Which Of The Following Equations Have Exactly One Solution

Pinpointing equations with exactly one solution is a fundamental skill in algebra, critical for solving various mathematical problems and understanding the nature of equations. An equation has exactly one solution when there is only one value for the variable that makes the equation true. Let's dive into different types of equations and how to identify those with a single, unique solution.

Linear Equations

Linear equations are equations where the highest power of the variable is one. They can be written in the form ax + b = 0, where a and b are constants, and x is the variable.

Characteristics of Linear Equations:

• Graph as a straight line.
• Have at most one solution.

How to Identify a Linear Equation with One Solution:

A linear equation has exactly one solution if the coefficient of x (i.e., a) is not zero. For example:

• 2x + 3 = 0 has one solution.
• 5x - 7 = 0 has one solution.

Example:

Solve the equation 3x + 6 = 0.

1. Subtract 6 from both sides: 3x = -6
2. Divide by 3: x = -2

Therefore, x = -2 is the only solution for the equation 3x + 6 = 0.

Quadratic Equations

Quadratic equations are equations where the highest power of the variable is two. They can be written in the form ax² + bx + c = 0, where a, b, and c are constants, and x is the variable.

Characteristics of Quadratic Equations:

• Graph as a parabola.
• Can have zero, one, or two real solutions.

The Discriminant: Determining the Number of Solutions

The discriminant, denoted as Δ (Delta), is a part of the quadratic formula that helps determine the nature of the solutions. The discriminant is given by:

• Δ = b² - 4ac

Based on the value of the discriminant:

• If Δ > 0, the equation has two distinct real solutions.
• If Δ = 0, the equation has exactly one real solution (a repeated or double root).
• If Δ < 0, the equation has no real solutions (two complex solutions).

How to Identify a Quadratic Equation with One Solution:

To identify a quadratic equation with exactly one solution, calculate the discriminant and check if it equals zero.

Example:

Consider the equation x² - 6x + 9 = 0. Here, a = 1, b = -6, and c = 9.

1. Calculate the discriminant: Δ = (-6)² - 4(1)(9) = 36 - 36 = 0

Since the discriminant is zero, the equation has exactly one real solution.

Finding the Solution:

When Δ = 0, the solution can be found using the formula:

• x = -b / 2a

In our example:

• x = -(-6) / (2 * 1) = 6 / 2 = 3

Thus, x = 3 is the only solution for the equation x² - 6x + 9 = 0.

Absolute Value Equations

Absolute value equations involve the absolute value of a variable expression. The absolute value of a number is its distance from zero on the number line.

General Form:

• |ax + b| = c

Characteristics of Absolute Value Equations:

• Can have zero, one, or two solutions.

How to Identify an Absolute Value Equation with One Solution:

An absolute value equation |ax + b| = c has one solution under specific conditions:

• Case 1: c = 0

  If c = 0, then the equation becomes |ax + b| = 0. The only way for the absolute value of an expression to be zero is if the expression itself is zero. Therefore, ax + b = 0. This is a linear equation that has exactly one solution if a ≠ 0.

• Case 2: Special Cases

  In some cases, an absolute value equation might simplify to a form that clearly has only one solution due to constraints or other conditions.

Example 1: c = 0

Solve the equation |2x - 4| = 0.

1. Set the expression inside the absolute value to zero: 2x - 4 = 0
2. Solve for x: 2x = 4
3. x = 2

Thus, x = 2 is the only solution for the equation |2x - 4| = 0.

Example 2: c > 0

Consider the equation |x - 3| = 5.

1. Split into two equations:
   • x - 3 = 5
   • x - 3 = -5
2. Solve each equation:
   • x = 8
   • x = -2

In this case, there are two solutions (x = 8 and x = -2), so this equation does not have exactly one solution.

Exponential Equations

Exponential equations involve variables in the exponents.

General Form:

• a^(f(x)) = b

Where a and b are constants, and f(x) is an expression involving x.

Characteristics of Exponential Equations:

• The number of solutions depends on the specific equation.
• Often solved using logarithms.

How to Identify an Exponential Equation with One Solution:

An exponential equation has exactly one solution when the equation can be simplified to a form where the exponents can be directly compared or when logarithmic properties lead to a single value for the variable.

Example 1:

Solve the equation 2^(x + 1) = 8.

1. Rewrite 8 as a power of 2: 2^(x + 1) = 2^3
2. Since the bases are equal, equate the exponents: x + 1 = 3
3. Solve for x: x = 2

Thus, x = 2 is the only solution for the equation 2^(x + 1) = 8.

