Unit 1 Algebra 2 Practice Quiz

To solve for x, add 4 to both sides to obtain 2x = 4, then divide by 2 to find x = 2. This uses basic algebraic manipulation.

The quadratic factors into (x + 2)(x + 3) because 2 and 3 add to 5 and multiply to 6. Recognizing the correct pair is key to factoring.

Using the exponent rule (a^b)^c = a^(b*c), (3²)³ becomes 3^(2� - 3) = 3❶, which equals 729. This reinforces exponent multiplication.

Substitute x = 3 into the function f(x) = 2x + 1 to get f(3) = 2(3) + 1 = 7. This is a simple function evaluation task.

Any nonzero number raised to the power of 0 is 1 by exponent rules. This fundamental property is essential in algebra.

Factor the quadratic as (x - 5)(x + 2) = 0 and use the zero product property to obtain x = 5 or x = -2. This demonstrates solving quadratics by factoring.

First compute g(2) = 2² = 4, then evaluate f(4) = 3(4) - 2 = 10. This tests understanding of function composition.

The equation log₂(x) = 5 means x = 2❵, which equals 32. This directly applies the definition of logarithms.

Since the fractions share the same denominator, add the numerators to get (2+3)/x = 5/x. This reaffirms operations with fractions.

To find the inverse, swap x and y in the equation y = (x - 3)/2 and solve for y, yielding f❻¹(x) = 2x + 3. This exercise reinforces the process of finding inverse functions.

Square both sides of the equation to remove the square root: (√(x + 5))² = 3² gives x + 5 = 9, so x = 4. Always check for extraneous solutions when dealing with radicals.

Substitute -3 into f(x) = x² to compute (-3)² = 9. Squaring any real number, whether negative or positive, yields a nonnegative result.

Factor the numerator as (x - 3)(x + 3) and then cancel the common factor (x + 3), leaving x - 3. This demonstrates factoring and simplifying rational expressions.

The equation is in vertex form y = (x - h)² + k, where h = 2 and k = 4, making the vertex (2, 4). This directly tests knowledge of vertex form.

Since 81 can be written as 3❴, equate the exponents to find x = 4. Recognizing exponent equivalences is key to solving such problems.

Combine the logarithms using the product rule to obtain log₃((x + 1)(x - 2)) = 1, then convert to exponential form resulting in (x + 1)(x - 2) = 3. Solve the quadratic x² - x - 5 = 0 and apply domain restrictions to choose the valid solution.

For √(2x - 7) to be defined, 2x - 7 must be non-negative (x ≥ 7/2). Also, the denominator x - 4 cannot be zero, so x � 4. Combining these conditions yields the domain [7/2, 4) ∪ (4, ∞).

First compute g(x) = x² - 1, then f(g(x)) = 2(x² - 1) + 3 = 2x² + 1. Setting 2x² + 1 equal to 11 gives x² = 5, so x = √5 or x = -√5.

Taking logarithms of both sides yields (x + 1)ln2 = (x - 1)ln3. Rearranging and solving for x gives x = (ln2 + ln3) / (ln3 - ln2). This problem demonstrates the use of logarithms to solve exponential equations.