Solving Equations to Find the Value of x

Algebra is a fundamental branch of mathematics that plays a crucial role in solving real-world problems. One of the most basic tasks in algebra is to solve equations for a variable, such as x. In this article, we will discuss several methods to find the value of x in different types of equations. We will also provide step-by-step solutions and verify the solutions to ensure their accuracy.

Equation Type 1: Linear Equations

Let's start with a simple linear equation:

(3x x 2)

To solve this equation:

Subtract x from both sides: 3x - x 2 This simplifies to: 2x 2 Divide both sides by 2: x 1

Therefore, the value of x in the equation (3x x 2) is x 1.

Equation Type 2: Square Root Equations

Next, let's solve an equation involving a square root:

(sqrt{x}^2 x)

Following the steps:

Take the square of both sides: (sqrt{x} x - 2) Take the square of both sides again: (x (x - 2)^2) Simplify the right side: (x x^2 - 4x 4) Move all terms to one side: (x - x^2 4x - 4 0) Combine like terms: (-x^2 5x - 4 0) Factor the left side: (-(x - 1)(x - 4) 0)

This gives us two solutions:

x - 1 0 or x - 4 0 x 1 or x 4

Verification:

For x 1: (sqrt{1}^2 1) works in the original equation For x 4: (sqrt{4}^2 4) works in the original equation

Therefore, the valid solution is (boxed{4}).

Equation Type 3: Fractional Exponent Equations

Let's consider an equation with a fractional exponent:

(f(x) - sqrt{x^2} x)

First, substitute the expression for (f(x)):

(f(x) x^{1/2} - 2)

Thus, the equation becomes:

(x^{1/2} - 2 x)

Next, simplify the left side:

(x^{3/2} x)

Divide both sides by x:

(x^{1/2} 1)

Square both sides:

(x 1^2)

Thus, the value of x is (boxed{1}).

Equation Type 4: Quadratic Equations

Solving a quadratic equation can be a bit more involved:

(sqrt{x^2} x)

Squaring both sides:

(x^2 x^2)

Subtract x from both sides:

(x^2 - x - 2 0)

Factor the equation:

((x - 2)(x 1) 0)

This gives us two solutions:

x - 2 0 or x 1 0 x 2 or x -1

Verification:

For x 2: (sqrt{x^2} 2) works in the original equation For x -1: (sqrt{x^2} 1) does not work in the original equation as it does not satisfy the equation (-1 1)

Therefore, the valid solution is (boxed{2}).

Conclusion

In this article, we discussed how to solve various types of equations for the variable x. We covered linear, square root, fractional exponent, and quadratic equations, providing step-by-step solutions for each type. It is essential to verify the solutions to ensure their accuracy. These methods are fundamental in algebra and are widely used in various fields, from physics to engineering.