WAEC 2023: How to simplify surds questions in waec
Table of Contents
paragraph
Introduction to the simplification of surds questions in Waec
paragraph
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.
paragraph
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).
paragraph
The table of content is carefully arranged so as enable you can comprehend how to WAEC surd questions.
paragraph
check my tutorial on how to solve linear equations in one variable here
paragraph
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.
paragraph
What are surds(how to simplify surd)
paragraph
Surds are irrational numbers expressed in their root form. It simply means that when the root of a particular number gives a number with nonrecurring decimals, that number with its root(take note) is a surd.
paragraph
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.
paragraph
Let's check out examples of root expressions that are surd:
paragraph
 $\sqrt{4}$ : Is this a surd or not? The square root of 4 is not a surd. The reason for this is that the square root of 4 is an integer(2). The square root of 4 gives a number (2) that does not have a nonrecurring decimal(meaning the number is not irrational). When the square root of an expression is irrational, meaning having decimals that are nonrecurring, only then is the expression a surd.
paragraphTherefore, $\sqrt{4}$ is not a surd.paragraph
 $\sqrt{3}$: 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 nonrecurring decimal numbers in the result.
paragraph
 $\sqrt{ \frac{1}{25} }$ 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 to take note of.
paragraph
 $\sqrt{ \frac{1}{16} }$ is not a surd.
paragraph
 $\sqrt{ \frac{1}{25} }$ is not a surd.
paragraph
 $\sqrt{8}$ is a surd.
paragraph
 $\sqrt{6}$ is a surd.
paragraph
 $\sqrt{9}$ is not a surd.
paragraph
 $\sqrt{100}$ is not a surd.
paragraph
 $\sqrt{0.25}$ is not a surd.
paragraph
How to simplify surd questions in Waec 2023
paragraph
Simplification of surds requires you to know the rules guiding surd reduction, addition, subtraction, multiplication and surd division.
paragraph
paragraph
1. Surd Reduction(How to simplify surd)
paragraph
Surd expression can be reduced by expressing the surd expression as the products of a perfect square and a number.
paragraph
Mathematically: $\sqrt{a} = \sqrt{b×c}$ where b must be a prefect square and c can be any integer
paragraph
Example: simplify $\sqrt{48}$,
paragraph
first, we check out the highest perfect square that is a factor of 48.
paragraph
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.
paragraph
Therefore, $\sqrt{48} = \sqrt{16×3}$,
paragraph
This can be simplified further to;
paragraph
$\sqrt{48} = \sqrt{16} × \sqrt{3}$
paragraph
Since the square root of 16 is 4, we can therefore write that:
paragraph
$\sqrt{48} = 4 × \sqrt{3}$
paragraph
Final answer becomes
paragraph
$\sqrt{48} = 4 \sqrt{3}$paragraph
2. Surd Addition(how to simplify surd)
paragraph
Only surds having the same root can be added together. Surds can be added by simply adding up the coefficient of the root.
paragraph
Mathematically,
paragraph
$c \sqrt{a} + b \sqrt{a} = (c + b) \sqrt{a}$, where c and b are the coefficient of the root. Please note that before you add up surds, the root must be the same.
paragraph
Simplify: $5 \sqrt{5} + 6 \sqrt{5}$
paragraph
This becomes; $5 \sqrt{5} + 6 \sqrt{5} = (5 + 6) \sqrt{5}$
paragraph
Finally, $5 \sqrt{5} + 6 \sqrt{5} = 11 \sqrt{5}$paragraph
3. Surd Subtraction(how to simplify surd)
paragraph
Only surds having the same roots can be subtracted from each expression. Surds can be subtracted by subtracting the coefficient of their roots.
paragraph
Mathematically,
paragraph
$c\sqrt{a}  b\sqrt{a} = (c  b) \sqrt{a}$, where c and b are the coefficient of the root. Please note that before you add up surds, the root must be the same.
