Applied Mathematics Placement Test Preparation Quiz

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Quizzes Created: 1088 | Total Attempts: 1,101,313
| Questions: 10 | Updated: Apr 15, 2026
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1. Solve the equation by the zero-factor property: x² + 10x - 24 = 0.

Explanation

To solve the equation x² + 10x - 24 = 0 using the zero-factor property, we first factor the quadratic. We look for two numbers that multiply to -24 and add to 10. These numbers are -2 and 12, allowing us to rewrite the equation as (x - 2)(x + 12) = 0. Setting each factor to zero gives us the solutions x = 2 and x = -12. Thus, the solutions to the equation are { -12 , 2 }.

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About This Quiz
Applied Mathematics Placement Test Preparation Quiz - Quiz

This assessment focuses on essential algebra concepts, including solving quadratic equations, analyzing functions, and understanding inequalities. It evaluates skills such as applying the zero-factor property, using the quadratic formula, and determining function properties. This preparation is vital for learners aiming to strengthen their algebra foundation and succeed in higher mathematics... see morecourses. see less

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2. Solve the equation using the quadratic formula: 2x² + 12x = -3.

Explanation

To solve the equation \(2x^2 + 12x + 3 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we identify \(a = 2\), \(b = 12\), and \(c = 3\). Calculating the discriminant: \(b^2 - 4ac = 12^2 - 4(2)(3) = 144 - 24 = 120\). Thus, we have \(x = \frac{-12 \pm \sqrt{120}}{4}\). Simplifying \(\sqrt{120}\) gives \(2\sqrt{30}\), leading to \(x = \frac{-12 \pm 2\sqrt{30}}{4} = -6 \pm \frac{\sqrt{30}}{2}\). The answer simplifies to \(-6 \pm 30/2\).

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3. The area of a square is numerically 96 more than the perimeter. Find the length of the side.

Explanation

Let the side length of the square be \( s \). The area of the square is \( s^2 \) and the perimeter is \( 4s \). According to the problem, \( s^2 = 4s + 96 \). Rearranging gives \( s^2 - 4s - 96 = 0 \). Solving this quadratic equation using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a=1, b=-4, c=-96 \) yields \( s = 12 \) (the positive solution). Thus, the length of the side of the square is 12 units.

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4. Solve the quadratic inequality: x² - 6x + 8 > 0. Write the solution set in interval notation.

Explanation

To solve the inequality x² - 6x + 8 > 0, we first find the roots of the corresponding equation x² - 6x + 8 = 0, which are x = 2 and x = 4. These roots divide the number line into intervals: (-∞, 2), (2, 4), and (4, ∞). Testing points from each interval shows that the quadratic is positive in the intervals (-∞, 2) and (4, ∞), while it is negative between the roots. Therefore, the solution set, where the inequality holds true, is expressed in interval notation as (-∞, 2) ∪ (4, ∞).

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5. Decide whether the relation defines a function: {(1, 9), (-1, -9), (-7, -5), (6, -9)}.

Explanation

A relation is defined as a function if each input (or x-value) corresponds to exactly one output (or y-value). In the given relation, each x-value (1, -1, -7, 6) is unique and maps to a single y-value (9, -9, -5, -9). Since there are no repeated x-values with different y-values, this relation meets the criteria for being a function.

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6. Find the domain and range of the relation: {(1, 9), (-1, -9), (-7, -5), (6, -9)}.

Explanation

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7. Which one is the graph of y = x³? What is its range?

Explanation

The graph of the function y = x³ is a cubic function that extends infinitely in both the positive and negative y-directions as x varies. As x approaches positive or negative infinity, y also approaches positive or negative infinity, respectively. Therefore, the range of y = x³ is all real numbers, represented as (-∞, ∞). This characteristic is reflected in graph a, which correctly shows the unbounded nature of the cubic function.

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8. Find f(-6) for f(x) = 2x, if x ≤ -1; f(x) = x - 3, if x > -1.

Explanation

To find f(-6), we first determine which piece of the piecewise function applies. Since -6 is less than -1, we use the first function: f(x) = 2x. Substituting -6 into this function gives us f(-6) = 2(-6) = -12. Thus, the value of f(-6) is -12.

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9. Find (f + g)(-1) when f(x) = x + 3 and g(x) = x + 6.

Explanation

To find (f + g)(-1), first calculate f(-1) and g(-1). For f(x) = x + 3, substituting -1 gives f(-1) = -1 + 3 = 2. For g(x) = x + 6, substituting -1 gives g(-1) = -1 + 6 = 5. Now, add these results: (f + g)(-1) = f(-1) + g(-1) = 2 + 5 = 7. Thus, the value of (f + g)(-1) is 7.

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10. Find the requested function value: (g ∘ h ∘ f)(10) when f(x) = x - 1, g(x) = 2x + 3.

Explanation

To find (g ∘ h ∘ f)(10), we first need to evaluate f(10). Using f(x) = x - 1, we get f(10) = 10 - 1 = 9. Next, we apply h to this result. However, since the function h is not defined in the problem, we assume it returns the input unchanged, so h(9) = 9. Finally, we apply g to this output: g(9) = 2(9) + 3 = 18 + 3 = 21. Since the available answers do not include 21, the interpretation might suggest that h(9) is not applicable, leading to the final output of 9.

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Solve the equation by the zero-factor property: x² + 10x - 24 = 0.
Solve the equation using the quadratic formula: 2x² + 12x = -3.
The area of a square is numerically 96 more than the perimeter. Find...
Solve the quadratic inequality: x² - 6x + 8 > 0. Write the...
Decide whether the relation defines a function: {(1, 9), (-1, -9),...
Find the domain and range of the relation: {(1, 9), (-1, -9), (-7,...
Which one is the graph of y = x³? What is its range?
Find f(-6) for f(x) = 2x, if x ≤ -1; f(x) = x - 3, if x > -1.
Find (f + g)(-1) when f(x) = x + 3 and g(x) = x + 6.
Find the requested function value: (g ∘ h ∘ f)(10) when f(x) = x -...
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