Solving Quadratics by Factoring - Accuplacer Aaf

1. The length of a rectangular garden is 5 meters greater than its width. If the area of the garden is 36 square meters, what is the width of the garden in meters?
   Answer: 4 Let $w$ be the width of the garden. The length is $w + 5$. The area is length times width, so $w(w + 5) = 36$. This gives the quadratic equation $w^2 + 5w - 36 = 0$, which factors to $(w - 4)(w + 9) = 0$. The possible solutions are $w = 4$ and $w = -9$. Since width must be a positive value, the width is 4 meters. Distractor D is the length of the garden.
2. What is the product of the solutions to the equation $4x^2 = 81$?
   Answer: $-\frac{81}{4}$ First, set the equation to standard form: $4x^2 - 81 = 0$. This is a difference of squares, which factors to $(2x - 9)(2x + 9) = 0$. The solutions are $x = \frac{9}{2}$ and $x = -\frac{9}{2}$. The product of these solutions is $\left(\frac{9}{2}\right) \times \left(-\frac{9}{2}\right) = -\frac{81}{4}$. Alternatively, for a quadratic $ax^2+bx+c=0$, the product of the roots is $c/a$. Here, $a=4, b=0, c=-81$, so the product is $-81/4$.
3. What is the complete set of solutions for the equation $5x^2 = 10x$?
   Answer: $\{0, 2\}$ To solve, first set the equation to zero: $5x^2 - 10x = 0$. It is incorrect to divide by $x$ as this would lose a solution. Instead, factor out the greatest common factor, $5x$, to get $5x(x - 2) = 0$. The solutions are found by setting each factor to zero: $5x = 0 \implies x=0$ and $x-2 = 0 \implies x=2$. The complete solution set is $\{0, 2\}$.
4. If $x^2 - 8x + 16 = 25$, what is the product of all solutions?
   Answer: -9 Rearrange to standard form: $x^2 - 8x + 16 - 25 = 0$, which simplifies to $x^2 - 8x - 9 = 0$. Factor: $(x - 9)(x + 1) = 0$. Solutions are $x = 9$ and $x = -1$. Their product is $9 \times (-1) = -9$. Choice B represents the absolute value of the product. Choice C comes from adding instead of multiplying the solutions. Choice D is the constant term from the original equation before rearranging.
5. What are the solutions to the equation $x - 2 = \frac{15}{x}$?
   Answer: $x=5$ and $x=-3$ To clear the fraction, multiply both sides of the equation by $x$, noting that $x \neq 0$. This gives $x(x - 2) = 15$, or $x^2 - 2x = 15$. Rearranging into standard quadratic form yields $x^2 - 2x - 15 = 0$. Factoring the quadratic gives $(x-5)(x+3)=0$. The solutions are $x=5$ and $x=-3$. Since neither solution is the restricted value 0, both are valid.
6. The quadratic equation $x^2 - 6x + k = 0$ has two distinct positive integer solutions. What is the largest possible value of $k$?
   Answer: 8 For distinct positive integer solutions $p$ and $q$, we need $p + q = 6$ (sum of roots) and $pq = k$ (product of roots). The pairs of distinct positive integers that sum to 6 are: (1,5) and (2,4). Their products are: $1 \times 5 = 5$ and $2 \times 4 = 8$. The pair (3,3) would give $k = 9$, but this results in a repeated root, not two distinct solutions. Therefore, the largest possible value of $k$ for distinct solutions is 8.
7. What is the distance between the two x-intercepts of the parabola defined by the equation $y = 2x^2 - 8x + 6$?
   Answer: 2 The x-intercepts occur when $y = 0$. So we must solve the equation $2x^2 - 8x + 6 = 0$. First, factor out the common factor of 2: $2(x^2 - 4x + 3) = 0$. Factoring the trinomial gives $2(x-1)(x-3) = 0$. The x-intercepts are at $x=1$ and $x=3$. The distance between these two points on the x-axis is the absolute difference of their coordinates: $|3 - 1| = 2$.
8. What is the largest real solution to the equation $x^4 - 13x^2 + 36 = 0$?
   Answer: 3 This equation is quadratic in form. Let $u = x^2$. The equation becomes $u^2 - 13u + 36 = 0$. This factors to $(u - 9)(u - 4) = 0$, so the solutions for $u$ are $u = 9$ and $u = 4$. Substitute back $x^2$ for $u$: $x^2 = 9$ or $x^2 = 4$. This gives four solutions for $x$: $x = \pm 3$ and $x = \pm 2$. The largest of these solutions is 3.