paragraph
Simplify: $\sqrt{2}  7\sqrt{2}$
paragraph
This becomes; $\sqrt{2}  7\sqrt{5} = (1  7) \sqrt{2}$
paragraph
Finally, $\sqrt{5}  7\sqrt{5} = 6\sqrt{5}$paragraph
4. Surd Multiplication:(how to simplify surd)
paragraph
(a) Multiplying surds having different roots and without coefficient;
paragraph
Mathematically $\sqrt{a} × \sqrt{b} = \sqrt{a×b}$
paragraph
Simplify: $\sqrt{5} × \sqrt{7} = \sqrt{5×7}$
paragraph
Finally, $\sqrt{5} × \sqrt{7} = \sqrt{35}$
paragraph
(b) Multiplying surds having coefficients:
paragraph
Mathematically, $a\sqrt{b} × c\sqrt{d} = (a×c) \sqrt{b×d}$; in this case, coefficient multiply coefficient and surd multiply surd
paragraph
Simplify: $2\sqrt{5} × 6\sqrt{11}$
paragraph
This becomes; $2\sqrt{5} × 6\sqrt{11} = (2×6)\sqrt{5×11}$
paragraph
Finally, $2\sqrt{5} × 6\sqrt{11} = 12\sqrt{55}$
paragraph
(c) Multiplying surds that are the same and without coefficient;
paragraph
Mathematically, $\sqrt{a} × \sqrt{a} = \sqrt{a×a} = a$
paragraph
Simplify; $\sqrt{7} × \sqrt{7}$
paragraph
Simplifying the above gives 7,
paragraph
Explanation: $\sqrt{7} × \sqrt{7} = \sqrt{7×7} = \sqrt{49} = 7$
paragraph
(d) Multiplying surds that are the same and with coefficient;
paragraph
Mathematically, $b\sqrt{a} × c\sqrt{a} = (b×c) \sqrt{a×a} = b×c×a$
paragraph
Simplify; $3\sqrt{2} × 4\sqrt{2}$
paragraph
Simplifying the above gives 24,
paragraph
Explanation: $3\sqrt{2} × 4\sqrt{2} = (3×4)\sqrt{2×2} = 12×\sqrt{4} = 12×2 = 24$
paragraph
5. Surd Division:(How to simplify surd)
paragraph
(a). Simplifying an expression with just a single surd as its denominator:
paragraph
Mathematically; $\dfrac{a\sqrt{c}}{\sqrt{b}} = \dfrac{a\sqrt{c}}{\sqrt{b}} \times \dfrac{b}{\sqrt{b}} = \dfrac{ab\sqrt{c}}{b}$
paragraph
Therefore; $a\sqrt{c}$
paragraph
Simplify; $\dfrac{4\sqrt{5}}{\sqrt{3}}$
paragraph
Using the principle above;
paragraph
$\dfrac{4\sqrt{5}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$
paragraph
Simplifying the numerator and denominator gives;
paragraph
$\dfrac{4\sqrt{5 \times 3}}{3}$
paragraph
Final answer = $\dfrac{4\sqrt{15}}{3}$
paragraph
(b). Simplifying expression with a more complex surd at its denominator:
paragraph
Suppose we are to simplify as expression like this;
paragraph
$\dfrac{c}{a + \sqrt{b}}$
paragraph
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.
paragraph
First thing to do is to get the conjugate of the denominator, the denominator for this expression is $a + \sqrt{b}$
paragraph
The conjugate of a surd expression is simply the expression with the reverse sign in place.
paragraph
Meaning for this expression, $a + \sqrt{b}$, the conjugate will be $a  \sqrt{b}$. its only the sign that will change.
