What Is The Value Of X To The Nearest Tenth

What is the Value of x to the Nearest Tenth? A Comprehensive Guide

Determining the value of 'x' to the nearest tenth is a fundamental skill in mathematics, applicable across various fields from basic algebra to complex calculus and beyond. This comprehensive guide will explore diverse methods for solving for 'x', examining different types of equations and providing practical examples to solidify your understanding. We'll cover linear equations, quadratic equations, trigonometric equations, and even touch upon more advanced methods where applicable. Understanding these techniques is crucial for anyone aiming to master mathematical problem-solving.

Understanding the Problem: What Does "to the Nearest Tenth" Mean?

Before diving into the solutions, let's clarify the phrase "to the nearest tenth." This signifies rounding the value of 'x' to one decimal place. For instance:

• 1.234 rounded to the nearest tenth is 1.2
• 3.789 rounded to the nearest tenth is 3.8
• 5.555 rounded to the nearest tenth is 5.6 (we round up if the second decimal place is 5 or greater)

This precision is important, as it provides a manageable level of accuracy in many applications where exact values aren't always necessary or attainable.

Solving for 'x' in Linear Equations

Linear equations are the simplest type of algebraic equation, characterized by a single variable raised to the power of one. They typically take the form of:

ax + b = c

Where 'a', 'b', and 'c' are constants, and 'x' is the variable we aim to solve for.

Example 1:

Solve for 'x' to the nearest tenth: 3x + 7 = 16

Solution:

1. Subtract 7 from both sides: 3x = 9
2. Divide both sides by 3: x = 3

In this case, 'x' is already a whole number, so rounding to the nearest tenth isn't necessary. The answer remains x = 3.0.

Example 2:

Solve for 'x' to the nearest tenth: 2.5x - 4 = 8.75

Solution:

1. Add 4 to both sides: 2.5x = 12.75
2. Divide both sides by 2.5: x = 5.1

The solution, 'x = 5.1', is already expressed to the nearest tenth.

Solving for 'x' in Quadratic Equations

Quadratic equations involve a variable raised to the power of two. They typically take the form:

ax² + bx + c = 0

Solving quadratic equations often requires using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

Example 3:

Solve for 'x' to the nearest tenth: x² + 3x - 10 = 0

Solution:

Here, a = 1, b = 3, and c = -10. Substituting these values into the quadratic formula:

x = (-3 ± √(3² - 4 * 1 * -10)) / 2 * 1

x = (-3 ± √49) / 2

x = (-3 ± 7) / 2

This gives us two possible solutions:

x = ( -3 + 7 ) / 2 = 2 x = ( -3 - 7 ) / 2 = -5

Both solutions are whole numbers, so the answers to the nearest tenth are x = 2.0 and x = -5.0.

Example 4:

Solve for 'x' to the nearest tenth: 2x² - 5x + 1 = 0

Solution:

Using the quadratic formula with a = 2, b = -5, and c = 1:

x = (5 ± √(25 - 4 * 2 * 1)) / 4

x = (5 ± √17) / 4

x ≈ (5 ± 4.123) / 4

This leads to two approximate solutions:

x ≈ (5 + 4.123) / 4 ≈ 2.28 x ≈ (5 - 4.123) / 4 ≈ 0.22

Rounding to the nearest tenth, we get x ≈ 2.3 and x ≈ 0.2.

Solving for 'x' in Trigonometric Equations

Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent. Solving these equations often requires using inverse trigonometric functions and understanding the unit circle.

Example 5:

Solve for 'x' to the nearest tenth (in degrees): sin(x) = 0.5

Solution:

Using the inverse sine function:

x = sin⁻¹(0.5) = 30°

There's another solution in the second quadrant (180° - 30° = 150°). Therefore, the solutions to the nearest tenth are x ≈ 30.0° and x ≈ 150.0°. Note that trigonometric equations often have multiple solutions.

Example 6:

Solve for 'x' to the nearest tenth (in radians): cos(x) = -0.7

Solution:

Using the inverse cosine function:

x = cos⁻¹(-0.7) ≈ 2.346 radians

Considering the symmetry of the cosine function, another solution exists in the third quadrant:

x ≈ 2π - 2.346 ≈ 3.937 radians

Rounding to the nearest tenth, the solutions are x ≈ 2.3 radians and x ≈ 3.9 radians.

Solving for 'x' in More Complex Equations

Solving for 'x' can become significantly more complex depending on the type of equation. Methods like logarithmic equations, exponential equations, and systems of equations often require more advanced techniques. These might include:

• Factoring: Breaking down complex expressions into simpler ones.
• Substitution: Replacing one variable with an equivalent expression.
• Elimination: Systematically removing variables from equations.
• Numerical Methods: Approximating solutions using iterative algorithms (like Newton-Raphson method).

Importance of Understanding Significant Figures and Rounding

When dealing with real-world applications, it is crucial to consider significant figures and rounding errors. The level of precision needed depends on the context. Rounding to the nearest tenth might be sufficient for some scenarios, while others require greater precision (e.g., to the nearest hundredth or thousandth). Always ensure that your rounding doesn't introduce significant errors into your final results.

Conclusion

Solving for 'x' to the nearest tenth encompasses a broad range of mathematical skills. Mastering this skill requires a firm understanding of different equation types and the appropriate solving techniques. From linear and quadratic equations to trigonometric and more complex forms, the ability to isolate 'x' and accurately round to the nearest tenth is essential for success in many areas of study and practical application. Remember to practice regularly to build your confidence and proficiency. The examples provided offer a starting point for further exploration and practice, allowing you to confidently tackle a wide range of problems.