Example 2:

Solve the equation 3^(2x) = 9.

1. Rewrite 9 as a power of 3: 3^(2x) = 3^2
2. Equate the exponents: 2x = 2
3. Solve for x: x = 1

Thus, x = 1 is the only solution for the equation 3^(2x) = 9.

Example 3:

Consider 4^x = 64.

1. Express 64 as a power of 4: 4^x = 4^3
2. Equate the exponents: x = 3

Here, x = 3 is the unique solution.

Logarithmic Equations

Logarithmic equations involve logarithms of variable expressions.

General Form:

• logₐ(f(x)) = b

Where a and b are constants, and f(x) is an expression involving x.

Characteristics of Logarithmic Equations:

• The number of solutions depends on the specific equation.
• Solved using properties of logarithms and exponential forms.

How to Identify a Logarithmic Equation with One Solution:

A logarithmic equation has exactly one solution when, after applying logarithmic properties and simplifying, you obtain a single value for the variable.

Example 1:

Solve the equation log₂(x + 3) = 4.

1. Convert to exponential form: x + 3 = 2⁴
2. Simplify: x + 3 = 16
3. Solve for x: x = 13

Thus, x = 13 is the only solution for the equation log₂(x + 3) = 4.

Example 2:

Solve the equation log₁₀(3x - 2) = 1.

1. Convert to exponential form: 3x - 2 = 10¹
2. Simplify: 3x - 2 = 10
3. Solve for x: 3x = 12
4. x = 4

Thus, x = 4 is the only solution for the equation log₁₀(3x - 2) = 1.

Example 3:

Consider log(x) + log(x - 3) = 1.

1. Combine the logarithms: log(x(x - 3)) = 1
2. Convert to exponential form: x(x - 3) = 10^1
3. Simplify: x^2 - 3x = 10
4. Rearrange: x^2 - 3x - 10 = 0
5. Factor: (x - 5)(x + 2) = 0
6. Solve for x: x = 5 or x = -2

However, since the logarithm of a negative number is undefined, x = -2 is not a valid solution. Thus, x = 5 is the only valid solution.

Rational Equations

Rational equations involve rational expressions (fractions with polynomials in the numerator and/or denominator).

General Form:

• P(x) / Q(x) = R(x) / S(x)

Where P(x), Q(x), R(x), and S(x) are polynomials.

Characteristics of Rational Equations:

• The number of solutions depends on the specific equation.
• Requires careful attention to excluded values (values of x that make the denominator zero).

How to Identify a Rational Equation with One Solution:

A rational equation has exactly one solution when, after clearing the fractions and simplifying, you obtain a single valid value for the variable that does not make any of the original denominators zero.

Example 1:

Solve the equation (x + 1) / (x - 2) = 3 / (x - 2).

1. Multiply both sides by (x - 2): x + 1 = 3
2. Solve for x: x = 2

However, x = 2 makes the denominator (x - 2) equal to zero, so it is an excluded value. Therefore, this equation has no solution.

Example 2:

Solve the equation 2 / x = 4 / (x + 1).

1. Cross-multiply: 2(x + 1) = 4x
2. Simplify: 2x + 2 = 4x
3. Solve for x: 2 = 2x
4. x = 1

Check that x = 1 does not make any denominator zero. Thus, x = 1 is the only solution for the equation 2 / x = 4 / (x + 1).

Example 3:

Solve 1/x = 2/(x+1)

1. Cross multiply: x + 1 = 2x
2. Simplify: 1 = x
3. Therefore, x = 1

Radical Equations

Radical equations involve variables inside radicals (square roots, cube roots, etc.).

General Form:

• √(f(x)) = g(x)

Where f(x) and g(x) are expressions involving x.

Characteristics of Radical Equations:

• The number of solutions depends on the specific equation.
• Requires squaring (or raising to the appropriate power) both sides, which can introduce extraneous solutions.

How to Identify a Radical Equation with One Solution:

A radical equation has exactly one solution when, after isolating the radical, raising both sides to the appropriate power, and solving for the variable, you obtain a single value that satisfies the original equation and does not introduce any inconsistencies.

Example 1:

Solve the equation √(x + 2) = 3.

1. Square both sides: x + 2 = 9
2. Solve for x: x = 7

Check that x = 7 satisfies the original equation: √(7 + 2) = √9 = 3. Thus, x = 7 is the only solution for the equation √(x + 2) = 3.

Example 2:

Solve the equation √(2x - 1) = x - 2.

