\sqrt{18}$
$= 2\sqrt{18} = 2\sqrt{9}\,\sqrt{2}$
$= 2 \cdot 3 \cdot \sqrt{2}$
$= 6\sqrt{2}$
Fast way — extract the largest square factor straight away:
$\sqrt{72} = \sqrt{36}\,\sqrt{2}$
$= 6\sqrt{2}$
Lesson: always pull out the largest square factor — fewer steps, less arithmetic.
You try
Simplify $\sqrt{48}$
Largest square factor of 48? Then use $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
$\sqrt{48} = \sqrt{16}\,\sqrt{3}$
$= 4\sqrt{3}$
$4\sqrt{3}$
Section 5 of 9
Add and Subtract
Rule
You can only add or subtract like surds (same number under the root). Method: simplify each surd first, then combine the like surds.
(i) $2 + 3\sqrt{5} + 7 + 2\sqrt{5}$
$= 9 + 5\sqrt{5}$
(ii) $\sqrt{3} + \sqrt{5}$
Three wrong moves students often try — none of them work:
$\sqrt{3} + \sqrt{5} = 3\sqrt{5} \quad\times$
$\sqrt{3} + \sqrt{5} = \sqrt{15} \quad\times$ (that's multiplication)
$\sqrt{3} + \sqrt{5} = \sqrt{2+3} = \sqrt{2} + \sqrt{3} \quad\times$ (property (iv))
Different surds — no shortcut exists. Leave as is:
$\sqrt{3} + \sqrt{5}$
(iii) $\sqrt{50} - \sqrt{32}$
$\sqrt{50} - \sqrt{32}$ ≠ $\sqrt{18}$
$\sqrt{50} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}$
$\sqrt{32} = \sqrt{16}\,\sqrt{2} = 4\sqrt{2}$
$5\sqrt{2} - 4\sqrt{2} = \sqrt{2}$
(iv) $\sqrt{12} - \sqrt{45}$
$\sqrt{12} = \sqrt{4}\,\sqrt{3} = 2\sqrt{3}$
$\sqrt{45} = \sqrt{9}\,\sqrt{5} = 3\sqrt{5}$
$2\sqrt{3} - 3\sqrt{5}$
Different surds → leave as is.
(v) $\sqrt{27} - \sqrt{300}$
$\sqrt{27} = \sqrt{9}\,\sqrt{3} = 3\sqrt{3}$
$\sqrt{300} = \sqrt{100}\,\sqrt{3} = 10\sqrt{3}$
$3\sqrt{3} - 10\sqrt{3} = -7\sqrt{3}$
You try
Simplify: $\sqrt{75} + \sqrt{48}$
Simplify each surd first (pull out the largest square factor). Then look for like surds and combine.
$\sqrt{75} = \sqrt{25}\,\sqrt{3} = 5\sqrt{3}$
$\sqrt{48} = \sqrt{16}\,\sqrt{3} = 4\sqrt{3}$
$5\sqrt{3} + 4\sqrt{3}$
$= 9\sqrt{3}$
$9\sqrt{3}$
Section 6 of 9
Multiplication
Key rule
$\sqrt{a}\,\sqrt{b} = \sqrt{ab}$
Basic examples
(i)$3 \cdot 2\sqrt{5} = 6\sqrt{5}$
(ii)$\sqrt{5}\,(\sqrt{5}) = \sqrt{25} = 5$
(iii)$\sqrt{3}\,(\sqrt{7}) = \sqrt{21}$
(iv)$2\sqrt{3}\,(7\sqrt{5}) = 14\sqrt{15}$
(v)$5\,(2 + 6\sqrt{3}) = 10 + 30\sqrt{3}$
(vi)$\sqrt{2}\,(5 + 3\sqrt{2}) = 5\sqrt{2} + 6$
(vii) $(2 + \sqrt{5})(3 + \sqrt{5})$
$= 6 + 2\sqrt{5} + 3\sqrt{5} + 5$
$= 11 + 5\sqrt{5}$
(viii) $(2 + 3\sqrt{3})(5 - 2\sqrt{3})$
$= 2(5-2\sqrt{3}) + 3\sqrt{3}(5-2\sqrt{3})$
$= 10 - 4\sqrt{3} + 15\sqrt{3} - 18$
$= -8 + 11\sqrt{3}$
(ix) $(5 + 2\sqrt{3})(4 - 3\sqrt{3})$
$= 5(4 - 3\sqrt{3}) + 2\sqrt{3}(4 - 3\sqrt{3})$
$= 20 - 15\sqrt{3} + 8\sqrt{3} - 6(3)$
$= 20 - 18 - 15\sqrt{3} + 8\sqrt{3}$
$= 2 - 7\sqrt{3}$
Square of a sum $(2 + 3\sqrt{5})^{2}$
$= (2 + 3\sqrt{5})(2 + 3\sqrt{5})$
$= 4 + 12\sqrt{5} + 9(5)$
$= 49 + 12\sqrt{5}$
Difference of two squares — $(x+y)(x-y) = x^{2} - y^{2}$
$(5 + \sqrt{3})(5 - \sqrt{3}) = 25 - 3 = 22$
$(7 + \sqrt{2})(7 - \sqrt{2}) = 49 - 2 = 47$
You try
Multiply out: $(3 + 5\sqrt{2})(4 + 2\sqrt{2})$
FOIL it out. Watch the last term — $\sqrt{2}\cdot\sqrt{2} = 2$.