paragraph
Now, we will rationalize the expression by multiplying the numerator and denominator by the conjugate, doing this we will have;
paragraph
$\dfrac{c}{a + \sqrt{b}} \times \dfrac{a  \sqrt{b}}{a  \sqrt{b}}$
paragraph
Simplifying that gives;
paragraph
$\dfrac{c(a  \sqrt{b})}{(a + \sqrt{b})(a  \sqrt{b})}$
paragraph
finally,
paragraph
$\dfrac{ca  c\sqrt{b})}{a^2  b}$
paragraph
Solved waec past questions on surd(how to simplify surd)
paragraph

Simplify: $\sqrt{108} + \sqrt{125}  \sqrt{75}$ (waec 2018)paragraphSolutionparagraph$\sqrt{108} + \sqrt{125}  \sqrt{75}$paragraphThe first step is to perform a surd reduction on each term in the expression.paragraphThe 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,paragraphTherefore, we can express $\sqrt{108}$ as $\sqrt{36×3}$paragraphThe same can be done to $\sqrt{125}$, the highest perfect square that can divide 125 is 25.paragraphTherefore, $\sqrt{125}$ as $\sqrt{25×5}$paragraphDoing the same for the third term in the expression. The highest perfect square that can divide 75 is 25paragraphTherefore, $\sqrt{75}$ as $\sqrt{25×3}$paragraphWe can now rewrite the expression:$\sqrt{108} + \sqrt{125}  \sqrt{75}$ asparagraph$\sqrt{36×3} + \sqrt{25×5}  \sqrt{25×3}$paragraphSimplifying this further gives;paragraph$\sqrt{36}×\sqrt{3} + \sqrt{25}×\sqrt{5} + \sqrt{25}×\sqrt{3}$paragraphFurther simplification gives;paragraph$6×\sqrt{3} + 5×\sqrt{5}  5×\sqrt{3}$paragraphWhich can also be written as;paragraph$6\sqrt{3} + 5\sqrt{5}  5\sqrt{3}$paragraphCollecting like terms ;paragraph$6\sqrt{3}  5\sqrt{3}+ 5\sqrt{5}$paragraphSince the first two terms have the same root, surd subtraction can be performed on the two terms by simply subtract their coefficient.paragraphTherefore;paragraph$(6  5)\sqrt{3} + 5\sqrt{5}$paragraphFinally,paragraph$\sqrt{3} + 5\sqrt{5}$paragraph

Simplify: $3\sqrt{75}  \sqrt{12} + \sqrt{108}$ (waec 2014)paragraphSolutionparagraphThe expression:paragraph$3\sqrt{75}  \sqrt{12} + \sqrt{108}$paragraphCan be written as thisparagraph$3\sqrt{25×3}  \sqrt{4×3} + \sqrt{36×3}$paragraphCan be reduced further to:paragraph$3\sqrt{25}×\sqrt{3}  \sqrt{4}×\sqrt{3} + \sqrt{36}×\sqrt{3}$paragraphSimplifying further gives;paragraph$3×5×\sqrt{3}  2×\sqrt{3} + 6×\sqrt{3}$paragraphNow becomes;paragraph$15\sqrt{3}  2\sqrt{3} + 6\sqrt{3}$paragraphSince the root are the same, we can simply the coefficient of each roots to give;paragraph$(15  2 + 6)\sqrt{3}$paragraphSimplifying the terms in the bracket gives;paragraph$19\sqrt{3}$paragraphFinal answer: $19\sqrt{3}$paragraph

Simplify : $\frac{3\sqrt{5}×4\sqrt{6}} {2\sqrt{2}×3\sqrt{3}}$(waec 2012)
Solution
paragraph
$\frac{3\sqrt{5}×4\sqrt{6}} {2\sqrt{2}×3\sqrt{3}}$
paragraph
We are to perform surd multiplication on the numerator and denominator.
paragraph
Performing this gives;
paragraph
$\frac{(3×4)\sqrt{5×6}}{(2×3)\sqrt{2×3}}$
paragraph
Simplifying the numerator and denominator gives;
paragraph
$\frac{12\sqrt{30}} {6\sqrt{6}}$
paragraph
To simplify further, we rationalize the denominator by multiplying the numerator and the denominator by the denominator.
