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# WAEC 2020: How to simplify surds questions in waec

##### Table of Content(How to simplify surd)
• Introduction to the simplification of surds questions in waec
• What are surds(how to simplify surd)
• How to simplify surd questions in waec 2020
• Solved waec past questions on surd
• A short quiz on surd(Try them out)

### INTRODUCTION(Waec 2020, surd)

In this tutorial, How to simplify waec surds questions will be explicitly explained here. Simplification of surds has always been the simplest topic in mathematics(my opinion tho). Surds questions in waec,  jamb(utme), putme are all the same and can be tackled by understanding some basic principles which I will be revealing as we progress.

Also, most of the questions i will be using as examples will all be waec past questions. This is to show that waec surd questions are very easy, you can even tackle them with your eyes closed(Lol) .

The table of content is carefully arranged so as enable you can comprehend how to WAEC surd questions.

check my tutorial on how to solve linear equations in one variable here

I also have a tutorial on solve simultaneous equations with the substitution method here’s

With that being said let’s dive right in but first get your writing materials and a glass of juice if you like.  MATH IS FUN.

# What are surds?(How to simplify surd)

Surds are irrational numbers expressed in their root form. It simply means that when the root of a particular number gives a number with non-recurring decimals, that number with its root(take note)  is a surd.

Also, when the root of a number can be simplified, it is not a surd. But when simplification seems impossible unless been done with a calculator, then it is a surd.

Let’s check out example of root expressions that are surd:

1. $\sqrt{4}$ : Is this a surd or not? The square root of 4 is not a surd. The reason for this is because the square root of 4 is an integer(2). The square root of 4 gives a number (2) that does not have a non-recurring decimal(meaning the number is not irrational). When the square root of an expression is irrational, meaning having decimals that are non-recurring,  only then is the expression a surd.

Therefore,  $\sqrt{4}$  is not a surd.

2. $\sqrt{3}$: Is this a surd or not? The square root of 3 is a surd. The reason for this is that the square root of 3 is 1.7320508…. This is an irrational number because of the presence of non-recurring decimal numbers in the result.

3.  $\sqrt{ \frac{1}{25} }$ Is this a surd or not? The square root of $\dfrac{1}{25}$  is not a surd. The reason for this is because the square root of $\sqrt{ \frac{1}{25} }$  is $\dfrac{1}{5}$  further simplification gives 0.2. This decimal is rational take note.

4. $\sqrt{ \frac{1}{16} }$ is not a surd.

5. $\sqrt{ \frac{1}{25} }$ is not a surd.

6. $\sqrt{8}$ is a surd.

7. L$\sqrt{6}$ is a surd.

8. $\sqrt{9}$ is not a surd.

9. $\sqrt{100}$ is not a surd.

10. $\sqrt{0.25}$ is not a surd.

Let me end this section with this. An expression is only a surd when;

i. You need a calculator to evaluate the expression, for example root 3, or the root of any numbers that aren’t perfect squares.

ii. The result from the evaluation gives an irrational number, that is, a number with an infinite non-recurring decimals.

# How to simplify surd questions in WAEC 2020

Simplification of surds requires you to know the rules guiding surd reduction, addition, subtraction, multiplication and surd division.

##### 1. Surd Reduction(How to simplify surd)

Surd expression can be reduced by expressing the surd expression as the products of a perfect square and a number.

Mathematically: $\sqrt{a} = \sqrt{b×c}$ where b must be a prefect square and c can be any integer

Example: simplify $\sqrt{48}$,

first, we check out the highest perfect square that is a factor of 48.

Perfect squares that are factors of 48 are 4 and 16. Since 16 is the greatest of the two, we express 48 as a product of 16 and 3.

Therefore, $\sqrt{48} = \sqrt{16×3}$,

This can be simplified further to;

$\sqrt{48} = \sqrt{16} × \sqrt{3}$

Since the square root of 16 is 4, we can therefore write that:

$\sqrt{48} = 4 × \sqrt{3}$

$\sqrt{48} = 4 \sqrt{3}$

##### 2. Surd Addition(how to simplify surd)

Only surds having the same root can be added together. Surds can be added by simply adding up the coefficient of the root.