$(3 + 5\sqrt{2})(4 + 2\sqrt{2})$
$= 12 + 6\sqrt{2} + 20\sqrt{2} + 10(2)$
$= 32 + 26\sqrt{2}$
$32 + 26\sqrt{2}$
You try
Expand and simplify: $(2 - \sqrt{3})^{2}$
Write as $(2-\sqrt{3})(2-\sqrt{3})$ and FOIL. The cross-term comes up twice.
$(2-\sqrt{3})^{2} = (2-\sqrt{3})(2-\sqrt{3})$
$= 4 - 2\sqrt{3} - 2\sqrt{3} + 3$
$= 7 - 4\sqrt{3}$
$= 7 - 4\sqrt{3}$
$7 - 4\sqrt{3}$
Section 7 of 9
Conjugates
Conjugate rule
Every compound surd $a + \sqrt{b}$ has a conjugate $a - \sqrt{b}$ such that
$(a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b \;\in\; \mathbb{Q}$
In short:
Surd:$\;\;a + \sqrt{b}$
Conjugate:$\;\;a - \sqrt{b}$
A note on signs
There's a real difference between a minus inside the root and a minus outside:
$\sqrt{-a}$ — impossible (no real value)
$-\sqrt{a}$ — perfectly fine, just a negative number
You try
Find the conjugate of $4 + \sqrt{7}$ and multiply them together.
Flip the sign of the surd part. Then multiply using $(x+y)(x-y) = x^{2} - y^{2}$ — your answer should be rational.
Conjugate of $4 + \sqrt{7}$ is $4 - \sqrt{7}$
$(4 + \sqrt{7})(4 - \sqrt{7}) = 4^{2} - (\sqrt{7})^{2}$
$= 16 - 7$
$= 9$
Conjugate: $4 - \sqrt{7}$ · Product: $9$
Section 8 of 9
Rationalising the Denominator
You cannot leave a surd in the bottom of a fraction. Multiply above and below by something that turns the bottom rational.
Simple denominator — Simplify $\dfrac{8}{\sqrt{2}}$
Multiply above and below by $\sqrt{2}$:
$\dfrac{8}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{8\sqrt{2}}{2}$
$= 4\sqrt{2}$
Simplify $\dfrac{20}{\sqrt{5}}$
$\dfrac{20}{\sqrt{5}} \cdot \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{20\sqrt{5}}{5}$
$= 4\sqrt{5}$
Simplify $\dfrac{14}{\sqrt{2}}$
$\dfrac{14}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{14\sqrt{2}}{2}$
$= 7\sqrt{2}$
Compound denominator — Rule
Multiply above and below by the conjugate of the bottom.
Simplify $\dfrac{3}{5 + \sqrt{2}}$
First instinct: just multiply above and below by $\sqrt{2}$. Watch why it doesn't work:
$\dfrac{3}{5+\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{3\sqrt{2}}{5\sqrt{2} + 2} \quad\times$
Still a surd in the bottom — that didn't help.
