Find The Value Of X Rounded To The Nearest Tenth

Finding the value of x often involves solving equations, and the process can vary significantly depending on the type of equation you're dealing with. Whether it's a simple linear equation, a more complex quadratic, or even a trigonometric function, the key is to isolate x on one side of the equation. This article will guide you through various methods to find the value of x, rounded to the nearest tenth where necessary, covering a range of common mathematical scenarios.

Solving Linear Equations

Linear equations are the simplest to solve and generally follow a straightforward process of isolating the variable.

Basic Approach:

1. Simplify both sides: Combine any like terms on each side of the equation.
2. Isolate the term with x: Use addition or subtraction to move all terms without x to the opposite side of the equation.
3. Solve for x: Divide both sides of the equation by the coefficient of x.

Example 1:

Solve for x: 3x + 5 = 14

• Subtract 5 from both sides: 3x = 9
• Divide by 3: x = 3

Example 2:

Solve for x: 2x - 7 = 5x + 2

• Subtract 2x from both sides: -7 = 3x + 2
• Subtract 2 from both sides: -9 = 3x
• Divide by 3: x = -3

Dealing with Fractions:

When linear equations involve fractions, it's often helpful to eliminate the fractions early on.

1. Find the Least Common Denominator (LCD): Determine the smallest multiple that all denominators divide into.
2. Multiply every term by the LCD: This will clear the fractions.
3. Solve the resulting equation: Proceed as with a standard linear equation.

Example 3:

Solve for x: (x/2) + (1/3) = (5/6)

• The LCD of 2, 3, and 6 is 6.
• Multiply every term by 6: 6(x/2) + 6(1/3) = 6(5/6)
• Simplify: 3x + 2 = 5
• Subtract 2 from both sides: 3x = 3
• Divide by 3: x = 1

Tackling Quadratic Equations

Quadratic equations are of the form ax² + bx + c = 0, where a, b, and c are constants. There are several methods to solve these.

Factoring:

1. Set the equation to zero: Ensure the quadratic equation is in the form ax² + bx + c = 0.
2. Factor the quadratic expression: Find two binomials that multiply to give the quadratic expression.
3. Set each factor equal to zero: Solve for x in each resulting linear equation.

Example 4:

Solve for x: x² - 5x + 6 = 0

• Factor: (x - 2)(x - 3) = 0
• Set each factor to zero:
  • x - 2 = 0 => x = 2
  • x - 3 = 0 => x = 3

Quadratic Formula:

The quadratic formula is a universal method that works for any quadratic equation, even those that are difficult to factor.

x = (-b ± √(b² - 4ac)) / (2a)

Example 5:

Solve for x: 2x² + 3x - 5 = 0

• Identify a = 2, b = 3, and c = -5.
• Apply the quadratic formula: x = (-3 ± √(3² - 4(2)(-5))) / (2(2)) x = (-3 ± √(9 + 40)) / 4 x = (-3 ± √49) / 4 x = (-3 ± 7) / 4
• Solve for both possibilities:
  • x = (-3 + 7) / 4 = 1
  • x = (-3 - 7) / 4 = -2.5

Completing the Square:

This method involves manipulating the quadratic equation into a perfect square trinomial.

1. Divide by a: If a is not 1, divide the entire equation by a.
2. Move the constant term to the right side: This isolates the x² and x terms.
3. Complete the square: Add (b/2)² to both sides of the equation.
4. Factor the left side: It should now be a perfect square.
5. Take the square root: Take the square root of both sides.
6. Solve for x: Isolate x.

Example 6:

Solve for x: x² + 6x - 7 = 0

• Move the constant to the right: x² + 6x = 7
• Complete the square: (6/2)² = 9. Add 9 to both sides: x² + 6x + 9 = 7 + 9
• Factor the left side: (x + 3)² = 16
• Take the square root: x + 3 = ±4
• Solve for x:
  • x = -3 + 4 = 1
  • x = -3 - 4 = -7

Navigating Radical Equations

Radical equations contain variables inside a radical, most commonly a square root.

Basic Approach:

1. Isolate the radical: Get the radical term alone on one side of the equation.
2. Raise both sides to the appropriate power: If it's a square root, square both sides. If it's a cube root, cube both sides, and so on.
3. Solve the resulting equation: Solve for x.
4. Check for extraneous solutions: Always plug your solutions back into the original equation to ensure they are valid. Radical equations often produce solutions that don't work in the original equation due to squaring.

Example 7:

Solve for x: √(2x + 3) = 5

• The radical is already isolated.
• Square both sides: 2x + 3 = 25
• Subtract 3: 2x = 22
• Divide by 2: x = 11
• Check: √(2(11) + 3) = √(25) = 5. The solution is valid.

Example 8:

Solve for x: √(3x - 2) + 4 = x

• Isolate the radical: √(3x - 2) = x - 4
• Square both sides: 3x - 2 = (x - 4)² = x² - 8x + 16
• Rearrange to form a quadratic: 0 = x² - 11x + 18
• Factor: 0 = (x - 2)(x - 9)
• Solve: x = 2 or x = 9
• Check:
  • For x = 2: √(3(2) - 2) + 4 = √4 + 4 = 2 + 4 = 6 ≠ 2. x = 2 is extraneous.
  • For x = 9: √(3(9) - 2) + 4 = √25 + 4 = 5 + 4 = 9. x = 9 is a valid solution.

Working with Absolute Value Equations

Absolute value equations involve the absolute value function, denoted by | |. The absolute value of a number is its distance from zero, so it's always non-negative.

Basic Approach:

1. Isolate the absolute value: Get the absolute value expression alone on one side of the equation.
2. Set up two equations: One where the expression inside the absolute value is equal to the positive value on the other side, and one where it's equal to the negative value.
3. Solve both equations: Solve for x in each equation.

