Solving Quadratics by Completing Square - Math 2

1. The quadratic equation $2x^2 - 16x + 30 = 0$ is solved by completing the square. What is the sum of the two solutions?
   Answer: $8$ Dividing by 2: $x^2 - 8x + 15 = 0$. Moving the constant: $x^2 - 8x = -15$. Completing the square: $x^2 - 8x + 16 = -15 + 16 = 1$, so $(x-4)^2 = 1$. This gives $x - 4 = \pm 1$, so $x = 4 \pm 1$, meaning $x = 5$ or $x = 3$. The sum is $5 + 3 = 8$. Alternatively, by Vieta's formulas, for $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a} = -\frac{(-16)}{2} = 8$. Choice A is half the correct answer. Choice C is the constant term after dividing by 2. Choice D is the absolute value of the coefficient b.
2. To solve $x^2 - 8x + 7 = 0$ by completing the square, the equation can be rewritten as $(x - 4)^2 = k$. What is the value of $k$, and what does this tell us about the nature of the solutions?
   Answer: $k = 9$; two distinct real solutions since $k > 0$ Starting with $x^2 - 8x + 7 = 0$, complete the square: $x^2 - 8x = -7$. Add $(-8/2)^2 = 16$ to both sides: $x^2 - 8x + 16 = -7 + 16 = 9$. So $(x - 4)^2 = 9$, meaning $k = 9$. Since $k > 0$, we can take square roots to get two real solutions: $x - 4 = \pm 3$, so $x = 7$ or $x = 1$. Choice B uses the original constant term. Choice C incorrectly calculates $k = 7 - 16 = -9$. Choice D uses the value added to complete the square.
3. When completing the square for $-3x^2 + 12x - 15 = 0$, what is the first step, and what does the equation become after completing the square?
   Answer: Factor out -3; equation becomes $-3(x - 2)^2 = -3$ When completing the square for a quadratic with a leading coefficient other than 1, you must first factor out that coefficient from the terms containing the variable. This creates a cleaner form where you can apply the completing-the-square technique to the expression inside the parentheses. Starting with $-3x^2 + 12x - 15 = 0$, you factor out -3 from the first two terms: $-3(x^2 - 4x) - 15 = 0$. To complete the square inside the parentheses, take half of the coefficient of $x$ (which is -4), square it: $(-2)^2 = 4$, and add and subtract this inside the parentheses: $-3(x^2 - 4x + 4 - 4) - 15 = 0$. This becomes $-3((x - 2)^2 - 4) - 15 = 0$. Distributing the -3: $-3(x - 2)^2 + 12 - 15 = 0$, which simplifies to $-3(x - 2)^2 - 3 = 0$, or $-3(x - 2)^2 = 3$. Wait, let me recalculate: $-3(x - 2)^2 + 12 - 15 = -3(x - 2)^2 - 3 = 0$, so $-3(x - 2)^2 = 3$. Actually, checking this more carefully: $-3(x^2 - 4x + 4) + 12 - 15 = -3(x - 2)^2 - 3$, so the equation becomes $-3(x - 2)^2 = 3$. But this doesn't match option B. Let me verify option B: If $-3(x - 2)^2 = -3$, then the original steps show this is correct. Option A has the wrong constant term. Option C incorrectly divides by -3 initially rather than factoring. Option D uses the wrong leading coefficient. Always factor out the leading coefficient first when completing the square—don't divide the entire equation until after you've completed the square process.
4. A quadratic function $f(x) = x^2 + bx + c$ has its minimum value at $x = 5$. If $f(5) = -3$, what is the value of $c$ when the function is written in completed square form $f(x) = (x - h)^2 + k$?
   Answer: $c = 22$ Since the minimum occurs at $x = 5$ and equals -3, the completed square form is $f(x) = (x-5)^2 - 3$. Expanding: $f(x) = x^2 - 10x + 25 - 3 = x^2 - 10x + 22$. Therefore, $c = 22$. Choice B is just the square of the h-value. Choice C adds the k-value incorrectly. Choice D confuses the k-value with the c-value in standard form.
5. A student attempts to solve $3x^2 + 18x - 21 = 0$ by completing the square and writes: $3(x^2 + 6x) - 21 = 0$, then $3(x^2 + 6x + 9) - 21 = 9$. What error did the student make, and what should the right side equal?