Use the conjugate instead — multiply above and below by $5 - \sqrt{2}$:
$\dfrac{3}{5+\sqrt{2}} \cdot \dfrac{5-\sqrt{2}}{5-\sqrt{2}}$
Top:$3(5-\sqrt{2}) = 15 - 3\sqrt{2}$
Bottom:$(5+\sqrt{2})(5-\sqrt{2}) = 25 - 2 = 23$
$= \dfrac{15 - 3\sqrt{2}}{23}$
(i) Simplify $\dfrac{2 + \sqrt{5}}{3 + \sqrt{5}}$
$\dfrac{2+\sqrt{5}}{3+\sqrt{5}} \cdot \dfrac{3-\sqrt{5}}{3-\sqrt{5}}$
Top:$(2+\sqrt{5})(3-\sqrt{5})$
$= 2(3-\sqrt{5}) + \sqrt{5}(3-\sqrt{5})$
$= 6 - 2\sqrt{5} + 3\sqrt{5} - 5 = 1 + \sqrt{5}$
Bottom:$(3+\sqrt{5})(3-\sqrt{5}) = 9 - 5 = 4$
$= \dfrac{1 + \sqrt{5}}{4}$
(ii) Simplify $\dfrac{3 + 2\sqrt{7}}{4 - \sqrt{7}}$
$\dfrac{3+2\sqrt{7}}{4-\sqrt{7}} \cdot \dfrac{4+\sqrt{7}}{4+\sqrt{7}}$
Top:$(3+2\sqrt{7})(4+\sqrt{7})$
$= 3(4+\sqrt{7}) + 2\sqrt{7}(4+\sqrt{7})$
$= 12 + 3\sqrt{7} + 8\sqrt{7} + 14 = 26 + 11\sqrt{7}$
Bottom:$(4-\sqrt{7})(4+\sqrt{7}) = 16 - 7 = 9$
$= \dfrac{26 + 11\sqrt{7}}{9}$
You try
Rationalise: $\dfrac{4}{3 - \sqrt{5}}$
Multiply top and bottom by the conjugate of the bottom — i.e. $3 + \sqrt{5}$.
$\dfrac{4}{3-\sqrt{5}} \cdot \dfrac{3+\sqrt{5}}{3+\sqrt{5}}$
Top:$\;4(3+\sqrt{5}) = 12 + 4\sqrt{5}$
Bottom:$\;9 - 5 = 4$
$= \dfrac{12 + 4\sqrt{5}}{4} = 3 + \sqrt{5}$
$3 + \sqrt{5}$
You try
Rationalise: $\dfrac{1}{\sqrt{5} - 1}$
Conjugate of $\sqrt{5} - 1$ is $\sqrt{5} + 1$. Multiply above and below by it.
$\dfrac{1}{\sqrt{5}-1} \cdot \dfrac{\sqrt{5}+1}{\sqrt{5}+1} = \dfrac{\sqrt{5}+1}{(\sqrt{5})^{2} - 1^{2}}$
$= \dfrac{\sqrt{5}+1}{5 - 1}$
$= \dfrac{\sqrt{5}+1}{4}$
$\dfrac{\sqrt{5}+1}{4}$
Section 9 of 9
Surd Equations
Rule
Isolate the square root and square both sides.
(i) $\sqrt{x} = 9$
$\sqrt{x} = 9$
$x = 81$
(ii) $\sqrt{x} + 3 = 5$
Tempting move: square both sides immediately. Watch what happens:
$\left(\sqrt{x} + 3\right)^{2} = 5^{2}$
$x + 6\sqrt{x} + 9 = 25 \quad\times$
Now there's a $\sqrt{x}$ in the middle — squaring made it worse. Squaring a SUM brings a cross-term you didn't ask for.
Isolate the surd first, then square:
$\sqrt{x} + 3 = 5$
$\sqrt{x} = 2$
$x = 4$
(iii) $\sqrt{x} + 1 = 9$
$\sqrt{x} + 1 = 9$
$\sqrt{x} = 8$
$x = 64$
(iv) $x + \sqrt{x} = 6$
Bring x across, then square both sides.
$\sqrt{x} = 6 - x$
$\left(\sqrt{x}\right)^{2} = (6 - x)^{2}$
$x = 36 - 12x + x^{2}$
$x^{2} - 13x + 36 = 0$
$(x - 4)(x - 9) = 0$
$x = 4 \quad$ or $\quad x = 9$
Check both answers in the original equation:
$x = 4: \;\; 4 + \sqrt{4} = 4 + 2 = 6 \;\checkmark$
$x = 9: \;\; 9 + \sqrt{9} = 9 + 3 = 12$ ≠ $6 \;\times$
$\therefore\;\; x = 4$
Important
When you square both sides you can create extra roots that do not satisfy the original equation. ALWAYS check your answers.
You try
Solve: $\sqrt{x+5} = x - 1$
Square both sides. Then check each answer back in the original — at least one will fail.
$\left(\sqrt{x+5}\right)^{2} = (x-1)^{2}$
$x + 5 = x^{2} - 2x + 1$
$x^{2} - 3x - 4 = 0$
$(x-4)(x+1) = 0$
$x = 4 \;$ or $\; x = -1$
Check $x = 4$: $\sqrt{9} = 3$, $\;4 - 1 = 3 \;\checkmark$
Check $x = -1$: $\sqrt{4} = 2$, $\;-1 - 1 = -2 \;\times$
$\therefore\;\; x = 4$
$x = 4$
That's Surds.
Next up on Mathslive.ie: Indices.