Solving Quadratics by Factoring - Math 2

1. A rectangular garden has dimensions where the length is 3 feet more than the width. If the area of the garden is 88 square feet, what are the possible dimensions of the garden?
   Answer: Width = 8 feet, Length = 11 feet Let w = width, then length = w + 3. Area equation: w(w + 3) = 88, which gives w² + 3w - 88 = 0. Factoring: (w + 11)(w - 8) = 0, so w = -11 or w = 8. Since width must be positive, w = 8 feet and length = 11 feet. Choice B switches the dimensions incorrectly. Choice C includes negative dimensions which are not physically meaningful. Choice D uses wrong factorization.
2. The solutions to $x^2 - 6x + 5 = 0$ are $x = a$ and $x = b$ where $a < b$. What is the value of $a^2 + b^2$?
   Answer: $26$ Factoring: $x^2 - 6x + 5 = (x - 1)(x - 5) = 0$, so $x = 1$ or $x = 5$. Since $a < b$, we have $a = 1$ and $b = 5$. Therefore $a^2 + b^2 = 1^2 + 5^2 = 1 + 25 = 26$. Choice B is $(a + b)^2 = 6^2 = 36$. Choice C is $5 \times 6 = 30$. Choice D is $a + b + ab = 1 + 5 + 5 = 11$.
3. If $(2x - 3)(x + k) = 2x^2 + 5x - 12$, what is the value of $k$?
   Answer: $k = 4$ Expanding the left side: $(2x - 3)(x + k) = 2x^2 + 2kx - 3x - 3k = 2x^2 + (2k - 3)x - 3k$. Comparing with $2x^2 + 5x - 12$: the coefficient of $x$ gives $2k - 3 = 5$, so $k = 4$. We can verify with the constant term: $-3k = -3(4) = -12$ ✓. Choice A gives coefficient $2(-4) - 3 = -11$. Choice C gives $2(\frac{8}{3}) - 3 = \frac{7}{3}$. Choice D gives $2(-\frac{8}{3}) - 3 = -\frac{25}{3}$.
4. The quadratic equation $x^2 + kx + k = 0$ has two distinct real solutions. If both solutions are negative, what is the range of possible values for $k$?
   Answer: $k > 4$ only When analyzing quadratic equations with constraints on their solutions, you need to consider three key conditions: the discriminant (for real solutions), the sum of roots, and the product of roots. For the equation $x^2 + kx + k = 0$ to have two distinct real solutions, the discriminant must be positive: $k^2 - 4(1)(k) > 0$, which simplifies to $k^2 - 4k > 0$ or $k(k-4) > 0$. This means $k < 0$ or $k > 4$. For both solutions to be negative, you need two additional conditions. Using Vieta's formulas: the sum of roots is $-k$ and the product of roots is $k$. If both roots are negative, their sum must be negative (so $-k < 0$, meaning $k > 0$) and their product must be positive (so $k > 0$). Combining all three constraints: $k > 4$ (from the discriminant condition) AND $k > 0$ (from both root conditions). The intersection is simply $k > 4$. Choice A ($k > 0$ and $k \neq 4$) includes values like $k = 1$, which gives complex solutions, not two distinct real ones. Choice B ($0 < k < 4$) fails the discriminant test entirely. Choice C ($k > 4$ or $k < 0$) includes negative values of $k$, which would make at least one root positive since the sum of roots would be positive. Remember: when quadratics have constraints on their roots, always check the discriminant first, then apply Vieta's formulas to test the sign conditions systematically.
5. For what value of $m$ does the equation $x^2 + mx + 16 = 0$ have exactly one solution?
   Answer: $m = \pm 8$ A quadratic has exactly one solution when it's a perfect square trinomial. For $x^2 + mx + 16 = 0$ to be a perfect square, we need $16 = (\frac{m}{2})^2$, so $\frac{m^2}{4} = 16$, giving $m^2 = 64$, thus $m = \pm 8$. When $m = 8$: $x^2 + 8x + 16 = (x + 4)^2 = 0$. When $m = -8$: $x^2 - 8x + 16 = (x - 4)^2 = 0$. Both give exactly one solution.
