Factorization, Exponential Equations, And Inequalities: A Step-by-Step Guide

This comprehensive guide delves into the intricacies of factorization and equation solving, providing detailed explanations and step-by-step solutions to help you master these fundamental mathematical concepts. We'll tackle a variety of problems, from factoring quadratic expressions to solving exponential equations and systems of linear equations. Get ready to enhance your mathematical prowess and build a solid foundation for advanced studies.

1. Factorize

Factorization is the process of breaking down a mathematical expression into its constituent factors. This is a crucial skill in algebra as it simplifies expressions and aids in solving equations. In this section, we will explore the factorization of two distinct expressions.

a. p^2 - 4124p = 27

This problem involves factoring a quadratic expression. The first step is to rearrange the equation into the standard quadratic form, which is ax^2 + bx + c = 0. In our case, the variable is 'p' instead of 'x', but the principle remains the same. Let's dive into the detailed solution.

Step 1: Rearrange the equation

Start by moving all terms to one side of the equation to set it equal to zero. This gives us:

p^2 - 4124p - 27 = 0

Step 2: Factor the quadratic expression

Now, we need to factor the quadratic expression p^2 - 4124p - 27. We are looking for two numbers that multiply to -27 and add up to -4124. This might seem daunting, but we can approach it systematically. Since the constant term is -27, the factors will have opposite signs. The factors of 27 are 1 and 27, and 3 and 9. Given the large coefficient of the 'p' term (-4124), we should look for factors that yield a significant difference. We observe that:

-27 * 1 = -27

-27 + 1 = -26

However, these factors do not give us the required sum of -4124. We might consider alternative approaches or realize there might be a mistake in the problem statement. Given the coefficient -4124, a more realistic factorization might involve larger numbers or a different constant term. Let's assume there's a typo and the equation should have been something factorizable with smaller integer values or can be solved using the quadratic formula.

Step 3: Reassessment and Alternative Approach

Since direct factorization with simple integers seems unlikely given the large coefficient, we can use the quadratic formula to find the roots of the equation. The quadratic formula is given by:

p = [-b ± √(b^2 - 4ac)] / (2a)

For our equation, a = 1, b = -4124, and c = -27. Plugging these values into the quadratic formula, we get:

p = [4124 ± √((-4124)^2 - 4 * 1 * -27)] / (2 * 1)

p = [4124 ± √(16990576 + 108)] / 2

p = [4124 ± √16990684] / 2

p ≈ [4124 ± 4121.97] / 2

This gives us two possible values for p:

p1 ≈ (4124 + 4121.97) / 2 ≈ 4122.985

p2 ≈ (4124 - 4121.97) / 2 ≈ 1.015

These roots suggest that the initial expression might not factorize neatly with integers. Thus, leaving it in this form or using these approximations would be the practical approach.

b. 4x^2 + 12x + 9

This expression is a quadratic trinomial, and we can try to factor it using the perfect square trinomial pattern. A perfect square trinomial is of the form a^2 + 2ab + b^2 or a^2 - 2ab + b^2, which factors into (a + b)^2 or (a - b)^2, respectively. Let's examine the given expression:

Step 1: Identify the pattern

We observe that 4x^2 is a perfect square, as it can be written as (2x)^2. Similarly, 9 is also a perfect square, as it can be written as 3^2. The middle term, 12x, might fit the 2ab part of the perfect square trinomial pattern. Let's verify:

If a = 2x and b = 3, then 2ab = 2 * (2x) * 3 = 12x

This matches the middle term in our expression.

Step 2: Factor the trinomial

Since the expression fits the perfect square trinomial pattern, we can factor it as follows:

4x^2 + 12x + 9 = (2x)^2 + 2 * (2x) * 3 + 3^2

Using the formula a^2 + 2ab + b^2 = (a + b)^2, we get:

(2x + 3)^2

Therefore, the factored form of the expression is (2x + 3)^2. This completes the factorization of the given quadratic trinomial.

2. Solve for x

Solving for x involves finding the value(s) of x that satisfy a given equation. This is a fundamental skill in algebra and is used extensively in various mathematical and scientific applications. We will now solve several different types of equations for x, each requiring a unique approach.

a. 4^x = 1

This is an exponential equation. To solve for x, we need to express both sides of the equation with the same base or use logarithms. Let's explore the solution.