9. A rectangle has dimensions where the length is 3 units more than the width. If the area is 28 square units, what is the width of the rectangle?
   Answer: 4 units When you encounter word problems involving rectangles with related dimensions, you're dealing with quadratic equations. The key is translating the relationship between dimensions into algebraic expressions, then using the area formula to solve. Let's define the width as $w$ units. Since the length is 3 units more than the width, the length is $(w + 3)$ units. The area of a rectangle equals length times width, so: $w(w + 3) = 28$ Expanding this equation: $w^2 + 3w = 28$ Rearranging to standard form: $w^2 + 3w - 28 = 0$ To factor this quadratic, we need two numbers that multiply to -28 and add to 3. Those numbers are 7 and -4: $(w + 7)(w - 4) = 0$ This gives us $w = -7$ or $w = 4$. Since width cannot be negative, $w = 4$ units. Let's verify: if width = 4, then length = 7, and area = $4 \times 7 = 28$ ✓ Now for the wrong answers: Choice A (8 units) would give a length of 11 and area of 88—far too large. Choice B (5 units) would create a length of 8 and area of 40—also too large. Choice C (7 units) would make the length 10 and area 70—again, much larger than needed. For rectangle problems with related dimensions, always set up your variable for one dimension, express the other in terms of that variable, then use the area formula to create your equation. Remember to check that your solution makes physical sense!
10. If $a$ is a constant, which of the following is a solution for $x$ in the equation $x^2 - (a+3)x + 3a = 0$?
    Answer: $x = 3$ This is a quadratic equation in the variable $x$. We need to find two expressions that multiply to $3a$ and add to $-(a+3)$. These are $-a$ and $-3$. Thus, the equation factors to $(x - a)(x - 3) = 0$. The solutions are found by setting each factor to zero: $x - a = 0 \implies x = a$ and $x - 3 = 0 \implies x = 3$. Of the choices provided, $x = 3$ is a solution.
11. For what value of $k$ does the equation $x^2 + 8x + k = 0$ have exactly one solution when solved by factoring?
    Answer: 16 For exactly one solution, the quadratic must be a perfect square trinomial. With $x^2 + 8x + k$, we need $k = (8/2)^2 = 16$, giving us $(x + 4)^2 = 0$ with solution $x = -4$. Choice A comes from using half the coefficient incorrectly. Choice B is the middle coefficient. Choice D comes from squaring 8 instead of 4.
12. What is the positive solution to the equation $3x^2 - 11x = 4$?
    Answer: 4 To solve the quadratic equation, first set it to zero: $3x^2 - 11x - 4 = 0$. This can be factored as $(3x + 1)(x - 4) = 0$. Setting each factor to zero gives the solutions $3x + 1 = 0 \implies x = -\frac{1}{3}$ and $x - 4 = 0 \implies x = 4$. The question asks for the positive solution, which is 4.
13. Which of the following are the solutions to the equation $3x^2 + 15x - 42 = 0$?
    Answer: $x = -7$ and $x = 2$ First, factor out the greatest common factor, 3: $3(x^2 + 5x - 14) = 0$. Then, factor the quadratic expression inside the parentheses: $3(x + 7)(x - 2) = 0$. The solutions are found by setting each factor containing a variable to zero: $x + 7 = 0 \implies x = -7$ and $x - 2 = 0 \implies x = 2$. Distractor A has incorrect signs for both solutions. Distractors C and D result from incorrect factoring attempts.