1. Square both sides: 2x - 1 = (x - 2)²
2. Expand: 2x - 1 = x² - 4x + 4
3. Rearrange: x² - 6x + 5 = 0
4. Factor: (x - 5)(x - 1) = 0
5. Solve for x: x = 5 or x = 1

Check each solution:

• For x = 5: √(2(5) - 1) = √(9) = 3 and 5 - 2 = 3. Thus, x = 5 is a valid solution.
• For x = 1: √(2(1) - 1) = √(1) = 1 and 1 - 2 = -1. Thus, x = 1 is an extraneous solution.

Therefore, x = 5 is the only solution for the equation √(2x - 1) = x - 2.

Example 3:

Consider √(x + 3) = x + 1.

1. Square both sides: x + 3 = (x + 1)^2
2. Expand: x + 3 = x^2 + 2x + 1
3. Rearrange: x^2 + x - 2 = 0
4. Factor: (x + 2)(x - 1) = 0
5. Solve: x = -2 or x = 1

Check:

• If x = -2: √(-2 + 3) = 1 and -2 + 1 = -1. So x = -2 is extraneous.
• If x = 1: √(1 + 3) = 2 and 1 + 1 = 2. So x = 1 is valid.

Thus, only x = 1 is a solution.

Trigonometric Equations

Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent.

General Form:

• sin(x) = a
• cos(x) = b
• tan(x) = c

Where a, b, and c are constants.

Characteristics of Trigonometric Equations:

• Trigonometric functions are periodic, so they typically have infinitely many solutions.
• However, when restricted to a specific interval, they can have a finite number of solutions, including exactly one.

How to Identify a Trigonometric Equation with One Solution Within a Given Interval:

A trigonometric equation can have exactly one solution within a specified interval if the trigonometric function takes on a particular value only once within that interval.

Example 1:

Solve the equation sin(x) = 0.5 for 0 ≤ x ≤ π/2.

1. Find the principal value: x = arcsin(0.5) = π/6

Since π/6 lies within the interval [0, π/2], it is a valid solution. Moreover, within this interval, sine is increasing, so there are no other solutions.

Thus, x = π/6 is the only solution for the equation sin(x) = 0.5 in the interval 0 ≤ x ≤ π/2.

Example 2:

Solve the equation cos(x) = 0 for 0 ≤ x ≤ π.

1. Find the value(s) of x for which cos(x) = 0: x = π/2

Since π/2 lies within the interval [0, π], it is a valid solution. Cosine is decreasing in this interval after 0 and until π, so there is only one point where cos(x) = 0.

Thus, x = π/2 is the only solution for the equation cos(x) = 0 in the interval 0 ≤ x ≤ π.

Example 3:

Find x such that tan(x) = 1 for 0 ≤ x ≤ π/2.

1. Since tan(x) = 1 when x = π/4, and π/4 is in the interval, then x = π/4.

System of Equations

A system of equations is a set of two or more equations with the same variables. The solution to a system of equations is the set of values for the variables that make all equations in the system true.

Linear Systems

• A system of linear equations may have one solution, no solution, or infinitely many solutions.

How to Identify a System of Equations with One Solution

For a system of two linear equations in two variables, there is exactly one solution if the lines represented by the equations intersect at one point. This occurs when the lines have different slopes.

Example

Consider the system of equations:

1. 2x + y = 5
2. x - y = 1

Solve the system:

• From equation (2), x = y + 1
• Substitute into equation (1): 2(y + 1) + y = 5
• Simplify: 2y + 2 + y = 5
• Combine terms: 3y = 3
• Solve for y: y = 1
• Substitute y = 1 into x = y + 1: x = 1 + 1 = 2

The solution to the system is x = 2 and y = 1. Since there is only one pair of values for x and y that satisfy both equations, the system has exactly one solution.

Practical Tips for Solving Equations

1. Simplify: Always simplify both sides of the equation as much as possible before attempting to solve for the variable.
2. Isolate the Variable: Use algebraic operations to isolate the variable on one side of the equation.
3. Check Solutions: When solving equations, especially radical and rational equations, always check your solutions by substituting them back into the original equation to ensure they are valid.
4. Consider the Domain: Be mindful of the domain of the functions involved. For example, the argument of a logarithm must be positive, and the denominator of a fraction cannot be zero.

Conclusion

Identifying which equations have exactly one solution requires understanding the characteristics of different types of equations, including linear, quadratic, absolute value, exponential, logarithmic, rational, radical, and trigonometric equations. By applying appropriate techniques and considering the properties of each type, you can accurately determine whether an equation has a unique solution. Always remember to check your solutions to avoid extraneous values and ensure they satisfy the original equation. Mastering these skills is essential for success in algebra and beyond.