Mathematically,

$c \sqrt{a} + b \sqrt{a} = (c + b) \sqrt{a}$, where c and b are the coefficient of the root. Please note that before you add up surds, the root must be the same.

Simplify: $5 \sqrt{5} + 6 \sqrt{5}$

This becomes; $5 \sqrt{5} + 6 \sqrt{5} = (5 + 6) \sqrt{5}$

Finally, $5 \sqrt{5} + 6 \sqrt{5} = 11 \sqrt{5}$

##### 3. Surd Subtraction(how to simplify surd)

Only surds having the same roots can be subtracted from each expression. Surds can be subtracted by subtracting the coefficient of their roots.

Mathematically,

$c\sqrt{a} – b\sqrt{a} = (c – b) \sqrt{a}$, where c and b are the coefficient of the root. Please note that before you add up surds, the root must be the same.

Simplify: $\sqrt{2} – 7\sqrt{2}$

This becomes; $\sqrt{2} – 7\sqrt{5} = (1 – 7) \sqrt{2}$

Finally, $\sqrt{5} – 7\sqrt{5} = -6\sqrt{5}$

##### 4. Surd Multiplication:(how to simplify surd)

(a) Multiplying surds having different roots and without coefficient;

Mathematically $\sqrt{a} × \sqrt{b} = \sqrt{a×b}$

Simplify: $\sqrt{5} × \sqrt{7} = \sqrt{5×7}$

Finally, $\sqrt{5} × \sqrt{7} = \sqrt{35}$

(b) Multiplying surds having coefficients:

Mathematically, $a\sqrt{b} × c\sqrt{d} = (a×c) \sqrt{b×d}$; in this case, coefficient multiply coefficient and surd multiply surd

Simplify: $2\sqrt{5} × 6\sqrt{11}$

This becomes; $2\sqrt{5} × 6\sqrt{11} = (2×6)\sqrt{5×11}$

Finally, $2\sqrt{5} × 6\sqrt{11} = 12\sqrt{55}$

(c) Multiplying surds that are the same and without coefficient;

Mathematically, $\sqrt{a} × \sqrt{a} = \sqrt{a×a} = a$

Simplify; $\sqrt{7} × \sqrt{7}$

Simplifying the above gives 7,

Explanation: $\sqrt{7} × \sqrt{7} = \sqrt{7×7} = \sqrt{49} = 7$

(d) Multiplying surds that are the same and with coefficient;

Mathematically, $b\sqrt{a} × c\sqrt{a} = (b×c) \sqrt{a×a} = b×c×a$

Simplify; $3\sqrt{2} × 4\sqrt{2}$

Simplifying the above gives 24,

Explanation: $3\sqrt{2} × 4\sqrt{2} = (3×4)\sqrt{2×2} = 12×\sqrt{4} = 12×2 = 24$

##### 5. Surd Division:(How to simplify surd)

(a). Simplifying an expression with just a single surd as its denominator:

Mathematically; $\dfrac{a\sqrt{c}}{\sqrt{b}}$ = $\dfrac{a\sqrt{c}}{\sqrt{b}} \times \dfrac{b}{\sqrt{b}} = \dfrac{ab\sqrt{c}}{b}$

Therefore; $a\sqrt{c}$

Simplify; $\dfrac{4\sqrt{5}}{\sqrt{3}}$

Using the principle above;

$\dfrac{4\sqrt{5}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

Simplifying the numerator and denominator gives;

$\dfrac{4\sqrt{5 \times 3}}{3}$

Final answer = $\dfrac{4\sqrt{15}}{3}$

(b). Simplifying expression with a more complex surd at its denominator:

Suppose we are to simplify as expression like this;

$\dfrac{c}{a + \sqrt{b}}$

since the denominator of this expression is not a single surd, we will rationalize the expression using the conjugate of the denominator, Let me break it down.

First thing to do is to get the conjugate of the denominator, the denominator for this expression is $a + \sqrt{b}$

The conjugate of a surd expression is simply the expression with the reverse sign in place.

