Last updated on October 18th, 2024 at 12:29 pm

Are you a student preparing for your SSCE (Senior Secondary Certificate Examination) or JSCE (Junior Secondary Certificate Examination)? If so, we’ve got great news! You can now access the **New General Mathematics textbooks** for both SSCE and JSCE for FREE in PDF format.

Table of Contents

Toggle### Why You Need These Mathematics Textbooks

Mathematics is a core subject in both the SSCE and JSCE, and having the right textbook is crucial for understanding the key concepts and solving problems efficiently. The **New General Mathematics** textbooks are well-known for their clear explanations, practical examples, and exercises that help students develop critical thinking skills necessary for success in these exams.

Whether you’re a teacher looking for resources for your class or a student aiming to improve your grades, these textbooks provide all the content you need to excel in your exams.

### What’s Inside the Textbooks?

The **New General Mathematics** textbooks for SSCE and JSCE cover all the essential topics that align with the national curriculum. Here’s a breakdown of what you can expect:

**SSCE New General Mathematics**:- Algebra, geometry, trigonometry
- Statistics and probability
- Calculus fundamentals
- Practice exercises with real exam-style questions

## Content Example 1

{Problem 1:} Solve the following equation for \( z \), where \( z \) is a complex number:

\[

z^2 + (3 – 4i)z + (13 + 8i) = 0

\]

{Solution:}

This is a quadratic equation in \( z \). The general solution to a quadratic equation \( az^2 + bz + c = 0 \) is given by the quadratic formula:

\[

z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}

\]

Here, \( a = 1 \), \( b = 3 – 4i \), and \( c = 13 + 8i \).

1. {Step 1: Compute \( b^2 \):}

\[

b^2 = (3 – 4i)^2 = 9 – 24i + 16i^2 = 9 – 24i – 16 = -7 – 24i

\]

2. {Step 2: Compute \( 4ac \):}

\[

4ac = 4(1)(13 + 8i) = 52 + 32i

\]

3. {Step 3: Compute the discriminant \( b^2 – 4ac \):}

\[

b^2 – 4ac = (-7 – 24i) – (52 + 32i) = -7 – 24i – 52 – 32i = -59 – 56i

\]

4.{Step 4: Compute the square root of the discriminant \( \sqrt{-59 – 56i} \):}

To find the square root, express the complex number in polar form and then take the square root. The result is approximately:

\[

\sqrt{-59 – 56i} \approx 2 – 28i

\]

5. {Step 5: Compute the two solutions for \( z \):}

\[

z = \frac{-(3 – 4i) \pm (2 – 28i)}{2}

\]

This gives two solutions:

\[

z_1 = \frac{-3 + 4i + 2 – 28i}{2} = \frac{-1 – 24i}{2} = -\frac{1}{2} – 12i

\]

\[

z_2 = \frac{-3 + 4i – 2 + 28i}{2} = \frac{-5 + 32i}{2} = -\frac{5}{2} + 16i

\]

Thus, the solutions are:

\[

z_1 = -\frac{1}{2} – 12i, \quad z_2 = -\frac{5}{2} + 16i

\]

## Content Example 2

{Problem 2:} Simplify the following expression and express the result in standard form \( a + bi \):

\[

\frac{3 + 4i}{1 – 2i}

\]

{Solution:}

1. {Step 1: Multiply by the conjugate of the denominator:}

\[

\frac{3 + 4i}{1 – 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(3 + 4i)(1 + 2i)}{(1 – 2i)(1 + 2i)}

\]

2. {Step 2: Simplify the denominator:}

\[

(1 – 2i)(1 + 2i) = 1^2 – (2i)^2 = 1 – (-4) = 1 + 4 = 5

\]

3. {Step 3: Expand the numerator:}

\[

(3 + 4i)(1 + 2i) = 3(1) + 3(2i) + 4i(1) + 4i(2i)

\]

\[

= 3 + 6i + 4i + 8i^2 = 3 + 10i + 8(-1) = 3 + 10i – 8 = -5 + 10i

\]

4. {Step 4: Divide by the denominator:}

\[

\frac{-5 + 10i}{5} = \frac{-5}{5} + \frac{10i}{5} = -1 + 2i

\]

Thus, the simplified expression is:

\[

-1 + 2i

\]

**JSCE New General Mathematics**:- Introduction to algebra and geometry
- Fractions, decimals, and percentages
- Basic statistics and probability
- Fun and interactive exercises to build foundational skills

Both textbooks are designed to help students grasp the necessary concepts and apply them in exam situations. They include worked-out examples, test-yourself questions, and revision sections that ensure a comprehensive learning experience.

## Sample content

{Fraction Question:}

If a recipe requires \(\frac{3}{4}\) cup of sugar, but you only have a \(\frac{1}{2}\) cup measure, how many \(\frac{1}{2}\) cups of sugar would you need to use to make the recipe?

{Answer:}

To solve this, divide the amount of sugar needed \(\left( \frac{3}{4} \right)\) by the size of the measuring cup \(\left( \frac{1}{2} \right)\):

\[

\text{Number of } \frac{1}{2} \text{ cups} = \frac{\frac{3}{4}}{\frac{1}{2}}

\]

Calculating further:

\[

= \frac{3}{4} \times \frac{2}{1} = \frac{6}{4} = 1 \frac{1}{2}

\]

So, you would need \(1 \frac{1}{2}\) cups of sugar to get \(\frac{3}{4}\) cup.

2. {Decimal Question:}

Convert \(\frac{7}{8}\) into a decimal.

{Answer:}

To convert \(\frac{7}{8}\) to a decimal, divide 7 by 8:

\[

\frac{7}{8} = 7 \div 8 = 0.875

\]

So, \(\frac{7}{8}\) as a decimal is \(0.875\).

### Benefits of Downloading the New General Mathematics Textbooks

**Updated Content**: These textbooks are up-to-date with the latest syllabus, ensuring that you study the most relevant topics.**PDF Format**: You can easily download and access the books on any device, be it a phone, tablet, or computer.**Free of Charge**: We provide the textbooks completely free, saving you money while giving you access to quality resources.

### How to Download the SSCE and JSCE Mathematics Textbooks

To download the **New General Mathematics** textbooks for both SSCE and JSCE, click on the links below:

*Download SSCE New General Mathematics Textbook PDF**Download JSCE New General Mathematics Textbook PDF*

These links will take you directly to the download page where you can access the textbooks in PDF format.

### Conclusion

Having the right study materials can make all the difference in your exam preparation. The **New General Mathematics** textbooks for SSCE and JSCE provide everything you need to succeed, from detailed explanations of complex concepts to ample practice exercises. Best of all, you can download them for free and start studying immediately!

Get ready to ace your exams with these valuable resources—download your copies today!