6. The equation $3x^2 - 12x - 15 = 0$ is equivalent to which of the following factored forms?
   Answer: $3(x - 5)(x + 1) = 0$ First factor out the GCF: $3x^2 - 12x - 15 = 3(x^2 - 4x - 5)$. Then factor the quadratic: $x^2 - 4x - 5 = (x - 5)(x + 1)$. So the complete factorization is $3(x - 5)(x + 1) = 0$. Choice B gives $3(x^2 - 2x - 3) = 3x^2 - 6x - 9$. Choice C gives $3x^2 - 12x - 15$ but isn't fully factored. Choice D gives $3(x^2 + 4x - 5) = 3x^2 + 12x - 15$.
7. The expression $4x^2 - 12x + k$ can be factored as $4(x - a)^2$ for some values of $k$ and $a$. What is the value of $k + a$?
   Answer: $k + a = \frac{21}{2}$ If 4x² - 12x + k = 4(x - a)², then expanding the right side: 4(x - a)² = 4(x² - 2ax + a²) = 4x² - 8ax + 4a². Comparing coefficients: -8a = -12, so a = 3/2; and 4a² = k, so k = 4(3/2)² = 4(9/4) = 9. Therefore k + a = 9 + 3/2 = 18/2 + 3/2 = 21/2. Choice A would result from miscalculating a or k. Choice C comes from assuming a = 3 (forgetting the factor of 2). Choice D comes from adding incorrectly or using wrong values.
8. The quadratic equation $3x^2 - kx + 12 = 0$ can be factored as $(3x + p)(x + q) = 0$ where $p$ and $q$ are integers. How many different integer values of $k$ make this possible?
   Answer: 8 different values of $k$ Expanding (3x + p)(x + q) = 3x² + (3q + p)x + pq. Comparing with 3x² - kx + 12: 3q + p = -k and pq = 12. So k = -(3q + p). We need integer factor pairs (p,q) where pq = 12. The pairs are: (1,12), (2,6), (3,4), (4,3), (6,2), (12,1), (-1,-12), (-2,-6), (-3,-4), (-4,-3), (-6,-2), (-12,-1). For each pair, k = -(3q + p): (1,12): k = -(36+1) = -37; (2,6): k = -(18+2) = -20; (3,4): k = -(12+3) = -15; (4,3): k = -(9+4) = -13; (6,2): k = -(6+6) = -12; (12,1): k = -(3+12) = -15; (-1,-12): k = -(-36-1) = 37; (-2,-6): k = -(-18-2) = 20; (-3,-4): k = -(-12-3) = 15; (-4,-3): k = -(-9-4) = 13; (-6,-2): k = -(-6-6) = 12; (-12,-1): k = -(-3-12) = 15. The distinct values are: -37, -20, -15, -13, -12, 37, 20, 15, 13, 12. That's 10 values, but some repeat. Actually: -37, -20, -15, -13, -12, 12, 13, 15, 20, 37. Removing duplicates: 8 distinct values.