Meaning for this expression, $a + \sqrt{b}$, the conjugate will be $a – \sqrt{b}$. its only the sign that will change.

Now, we will rationalize the expression by multiplying the numerator and denominator by the conjugate, doing this we will have;

$\dfrac{c}{a + \sqrt{b}} \times \dfrac{a – \sqrt{b}}{a – \sqrt{b}}$

Simplifying that gives;

$\dfrac{c(a – \sqrt{b})}{(a + \sqrt{b})(a – \sqrt{b})}$

finally,

$\dfrac{ca – c\sqrt{b})}{a^2 – b}$

## Solved waec past questions on surd(how to simplify surd)

###### 1. Simplify: $\sqrt{108} + \sqrt{125} – \sqrt{75}$ (waec 2018)

Solution

$\sqrt{108} + \sqrt{125} – \sqrt{75}$

The first step is to perform a surd reduction on each term in the expression.

The first term in the expression is $\sqrt{108}$, we need to get the highest perfect square that can divide 108 with a remainder, the perfect square for this is 36,

Therefore, we can express $\sqrt{108}$ as $\sqrt{36×3}$

The same can be done to $\sqrt{125}$, the highest perfect square that can divide 125 is 25.

Therefore, $\sqrt{125}$ as $\sqrt{25×5}$

Doing the same for the third term in the expression. The highest perfect square that can divide 75 is 25

Therefore, $\sqrt{75}$ as $\sqrt{25×3}$

We can now rewrite the expression:$\sqrt{108} + \sqrt{125} – \sqrt{75}$ as

$\sqrt{36×3} + \sqrt{25×5} – \sqrt{25×3}$

Simplifying this further gives;

$\sqrt{36}×\sqrt{3} + \sqrt{25}×\sqrt{5} + \sqrt{25}×\sqrt{3}$

Further simplification gives;

$6×\sqrt{3} + 5×\sqrt{5} – 5×\sqrt{3}$

Which can also be written as;

$6\sqrt{3} + 5\sqrt{5} – 5\sqrt{3}$

Collecting like terms ;

$6\sqrt{3} – 5\sqrt{3}+ 5\sqrt{5}$

Since the first two terms have the same root,  surd subtraction can be performed on the two terms by simply subtract their coefficient.

Therefore;

$(6 – 5)\sqrt{3} + 5\sqrt{5}$

Finally,

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

###### 2. Simplify: $3\sqrt{75} – \sqrt{12} + \sqrt{108}$ (waec 2014)

Solution

The expression:

$3\sqrt{75} – \sqrt{12} + \sqrt{108}$

Can be written as this

$3\sqrt{25×3} – \sqrt{4×3} + \sqrt{36×3}$

Can be reduced further to:

$3\sqrt{25}×\sqrt{3} – \sqrt{4}×\sqrt{3} + \sqrt{36}×\sqrt{3}$

Simplifying further gives;

$3×5×\sqrt{3} – 2×\sqrt{3} + 6×\sqrt{3}$

Now becomes;

$15\sqrt{3} – 2\sqrt{3} + 6\sqrt{3}$

Since the root are the same, we can simply the coefficient of each roots to give;

$(15 – 2 + 6)\sqrt{3}$

Simplifying the terms in the bracket gives;

$19\sqrt{3}$

Final answer: $19\sqrt{3}$

###### 3. Simplify : $\frac{3\sqrt{5}×4\sqrt{6}} {2\sqrt{2}×3\sqrt{3}}$(waec 2012)

Solution

$\frac{3\sqrt{5}×4\sqrt{6}} {2\sqrt{2}×3\sqrt{3}}$

We are to perform surd multiplication on the numerator and denominator.

Performing this gives;

$\frac{(3×4)\sqrt{5×6}}{(2×3)\sqrt{2×3}}$

Simplifying the numerator and denominator gives;

$\frac{12\sqrt{30}} {6\sqrt{6}}$

To simplify further, we rationalize the denominator by multiplying the numerator and the denominator by the denominator.