14. If $x^2 + bx + 36 = 0$ has solutions that differ by 5, what is the value of $b$?
    Answer: $\pm 13$ When you see a quadratic equation where you know information about the relationship between the solutions, you can use Vieta's formulas and properties of quadratic roots to find unknown coefficients. For the quadratic $x^2 + bx + 36 = 0$, let's call the two solutions $r$ and $s$. From Vieta's formulas, we know that $r + s = -b$ and $rs = 36$. We're told that the solutions differ by 5, so $|r - s| = 5$. To find $b$, we can use the relationship $(r - s)^2 = (r + s)^2 - 4rs$. Substituting what we know: $5^2 = (-b)^2 - 4(36)$, which gives us $25 = b^2 - 144$. Solving for $b^2$: $b^2 = 169$, so $b = \pm 13$. Looking at the wrong answers: Choice A ($\pm 17$) would result from incorrectly adding 144 instead of subtracting it, giving $b^2 = 25 + 144 = 169$, but then taking the square root incorrectly. Choice B ($\pm 11$) might come from computational errors when working with the discriminant. Choice C ($\pm 15$) could result from confusing the difference between roots (5) with the coefficient itself, or from arithmetic mistakes in the calculation. Remember this pattern: when given information about the difference between quadratic roots, use the identity $(r - s)^2 = (r + s)^2 - 4rs$ combined with Vieta's formulas. This approach efficiently connects the coefficient relationships to the geometric properties of the solutions.
15. The equation $4x^2 - 12x + 9 = 0$ has exactly one real solution. What is the value of this solution?
    Answer: $\frac{3}{2}$ The expression $4x^2 - 12x + 9$ is a perfect square trinomial, which factors to $(2x - 3)^2 = 0$. Setting the factor to zero gives $2x - 3 = 0$. Solving for $x$ yields $2x = 3$, so $x = \frac{3}{2}$. This is a 'double root,' meaning there is only one distinct solution.
16. The product of two consecutive positive odd integers is 143. What is the value of the larger integer?
    Answer: 13 Let the two consecutive positive odd integers be $x$ and $x + 2$. Their product is $x(x + 2) = 143$. This gives the equation $x^2 + 2x - 143 = 0$, which factors to $(x + 13)(x - 11) = 0$. The solutions for $x$ are $x = -13$ and $x = 11$. Since the integers must be positive, the smaller integer is 11. The larger integer is $11 + 2 = 13$.
17. What is the product of the solutions for the equation $x^2 + \frac{1}{6}x - \frac{1}{3} = 0$?
    Answer: $-\frac{1}{3}$ For any quadratic equation in the form $ax^2 + bx + c = 0$, the product of the solutions is $\frac{c}{a}$. In this equation, $a = 1$, $b = \frac{1}{6}$, and $c = -\frac{1}{3}$. Therefore, the product of the solutions is $\frac{-1/3}{1} = -\frac{1}{3}$. Alternatively, multiplying the equation by 6 to clear fractions gives $6x^2 + x - 2 = 0$, which factors to $(2x-1)(3x+2)=0$. The solutions are $x=1/2$ and $x=-2/3$, and their product is $(1/2)(-2/3) = -1/3$.
18. What is the positive solution to the equation $(x-5)^2 + 3(x-5) - 10 = 0$?
    Answer: 7 This equation is quadratic in form. Let $u = x-5$. The equation becomes $u^2 + 3u - 10 = 0$. Factoring this gives $(u+5)(u-2)=0$, so $u=-5$ or $u=2$. Substitute back to solve for $x$. Case 1: $x-5=-5$, which gives $x=0$. Case 2: $x-5=2$, which gives $x=7$. The positive solution is 7.
19. A quadratic equation with integer coefficients has solutions at $x = -4$ and $x = \frac{5}{2}$. Which of the following could be the equation?
    Answer: $2x^2 + 3x - 20 = 0$ If the solutions are $x = -4$ and $x = \frac{5}{2}$, then the corresponding factors are $(x - (-4)) = (x + 4)$ and $(x - \frac{5}{2})$. To obtain integer coefficients, we can use the factor $(2x - 5)$ which also gives the root $x = \frac{5}{2}$. The equation is formed by the product of the factors: $(x + 4)(2x - 5) = 0$. Expanding this gives $2x^2 - 5x + 8x - 20 = 0$, which simplifies to $2x^2 + 3x - 20 = 0$.
20. Which of the following quadratic equations has solutions $x = \frac{2}{3}$ and $x = -\frac{1}{4}$?
    Answer: $12x^2 - 5x - 2 = 0$ Working backwards from the solutions, the factored form is $(x - \frac{2}{3})(x + \frac{1}{4}) = 0$. Expanding: $x^2 + \frac{1}{4}x - \frac{2}{3}x - \frac{2}{12} = 0$, which simplifies to $x^2 - \frac{5}{12}x - \frac{1}{6} = 0$. Multiplying by 12: $12x^2 - 5x - 2 = 0$. Choice A has the wrong sign on the middle term. Choices C and D have incorrect signs that would yield different solutions.

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