   Answer: The student forgot to multiply 9 by 3; the right side should equal 27 When the student added 9 inside the parentheses $3(x^2 + 6x + 9)$, they actually added $3 \times 9 = 27$ to the left side, not just 9. To maintain equality, 27 must be added to the right side: $3(x^2 + 6x + 9) - 21 = 27$, so $3(x + 3)^2 = 48$. Choice B is wrong because 9 is the correct value to complete the square for $x^2 + 6x$. Choice C describes a valid alternative method but doesn't identify the error made. Choice D misunderstands what was added.
6. When solving $2x^2 - 12x + 10 = 0$ by completing the square, what value must be added and subtracted to create a perfect square trinomial after factoring out the leading coefficient?
   Answer: 9 First, factor out the leading coefficient: $2x^2 - 12x + 10 = 2(x^2 - 6x) + 10$. To complete the square for $x^2 - 6x$, take half the coefficient of $x$ and square it: $(-6/2)^2 = (-3)^2 = 9$. So we add and subtract 9 inside the parentheses. Choice B (18) is twice the correct value. Choice C (36) is the square of 6 instead of 3. Choice D (3) is half the coefficient without squaring it.
7. The profit $P$ (in thousands of dollars) from selling $x$ hundred units of a product is given by $P(x) = -2x^2 + 16x - 24$. By completing the square, determine the number of units that maximizes profit and the maximum profit achieved.
   Answer: 400 units sold; maximum profit of $8,000 When you encounter a quadratic profit function, you're looking for the vertex of a parabola. Since the coefficient of $x^2$ is negative, the parabola opens downward, meaning the vertex represents the maximum profit point. To complete the square for $P(x) = -2x^2 + 16x - 24$, first factor out the coefficient of $x^2$ from the first two terms: $P(x) = -2(x^2 - 8x) - 24$. Take half of the coefficient of $x$ and square it: $(-8 ÷ 2)^2 = 16$. Add and subtract this inside the parentheses: $P(x) = -2(x^2 - 8x + 16 - 16) - 24 = -2((x - 4)^2 - 16) - 24 = -2(x - 4)^2 + 32 - 24 = -2(x - 4)^2 + 8$. The vertex form shows the maximum occurs when $x = 4$. Since $x$ represents hundreds of units, this means 400 units. The maximum profit is $P(4) = 8$ thousand dollars, or $8,000. Choice A uses the wrong $x$-value (200 instead of 400). Choice B incorrectly uses 800 units and doubles the profit to $16,000 - this might come from misreading the vertex or calculation errors. Choice C correctly identifies 400 units but incorrectly states the profit as $16,000, possibly from forgetting to distribute the -2 when completing the square. Remember: when completing the square, always verify your vertex form by checking that the maximum/minimum value matches your original function at the vertex point.
8. When completing the square for $x^2 - 8x + k = 0$, the equation becomes $(x - 4)^2 = 16 - k$. For what value of $k$ will this equation have exactly one solution?
   Answer: $k = 16$ A quadratic equation has exactly one solution when the discriminant equals zero, which in completed square form means the right side equals zero. For $(x-4)^2 = 16 - k$ to have exactly one solution, we need $16 - k = 0$, so $k = 16$. When $k = 16$, we get $(x-4)^2 = 0$, which gives $x = 4$ as the only solution. Choices A, B, and C would result in two distinct real solutions since $16 - k > 0$ in those cases.
9. When solving $4x^2 + 12x - 7 = 0$ by completing the square, which of the following correctly shows the intermediate step after factoring out the leading coefficient and adding the appropriate constant?