Therefore,  we have;

$\frac{12\sqrt{30}} {6\sqrt{6}} × \frac{6\sqrt{6}} {6\sqrt{6}}$

This can still be written as;

$\frac{12\sqrt{30}×6\sqrt{6}} {6\sqrt{6}×6\sqrt{6}}$

We can simplify the numerator and denominator by performing surd multiplication :

$\frac{(12×6)\sqrt{30×6}} {(6×6)\sqrt{6×6}}$

Simplifying further;

$\frac{72\sqrt{180}} {36\sqrt{36}}$

Performing a surd reduction on root 180 and reducing the denominator gives;

$\frac{72\sqrt{36×5}} {36×6}$

Simplify the numerator;

$\frac{72\sqrt{36}×\sqrt{5}} {36×6}$

Further simplification gives;

$\frac{72×6×\sqrt{5}} {36×6}$

Since 6 is present at the numerator and Denominator they cancel each other out to give;

$\frac{72×\sqrt{5}} {36}$

Since 36 can divide 72 to give 2, the final answer becomes;

$2\sqrt{5}$

###### 4. Simplify $2\sqrt{3} – \dfrac{6}{\sqrt{3}} + \dfrac{3}{\sqrt{27}}$ (waec 2010)

Solution

Surd division should be performed on the second and third term;

Doing this gives;

$2\sqrt{3} – \dfrac{6}{\sqrt{3}}×\dfrac{\sqrt{3}} {\sqrt{3}} + \dfrac{3}{\sqrt{27}}×\dfrac{\sqrt{27}}{\sqrt{27}}$

This becomes;

Doing this gives;

$2\sqrt{3} – \dfrac{6×\sqrt{3}}{\sqrt{3}×\sqrt{3}} + \dfrac{3×\sqrt{27}}{\sqrt{27}×\sqrt{27}}$

Simplifying the numerator and denominator gives;

$2\sqrt{3} – \dfrac{6×\sqrt{3}}{3} + \dfrac{3×\sqrt{9×3}}{27}$

For the second term 3 can divide 6 to give 2, and the third term can be simplified further to give;

$2\sqrt{3} – 2\sqrt{3} + \dfrac{3×\sqrt{9}×\sqrt{3}}{27}$

The first and second can be simplified to give 0;

As (2 – 2) is zero,  therefore,  we are left with the third term;

$\dfrac{3×3×\sqrt{3}}{27}$

Simplifying further gives;

$\dfrac{9\sqrt{3}}{27}$

Since 9 will go in 27 three times,  we have the final answer as ;

$\dfrac{\sqrt{3}}{3}$

###### 5. Given that $(\sqrt{3} – 5\sqrt{2}) (\sqrt{3} – \sqrt{2} = a + b\sqrt{6}$ (waec 2009)

Solution

This question is quite different from the ones we have been solving but all the same, we are to use the basic principles of surd.

$(\sqrt{3} – 5\sqrt{2}) (\sqrt{3} + \sqrt{2} = a + b\sqrt{6}$

Let’s focus on reducing the left hand side of the expression;

$(\sqrt{3} – 5\sqrt{2}) (\sqrt{3} + \sqrt{2})$

To do bracket of parentheses expansion, we use each terms in the first bracket to multiply the second bracket as shown below;

$\sqrt{3}(\sqrt{3} + \sqrt{2}) – 5\sqrt{2}(\sqrt{3} + \sqrt{2})$

Clearing the brackets gives:

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

Performing a surd multiplication on each term gives;

$3 + \sqrt{6} – 5\sqrt{6} – 5×2$

$3 + \sqrt{6} – 5\sqrt{6} – 10$

Collecting like terms

$3 – 10 + \sqrt{6} – 5\sqrt{6}$

Simplifying the above expression gives;

$– 7 – 4\sqrt{6}$

Comparing the above with $a + b\sqrt{6}$

Shows that;

a = -7

###### 6. Simplify : $\sqrt{2}(\sqrt{6} + 2\sqrt{2}) – 2\sqrt{3}$ (waec 2017)

Solution:

The first thing to do is clear the brackets.