Example 9:

Solve for x: |2x - 1| = 7

• The absolute value is already isolated.
• Set up two equations:
  • 2x - 1 = 7 => 2x = 8 => x = 4
  • 2x - 1 = -7 => 2x = -6 => x = -3

Example 10:

Solve for x: |x + 3| + 2 = 5

• Isolate the absolute value: |x + 3| = 3
• Set up two equations:
  • x + 3 = 3 => x = 0
  • x + 3 = -3 => x = -6

Trigonometric Equations

Trigonometric equations involve trigonometric functions like sine, cosine, and tangent. Solving these often requires understanding the unit circle and the periodic nature of these functions.

Basic Approach:

1. Isolate the trigonometric function: Get the trigonometric function alone on one side of the equation.
2. Find the reference angle: Determine the angle whose trigonometric function value matches the one you have isolated.
3. Find all solutions within the given interval: Use the unit circle and the properties of trigonometric functions to find all angles that satisfy the equation within the specified interval (usually 0 to 2π or 0 to 360°). Remember to consider the signs of the trigonometric functions in different quadrants.
4. General solutions: If no specific interval is given, express the general solutions by adding integer multiples of the period of the function.

Example 11:

Solve for x: sin(x) = 0.5, 0 ≤ x < 2π

• The sine function is already isolated.
• The reference angle for sin(x) = 0.5 is π/6 (30°).
• Sine is positive in the first and second quadrants.
  • First quadrant: x = π/6
  • Second quadrant: x = π - π/6 = 5π/6

Example 12:

Solve for x: 2cos(x) - 1 = 0, 0 ≤ x < 2π

• Isolate the cosine function: cos(x) = 0.5
• The reference angle for cos(x) = 0.5 is π/3 (60°).
• Cosine is positive in the first and fourth quadrants.
  • First quadrant: x = π/3
  • Fourth quadrant: x = 2π - π/3 = 5π/3

Using Inverse Trigonometric Functions:

When the value of the trigonometric function isn't a standard value (like 0, 0.5, 1, √2/2, √3/2), you'll need to use inverse trigonometric functions (arcsin, arccos, arctan).

Example 13:

Solve for x: tan(x) = 2, 0 ≤ x < 2π

• x = arctan(2) ≈ 1.107 radians (using a calculator). This is the solution in the first quadrant.
• Tangent is also positive in the third quadrant, so we need to find the corresponding angle in the third quadrant.
• x = π + arctan(2) ≈ π + 1.107 ≈ 4.249 radians

Exponential and Logarithmic Equations

Exponential equations involve variables in the exponent, while logarithmic equations involve logarithms.

Exponential Equations:

1. Isolate the exponential term: Get the exponential term alone on one side of the equation.
2. Take the logarithm of both sides: Use either the common logarithm (base 10) or the natural logarithm (base e). The natural logarithm is often preferred when the base of the exponential term is e.
3. Use logarithm properties to bring the exponent down: log(a^b) = b*log(a)
4. Solve for x: Isolate x.

Example 14:

Solve for x: 2^(x+1) = 7

• The exponential term is already isolated.
• Take the natural logarithm of both sides: ln(2^(x+1)) = ln(7)
• Use the logarithm property: (x+1)ln(2) = ln(7)
• Divide by ln(2): x + 1 = ln(7)/ln(2)
• Subtract 1: x = ln(7)/ln(2) - 1 ≈ 1.807

Logarithmic Equations:

1. Isolate the logarithmic term: Get the logarithmic term alone on one side of the equation.
2. Rewrite the equation in exponential form: Use the definition of logarithms: log_b(a) = c is equivalent to b^c = a.
3. Solve for x: Isolate x.
4. Check for extraneous solutions: Make sure the argument of the logarithm is positive.

Example 15:

Solve for x: log_2(3x - 1) = 3

• The logarithmic term is already isolated.
• Rewrite in exponential form: 2^3 = 3x - 1
• Simplify: 8 = 3x - 1
• Add 1: 9 = 3x
• Divide by 3: x = 3
• Check: log_2(3(3) - 1) = log_2(8) = 3. The solution is valid.

Example 16:

Solve for x: ln(x + 2) = 1

• Rewrite in exponential form: e^1 = x + 2
• Simplify: e = x + 2
• Subtract 2: x = e - 2 ≈ 0.718

Rounding to the Nearest Tenth

Once you've found the value of x, you might need to round it to the nearest tenth. This means rounding to one decimal place.

Rules for Rounding:

• Look at the digit in the hundredths place: This is the digit immediately to the right of the tenths place.
• If the digit is 5 or greater, round up the digit in the tenths place.
• If the digit is less than 5, leave the digit in the tenths place as it is.

Examples:

• 3.14159 rounded to the nearest tenth is 3.1 (because the digit in the hundredths place is 4, which is less than 5).
• 12.789 rounded to the nearest tenth is 12.8 (because the digit in the hundredths place is 8, which is greater than or equal to 5).
• -5.65 rounded to the nearest tenth is -5.7 (because the digit in the hundredths place is 5, which is greater than or equal to 5).
• 0.049 rounded to the nearest tenth is 0.0 (because the digit in the hundredths place is 4, which is less than 5).
• 0.051 rounded to the nearest tenth is 0.1 (because the digit in the hundredths place is 5, which is greater than or equal to 5).

By understanding these different types of equations and the methods to solve them, you'll be well-equipped to find the value of x in a variety of mathematical contexts, and confidently round your answers to the nearest tenth. Remember to always check your answers, especially when dealing with radical and logarithmic equations, to avoid extraneous solutions.