9. If $p$ and $q$ are the solutions to $x^2 - 7x + 12 = 0$, what is the value of $\frac{1}{p} + \frac{1}{q}$?
   Answer: $\frac{7}{12}$ When you encounter a quadratic equation and need to find an expression involving the roots, think about Vieta's formulas—they provide elegant shortcuts without requiring you to solve for the individual roots. For the quadratic $x^2 - 7x + 12 = 0$, Vieta's formulas tell us that if $p$ and $q$ are the roots, then $p + q = 7$ (the coefficient of $x$ with opposite sign) and $pq = 12$ (the constant term). To find $\frac{1}{p} + \frac{1}{q}$, we need a common denominator: $\frac{1}{p} + \frac{1}{q} = \frac{q + p}{pq}$. Substituting our values: $\frac{7}{12}$. Let's examine why the other answers are incorrect. Choice A gives $\frac{1}{7}$, which you'd get if you mistakenly used $\frac{1}{p + q}$ instead of $\frac{p + q}{pq}$. Choice B gives $\frac{12}{7}$, which results from incorrectly flipping the fraction to $\frac{pq}{p + q}$. Choice C gives $\frac{1}{12}$, which you'd get if you used $\frac{1}{pq}$ instead of the correct expression. The correct answer is D: $\frac{7}{12}$. Study tip: Memorize that for any quadratic $ax^2 + bx + c = 0$ with roots $r$ and $s$, we have $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$. This saves time and reduces computational errors compared to solving for individual roots.
10. The equation $4x^2 - 4x - 3 = 0$ can be solved by factoring. Which of the following shows the correct factorization?
    Answer: $(2x - 3)(2x + 1) = 0$ To factor $4x^2 - 4x - 3$, we need two terms whose product is $4x^2 \cdot (-3) = -12x^2$ and whose sum is $-4x$. These are $-6x$ and $2x$. So $4x^2 - 4x - 3 = 4x^2 - 6x + 2x - 3 = 2x(2x - 3) + 1(2x - 3) = (2x - 3)(2x + 1)$. We can verify: $(2x - 3)(2x + 1) = 4x^2 + 2x - 6x - 3 = 4x^2 - 4x - 3$ ✓. Choice A gives $4x^2 - 4x + 3x - 3 = 4x^2 - x - 3$. Choice C gives $4x^2 + 4x - 3x - 3 = 4x^2 + x - 3$. Choice D gives $4x^2 - 2x + 6x - 3 = 4x^2 + 4x - 3$.
11. A manufacturer determines that the profit $P$ (in thousands of dollars) from producing $x$ thousand units is given by $P = -2x^2 + 16x - 30$. For what production levels will the company break even (profit equals zero)?
    Answer: $x = 3$ thousand units and $x = 5$ thousand units This problem tests your ability to find the zeros of a quadratic function, which represents break-even points in a business context. When a company "breaks even," its profit equals zero, so you need to solve $P = 0$. Set the profit equation equal to zero: $-2x^2 + 16x - 30 = 0$. You can solve this by factoring. First, factor out the common factor of -2: $-2(x^2 - 8x + 15) = 0$. This simplifies to $x^2 - 8x + 15 = 0$. To factor $x^2 - 8x + 15$, you need two numbers that multiply to 15 and add to -8. Those numbers are -3 and -5, giving you $(x - 3)(x - 5) = 0$. Therefore, $x = 3$ or $x = 5$. The company breaks even at 3 thousand units and 5 thousand units. Choice A ($x = 5$ and $x = 6$) includes one correct value but substitutes 6 for the actual solution of 3. Choice B ($x = 2$ and $x = 6$) contains neither correct solution and likely results from factoring errors. Choice C ($x = 3$ and $x = 10$) includes one correct value but 10 is far from any reasonable solution to this quadratic. When solving quadratic break-even problems, always verify your solutions by substituting back into the original equation. Both $x = 3$ and $x = 5$ should yield $P = 0$. This double-check catches algebraic mistakes and confirms you've found the true break-even points.
12. If $x^2 - 6x + 8 = 0$ and $x^2 - 6x + k = 5$, what is the value of $k$?
    Answer: $k = 13$ From the first equation: x² - 6x + 8 = 0, so x² - 6x = -8. Substituting into the second equation: x² - 6x + k = 5 becomes -8 + k = 5, so k = 13. We can verify: the first equation factors as (x-2)(x-4) = 0, giving x = 2 or x = 4. For x = 2: 4 - 12 + k = 5, so k = 13. For x = 4: 16 - 24 + k = 5, so k = 13. Choice A would come from incorrectly computing 5 - (-8) = 5 + 8 = 13, then making an error. Choice B assumes k equals the constant term from the first equation. Choice D comes from computing 5 + 8 + 5 = 18, misunderstanding the relationship.