Step 1: Express both sides with the same base

We can rewrite 1 as 4^0 since any non-zero number raised to the power of 0 is 1. Therefore, the equation becomes:

4^x = 4^0

Step 2: Equate the exponents

Since the bases are the same, we can equate the exponents:

x = 0

Thus, the solution for the equation 4^x = 1 is x = 0. This result is straightforward and showcases a basic property of exponents.

b. 2.4^(x-1) = 8^(-x)

This is another exponential equation, but it is slightly more complex than the previous one. To solve it, we need to express all terms with the same base and then equate the exponents. Let's walk through the solution step by step.

Step 1: Express all terms with the same base

We can express 4 and 8 as powers of 2: 4 = 2^2 and 8 = 2^3. Substituting these into the equation, we get:

2 * (2²⁾(x-1) = (2³⁾(-x)

Using the power of a power rule, (a^m)n = a^(mn), we can simplify the equation:

2 * 2^(2(x-1)) = 2^(-3x)

2 * 2^(2x-2) = 2^(-3x)

Now, we can rewrite the left side using the product of powers rule, a^m * a^n = a^(m+n):

2^1 * 2^(2x-2) = 2^(1 + 2x - 2)

2^(2x - 1) = 2^(-3x)

Step 2: Equate the exponents

Since the bases are the same, we can equate the exponents:

2x - 1 = -3x

Step 3: Solve for x

Now, we solve the linear equation for x:

2x + 3x = 1

5x = 1

x = 1/5

Therefore, the solution for the equation 2 * 4^(x-1) = 8^(-x) is x = 1/5. This solution involves a few key steps in manipulating exponential expressions and then solving a linear equation.

c. 3x + 4y = -545

d. 3x + y = 8

These are two linear equations forming a system of equations. To solve for x, we need to find the value of x that satisfies both equations simultaneously. There are several methods to solve a system of linear equations, including substitution, elimination, and graphical methods. Here, we will use the substitution method. Let's break down the solution.

Step 1: Solve one equation for y

We can solve the second equation for y:

3x + y = 8

y = 8 - 3x

Step 2: Substitute into the other equation

Now, substitute this expression for y into the first equation:

3x + 4(8 - 3x) = -545

Step 3: Simplify and solve for x

Expand and simplify the equation:

3x + 32 - 12x = -545

Combine like terms:

-9x + 32 = -545

Subtract 32 from both sides:

-9x = -577

Divide by -9:

x = -577 / -9

x = 64.111...

Rounding x to two decimal places, we have:

x ≈ 64.11

Step 4: Find y

Substitute the value of x back into the expression for y:

y = 8 - 3x

y = 8 - 3(64.11)

y = 8 - 192.33

y ≈ -184.33

Therefore, the approximate solution for the system of equations is x ≈ 64.11 and y ≈ -184.33. This method demonstrates a standard approach to solving systems of linear equations using substitution.

3. Solve for x

a. 3x < 9.14

This is a linear inequality. Solving for x involves isolating x on one side of the inequality. We will also represent the solution on a number line. Let's go through the steps.

Step 1: Isolate x

Divide both sides of the inequality by 3:

3x < 9.14

x < 9.14 / 3

x < 3.04666...

Rounding to two decimal places, we get:

x < 3.05

Step 2: Represent the solution on a number line

To represent this on a number line, we draw a number line and mark the point 3.05. Since x is less than 3.05, we use an open circle (to indicate that 3.05 is not included in the solution) and shade the region to the left of 3.05. This shaded region represents all values of x that satisfy the inequality.

[Number line illustration: An open circle at 3.05 with a shaded line extending to the left, indicating all values less than 3.05]

This number line representation visually demonstrates the set of all solutions to the inequality x < 3.05. This completes the solution for the linear inequality.

Conclusion

In this guide, we've covered a range of mathematical problems, including factorization, solving exponential equations, systems of linear equations, and linear inequalities. Each problem type requires a specific approach and set of techniques. By understanding these methods and practicing regularly, you can significantly improve your mathematical problem-solving skills. Remember, mathematics is a journey of continuous learning and practice, so keep exploring and challenging yourself with new problems.