End of Semester Algebra 2B Practice Quiz

Subtracting 5 from both sides gives 2x = 8. Dividing by 2 yields x = 4. This is a straightforward linear equation.

Combining like terms, add the coefficients: 3 + 4 gives 7. The variable remains as x. This exercise tests basic algebraic simplification.

The expression x² - 9 is a difference of two squares. It factors into (x - 3)(x + 3). Recognizing this pattern is key in algebra.

Multiplying both sides of the equation by 2 isolates x, giving x = 6. This is an introductory exercise on solving simple equations.

The slope formula is (y₂ - y₝) / (x₂ - x₝). Using the points (1, 2) and (3, 6), the slope is (6 - 2) / (3 - 1) = 4/2 = 2. This concept is fundamental in coordinate geometry.

Factoring the quadratic gives (x - 2)(x - 3) = 0. Setting each factor equal to zero yields x = 2 and x = 3. This is a standard method for solving quadratic equations.

Distribute to get 2x + 6 and -4x + 4, then combine like terms: 2x - 4x equals -2x and 6 + 4 equals 10. The simplified expression is -2x + 10.

Adding the two equations cancels y and gives 2x = 12, so x = 6. Substituting back into one of the original equations results in y = 4. This method of elimination is effective for solving linear systems.

Canceling the common factors (one x and one y) from the numerator and denominator simplifies the expression to 3x divided by y. This assesses understanding of exponent rules and factor cancellation.

Cross multiplying gives x + 2 = 2(x - 2). Simplifying leads to x = 6 after isolating the variable. It is important to check that the solution does not make any denominator zero.

The vertex x-coordinate is found using -b/(2a), which gives 2. Substituting x = 2 into the function yields y = 3, so the vertex is (2, 3). This is a common method for finding the vertex of a parabola.

Using the law of exponents, add the exponents: 2³ * 2❴ = 2^(3+4) = 2❷. Calculating 2❷ gives 128. This underlines the fundamental rules governing exponents.

Since 81 can be written as 3❴, the equation becomes 3ˣ = 3❴, yielding x = 4. Rewriting numbers with the same base allows direct comparison of exponents.

To find the inverse, swap x and y in the equation y = 2x + 5 and solve for y. This process leads to f❻¹(x) = (x - 5)/2. Understanding inverses is crucial for many algebraic applications.

Factor the numerator as (x - 3)(x + 3). Cancelling the common factor (x + 3) results in x - 3, with the stipulation that x � -3. This tests factoring and simplification skills.

Using the quadratic formula, the discriminant is 49, and the solutions are x = (3 ± 7)/4, which simplify to x = 5/2 and x = -1. This problem emphasizes the use of the quadratic formula. Always check the discriminant for real solutions.

By grouping or testing factors, the expression factors as (2x - 5)(3x + 2). Expanding these factors confirms the original expression. This reinforces methods of factoring quadratic expressions.

Combine the logarithms using the product rule: log₂[x(x - 2)] = 3, which implies x(x - 2) = 8. Solving the quadratic and considering the domain of logarithms yields x = 4 as the only valid solution.

Assume the function is f(x) = a(x - 2)(x + 3). With f(0) = -6a equated to 12, solving for a gives a = -2. Thus, the quadratic function is f(x) = -2(x - 2)(x + 3). This requires combining knowledge of roots and y-intercepts.

Squaring both sides produces the quadratic equation x² - 4x - 5 = 0, which factors to (x - 5)(x + 1) = 0. Testing both solutions in the original equation shows that only x = 5 is valid because the square root function returns a nonnegative value. This problem highlights the importance of checking for extraneous solutions.