paragraph
Therefore, we have;
paragraph
$\frac{12\sqrt{30}} {6\sqrt{6}} × \frac{6\sqrt{6}} {6\sqrt{6}}$
paragraph
This can still be written as;
paragraph
$\frac{12\sqrt{30}×6\sqrt{6}} {6\sqrt{6}×6\sqrt{6}}$
paragraph
We can simplify the numerator and denominator by performing surd multiplication :
paragraph
$\frac{(12×6)\sqrt{30×6}} {(6×6)\sqrt{6×6}}$
paragraph
Simplifying further;
paragraph
$\frac{72\sqrt{180}} {36\sqrt{36}}$
paragraph
Performing a surd reduction on root 180 and reducing the denominator gives;
paragraph
$\frac{72\sqrt{36×5}} {36×6}$
paragraph
Simplify the numerator;
paragraph
$\frac{72\sqrt{36}×\sqrt{5}} {36×6}$
paragraph
Further simplification gives;
paragraph
$\frac{72×6×\sqrt{5}} {36×6}$
paragraph
Since 6 is present at the numerator and Denominator they cancel each other out to give;
paragraph
$\frac{72×\sqrt{5}} {36}$
paragraph
Since 36 can divide 72 to give 2, the final answer becomes;
paragraph
$2\sqrt{5}$
paragraph
4. Simplify $2\sqrt{3}  \dfrac{6}{\sqrt{3}} + \dfrac{3}{\sqrt{27}}$ (waec 2010)
paragraph
Solution
paragraph
Surd division should be performed on the second and third term;
paragraph
Doing this gives;
paragraph
$2\sqrt{3}  \dfrac{6}{\sqrt{3}}×\dfrac{\sqrt{3}} {\sqrt{3}} + \dfrac{3}{\sqrt{27}}×\dfrac{\sqrt{27}}{\sqrt{27}}$
paragraph
This becomes;
paragraph
Doing this gives;
paragraph
$2\sqrt{3}  \dfrac{6×\sqrt{3}}{\sqrt{3}×\sqrt{3}} + \dfrac{3×\sqrt{27}}{\sqrt{27}×\sqrt{27}}$
paragraph
Simplifying the numerator and denominator gives;
paragraph
$2\sqrt{3}  \dfrac{6×\sqrt{3}}{3} + \dfrac{3×\sqrt{9×3}}{27}$
paragraph
For the second term 3 can divide 6 to give 2, and the third term can be simplified further to give;
paragraph
$2\sqrt{3}  2\sqrt{3} + \dfrac{3×\sqrt{9}×\sqrt{3}}{27}$
paragraph
The first and second can be simplified to give 0;
paragraph
As (2  2) is zero, therefore, we are left with the third term;
paragraph
$\dfrac{3×3×\sqrt{3}}{27}$
paragraph
Simplifying further gives;
paragraph
$\dfrac{9\sqrt{3}}{27}$
paragraph
Since 9 will go in 27 three times, we have the final answer as ;
paragraph
$\dfrac{\sqrt{3}}{3}$
paragraph
5. Given that $(\sqrt{3}  5\sqrt{2}) (\sqrt{3}  \sqrt{2}) = a + b\sqrt{6}$ (waec 2009)
paragraph
Solution
paragraph
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.
paragraph
$(\sqrt{3}  5\sqrt{2}) (\sqrt{3} + \sqrt{2}) = a + b\sqrt{6}$
paragraph
Let's focus on reducing the left hand side of the expression;
paragraph
$(\sqrt{3}  5\sqrt{2}) (\sqrt{3} + \sqrt{2})$
paragraph
To do bracket of parentheses expansion, we use each terms in the first bracket to multiply the second bracket as shown below;
paragraph
$\sqrt{3}(\sqrt{3} + \sqrt{2})  5\sqrt{2}(\sqrt{3} + \sqrt{2})$
paragraph
Clearing the brackets gives:
paragraph
$\sqrt{3}×\sqrt{3} + \sqrt{3}×\sqrt{2}  5\sqrt{2}×\sqrt{3}  5\sqrt{2}×\sqrt{2}$
paragraph
Performing a surd multiplication on each term gives;
paragraph
$3 + \sqrt{6}  5\sqrt{6}  5×2$
paragraph
$3 + \sqrt{6}  5\sqrt{6}  10$
paragraph
Collecting like terms
paragraph
$3  10 + \sqrt{6}  5\sqrt{6}$
paragraph
Simplifying the above expression gives;
paragraph
$ 7  4\sqrt{6}$
paragraph
Comparing the above with $a + b\sqrt{6}$
paragraph
Shows that;a = 7
b = 4
paragraph
6. Simplify : $\sqrt{2}(\sqrt{6} + 2\sqrt{2})  2\sqrt{3}$ (waec 2017)
paragraph
Solution:
paragraph
The first thing to do is clear the brackets.