13. Which equation has solutions $x = -\frac{2}{3}$ and $x = \frac{1}{4}$?
    Answer: $12x^2 + 5x - 2 = 0$ When you're given the roots of a quadratic equation and need to find the equation itself, you can work backwards using the relationship between roots and coefficients. If a quadratic has roots $r_1$ and $r_2$, it can be written as $a(x - r_1)(x - r_2) = 0$. Let's substitute our given roots: $x = -\frac{2}{3}$ and $x = \frac{1}{4}$. Starting with $a(x - (-\frac{2}{3}))(x - \frac{1}{4}) = 0$, we get $a(x + \frac{2}{3})(x - \frac{1}{4}) = 0$. Expanding: $(x + \frac{2}{3})(x - \frac{1}{4}) = x^2 - \frac{1}{4}x + \frac{2}{3}x - \frac{2}{12} = x^2 + \frac{5}{12}x - \frac{1}{6}$ To match the answer choices (which have integer coefficients), multiply by 12: $12x^2 + 5x - 2 = 0$. This is choice C. Let's verify by checking why the other options fail: Choice A ($12x^2 - 5x - 2 = 0$) has the wrong sign on the middle term. Using the quadratic formula would give different roots. Choice B ($12x^2 - 5x + 2 = 0$) has both wrong signs for the middle and constant terms, leading to entirely different solutions. Choice D ($12x^2 + 5x + 2 = 0$) has the wrong sign on the constant term. The discriminant $25 - 96 = -71$ is negative, meaning no real solutions exist. Study tip: When working backwards from roots to equations, remember that the sum of roots equals $-\frac{b}{a}$ and the product equals $\frac{c}{a}$. This can help you quickly check your work or eliminate wrong answers.
14. If $x^2 - 8x + k = 0$ has solutions that differ by 6, what is the value of $k$?
    Answer: $k = 7$ If the solutions are $r$ and $s$ with $r - s = 6$, then by Vieta's formulas: $r + s = 8$ and $rs = k$. Solving the system: $r = 7, s = 1$. Therefore $k = rs = 7(1) = 7$. We can verify: $x^2 - 8x + 7 = (x-7)(x-1) = 0$, giving solutions $x = 7$ and $x = 1$, which differ by 6.
15. Consider the equation $(x-1)^2 + (x-1) - 6 = 0$. If we let $u = x-1$, then after solving for $u$ and substituting back, what are the solutions for $x$?
    Answer: $x = -2$ and $x = 3$ When you encounter an equation with repeated expressions like $(x-1)^2 + (x-1) - 6 = 0$, substitution is your best friend. This technique transforms a complex equation into a simpler quadratic form. Let's substitute $u = x-1$, which converts our equation to $u^2 + u - 6 = 0$. This is now a standard quadratic that we can factor. We need two numbers that multiply to $-6$ and add to $1$. Those numbers are $3$ and $-2$, so we get $(u+3)(u-2) = 0$. This gives us $u = -3$ or $u = 2$. Now we substitute back: since $u = x-1$, we have $x-1 = -3$ or $x-1 = 2$. Solving these gives us $x = -2$ or $x = 3$. Let's check why the other answers are wrong. Choice A ($x = 1, 5$) would come from incorrectly solving $u = 2$ as $x = 1+2 = 3$ and making an error with the other root. Choice B ($x = -1, 4$) suggests errors in the back-substitution step, possibly confusing signs. Choice C ($x = 0, 2$) might result from factoring the quadratic incorrectly or making arithmetic mistakes in the substitution process. Strategy tip: Always verify your substitution method by checking one solution in the original equation. For example, when $x = -2$: $(-2-1)^2 + (-2-1) - 6 = 9 - 3 - 6 = 0$ ✓. This confirms you've handled the substitution correctly throughout the entire process.

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