Clearing the bracket gives;

$\sqrt{2}×\sqrt{6} + \sqrt{2}×2\sqrt{2} – 2\sqrt{3}$

To simplify the surd expression further,  we perform surd multiplication on the first and second term

$\sqrt{12} + 2\sqrt{4} – 2\sqrt{3}$

Performing a surd reduction on the first term while simplifying the second term gives;

$\sqrt{4×3} + 2×2 – 2\sqrt{3}$

Reducing the above further gives;

$\sqrt{4}×\sqrt{3} + 4 – 2\sqrt{3}$

Therefore;

$2\sqrt{3} + 4 – 2\sqrt{3}$

Collecting like terms

$2\sqrt{3} – 2\sqrt{3} + 4$

The first and the second term can be reduced by performing a surd subtraction on them.

Subtracting their coefficient (2  –  2) gives 0

Therefore,  the final answer is :

$4$

7. Simplify : $(\dfrac{10\sqrt{3}} {\sqrt{5}} – \sqrt{15}) ^2$ (waec 2016)

Solution:

The first thing to do is to reduce the expression inside the bracket by performing a surd division(rationalizing with the denominator)  on the first term.

Doing that gives;

$(\dfrac{10\sqrt{3}} {\sqrt{5}}×\dfrac{\sqrt{5}} {\sqrt{5}} – \sqrt{15}) ^2$

Simplifying further gives;

$(\dfrac{10\sqrt{3}×\sqrt{5}} {\sqrt{5}×\sqrt{5}} – \sqrt{15}) ^2$

Performing a surd multiplication on the numerator and the denominator of the  first term gives;

$(\dfrac{10\sqrt{15}} {5} – \sqrt{15}) ^2$

Reducing the above further gives;

$(2\sqrt{15} – \sqrt{15}) ^2$

Therefore,  since the surd have te same root,  surd subtraction can be performed

Subtracting the coefficient of the two terms gives (2 – 1) = 1

Therefore;

$(\sqrt{15}) ^2$

Squaring this expression ;

$\sqrt{15} × \sqrt{15}$

This gives a final answer of

$15$

###### 8. If : $(\dfrac{\sqrt{2} + \sqrt{3}} {\sqrt{3}}$ is simplified as $m + n\sqrt{6}$ find (m + n) (waec 2015)

Solution:

The first thing to do is to rationalize the expression using the Denominator.

Doing that gives;

$(\dfrac{\sqrt{2} + \sqrt{3}} {\sqrt{3}} × \dfrac{\sqrt{3}} {\sqrt{3}}$

Simplifying further gives;

$(\dfrac{(\sqrt{2} + \sqrt{3}) × \sqrt{3}} {\sqrt{3}×\sqrt{3}}$

Performing a surd multiplication on the numerator by clearing the bracket and simplifying the denominator gives;

$\dfrac{\sqrt{2}×\sqrt{3} + \sqrt{3}×\sqrt{3}} {3}$

Reducing the above further gives;

$\dfrac{\sqrt{6} + 3} {3}$

Reducing the expression further;

$\dfrac{\sqrt{6}} {3} + \dfrac{3} {3}$

Rearranging this surd to be in the form;

$m + n\sqrt{6}$

Will give;

$1 + \dfrac{1}{3}\sqrt{6}$

Therefore;

Comparing the two expression shows that

m = 1 and n = $\dfrac{1}{3}$

Calculating (m + n)

( 1 + $\dfrac{1}{3}$

The answer (m + n)   = $1\dfrac{1}{3}$

Final answer $1\dfrac{1}{3}$

click here to take a short quiz on waec mathematics past questions based on surd ###### Author
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My name is John Philip Obhenimen, i am very passionate about educating young people this is because the future is in our hands and our performance is closely tied to the information we consume as young people

### 1 throught on "WAEC 2020: A revision on surd, How to simplify surd"

1. Adewunmi says:

Am also a student.SS3 Student:HOW TO GET A1 IN THE WASSCE EXAM PHYSICS&BIOLOGY&CHEMISTRY&MATHS 2021

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