paragraph
Clearing the bracket gives;
paragraph
$\sqrt{2}×\sqrt{6} + \sqrt{2}×2\sqrt{2}  2\sqrt{3}$
paragraph
To simplify the surd expression further, we perform surd multiplication on the first and second term
paragraph
$\sqrt{12} + 2\sqrt{4}  2\sqrt{3}$
paragraph
Performing a surd reduction on the first term while simplifying the second term gives;
paragraph
$\sqrt{4×3} + 2×2  2\sqrt{3}$
paragraph
Reducing the above further gives;
paragraph
$\sqrt{4}×\sqrt{3} + 4  2\sqrt{3}$
paragraph
Therefore;
paragraph
$2\sqrt{3} + 4  2\sqrt{3}$
paragraph
Collecting like terms
paragraph
$2\sqrt{3}  2\sqrt{3} + 4$
paragraph
The first and the second term can be reduced by performing a surd subtraction on them.
paragraph
Subtracting their coefficient (2  2) gives 0
paragraph
Therefore, the final answer is :
paragraph
$4$
paragraph
7. Simplify : $(\dfrac{10\sqrt{3}} {\sqrt{5}}  \sqrt{15}) ^2$ (waec 2016)
paragraph
Solution:
paragraph
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.
paragraph
Doing that gives;
paragraph
$(\dfrac{10\sqrt{3}} {\sqrt{5}}×\dfrac{\sqrt{5}} {\sqrt{5}}  \sqrt{15}) ^2$
paragraph
Simplifying further gives;
paragraph
$(\dfrac{10\sqrt{3}×\sqrt{5}} {\sqrt{5}×\sqrt{5}}  \sqrt{15}) ^2$
paragraph
Performing a surd multiplication on the numerator and the denominator of the first term gives;
paragraph
$(\dfrac{10\sqrt{15}} {5}  \sqrt{15}) ^2$
paragraph
Reducing the above further gives;
paragraph
$(2\sqrt{15}  \sqrt{15}) ^2$
paragraph
Therefore, since the surd have te same root, surd subtraction can be performed
paragraph
Subtracting the coefficient of the two terms gives (2  1) = 1
paragraph
Therefore;
paragraph
$(\sqrt{15}) ^2$
paragraph
Squaring this expression ;
paragraph
$\sqrt{15} × \sqrt{15}$
paragraph
This gives a final answer of$15$
paragraph
8. If : $(\dfrac{\sqrt{2} + \sqrt{3}} {\sqrt{3}})$ is simplified as $m + n\sqrt{6}$ find (m + n) (waec 2015)
paragraph
Solution:
paragraph
The first thing to do is to rationalize the expression using the Denominator.
paragraph
Doing that gives;
paragraph
$(\dfrac{\sqrt{2} + \sqrt{3}} {\sqrt{3}} × \dfrac{\sqrt{3}} {\sqrt{3}} )$
paragraph
Simplifying further gives;
paragraph
$(\dfrac{(\sqrt{2} + \sqrt{3}) × \sqrt{3}} {\sqrt{3}×\sqrt{3}})$
paragraph
Performing a surd multiplication on the numerator by clearing the bracket and simplifying the denominator gives;
paragraph
$\dfrac{\sqrt{2}×\sqrt{3} + \sqrt{3}×\sqrt{3}} {3}$
paragraph
Reducing the above further gives;
paragraph
$\dfrac{\sqrt{6} + 3} {3}$
paragraph
Reducing the expression further;
paragraph
$\dfrac{\sqrt{6}} {3} + \dfrac{3} {3}$
paragraph
Rearranging this surd to be in the form;
paragraph
$m + n\sqrt{6}$
paragraph
Will give;
paragraph
$1 + \dfrac{1}{3}\sqrt{6}$
paragraph
Therefore;
paragraph
Comparing the two expression shows that
paragraph
m = 1 and n = $\dfrac{1}{3}$
paragraph
Calculating (m + n)
paragraph
$( 1 + \dfrac{1}{3} )$
paragraph
The answer $(m + n) = 1\dfrac{1}{3}$
paragraph
Final answer $1\dfrac{1}{3}$
paragraph
I recommend you check my article on the following:
 Check my tutorial on indices
 How to pass chemistry in the wassce 2020 exams with an A1
 How to blast mathematics in the wassce 2020 exams with an A
 How to get an A in the wassce biology 2020 exams
 How to blast physics in the wassce 2023 exams with an A.
 How to pass the wassce further mathematics exam excellently well in 2020
 How to manage time efficiently as a student.
 Seven mistakes to avoid as a year one students in the university.
paragraphThis is all we can take on “How to pass waec biology exams with distinctions“.paragraph
POSCHOLARS TEAM.