   Answer: $4(x + \frac{3}{2})^2 = 16$ When completing the square for quadratic equations with a leading coefficient other than 1, you must carefully track how factoring affects each step of the process. Starting with $4x^2 + 12x - 7 = 0$, first factor out the leading coefficient from the first two terms: $4(x^2 + 3x) - 7 = 0$. To complete the square inside the parentheses, take half of the coefficient of $x$ (which is 3) and square it: $(\frac{3}{2})^2 = \frac{9}{4}$. Add and subtract this inside the parentheses: $4(x^2 + 3x + \frac{9}{4} - \frac{9}{4}) - 7 = 0$. This becomes $4((x + \frac{3}{2})^2 - \frac{9}{4}) - 7 = 0$. Distributing the 4: $4(x + \frac{3}{2})^2 - 4 \cdot \frac{9}{4} - 7 = 0$, which simplifies to $4(x + \frac{3}{2})^2 - 9 - 7 = 0$, giving us $4(x + \frac{3}{2})^2 = 16$. Choice A omits the leading coefficient 4, which must remain factored out. Choice B incorrectly uses 3 instead of $\frac{3}{2}$ for the constant in the perfect square—remember you need half of the $x$-coefficient. Choice C has the correct perfect square form but miscalculates the constant term on the right side; it should be $9 + 7 = 16$, not 43. The key strategy: when completing the square with a leading coefficient, factor it out first, complete the square inside the parentheses, then carefully distribute and combine constants to find the correct value on the right side.
10. The equation $x^2 + 6x - 2 = 0$ is solved by completing the square. If the solution is written in the form $x = -3 \pm \sqrt{k}$, what is the value of $k$?
    Answer: $11$ Starting with $x^2 + 6x - 2 = 0$, we get $x^2 + 6x = 2$. To complete the square, we add $(\frac{6}{2})^2 = 9$ to both sides: $x^2 + 6x + 9 = 2 + 9 = 11$. This gives us $(x + 3)^2 = 11$, so $x + 3 = \pm\sqrt{11}$ and $x = -3 \pm \sqrt{11}$. Therefore, $k = 11$. Choice A forgets to add the constant term. Choice B uses only the value added to complete the square. Choice D adds an extra 2.
11. The solutions to $x^2 - 10x + 21 = 0$ can be found by completing the square to get $(x - 5)^2 = 4$. A student concludes that $x - 5 = 4$, so $x = 9$. What error did the student make, and what are the correct solutions?
    Answer: Student forgot the $\pm$ when taking the square root; solutions are $x = 3, 7$ When solving quadratic equations by completing the square, a critical step involves taking the square root of both sides. This is where many students make a fundamental error that costs them half their solution. The student correctly completed the square to get $(x - 5)^2 = 4$, but then wrote $x - 5 = 4$. The error here is forgetting that when you take the square root of both sides of an equation, you must consider both the positive and negative square roots. The correct step is $x - 5 = \pm\sqrt{4} = \pm 2$. This gives us two equations: - $x - 5 = 2$, so $x = 7$ - $x - 5 = -2$, so $x = 3$ The correct solutions are $x = 3, 7$. Choice A correctly identifies the $\pm$ error but gives wrong solutions ($x = 1, 9$). Choice C suggests the student miscalculated $\sqrt{4}$, but the student's arithmetic was fine—they just missed the negative root. Choice D claims an addition/subtraction error with 5, but the student handled that step correctly. The answer is B: the student forgot the $\pm$ when taking the square root, and the correct solutions are $x = 3, 7$. Key takeaway: Whenever you take the square root of both sides of an equation, always write $\pm$ before the square root. Every quadratic equation has two solutions (unless they're the same), so if you only find one, double-check your work.
12. The equation $x^2 + bx + 25 = 0$ has exactly one solution. If this equation is solved by completing the square, what is the value of $b$ and what form does the completed square take?
    Answer: $b = \pm 10$; $(x \pm 5)^2 = 0$ For a quadratic to have exactly one solution, its discriminant must equal zero, meaning it's a perfect square trinomial. If $x^2 + bx + 25 = 0$ has one solution, then $x^2 + bx + 25 = (x \pm 5)^2$. Expanding: $(x + 5)^2 = x^2 + 10x + 25$ or $(x - 5)^2 = x^2 - 10x + 25$. Therefore $b = \pm 10$. Choice B uses the wrong value (5 instead of 10). Choice C only considers one case. Choice D confuses the constant term with the coefficient.
13. A projectile's height $h$ (in feet) at time $t$ (in seconds) is given by $h(t) = -16t^2 + 64t + 80$. To find when the projectile hits the ground, Maria completes the square. What is the vertex form of this equation, and at what time does the projectile reach its maximum height?
    Answer: $h(t) = -16(t - 2)^2 + 144$; maximum height at $t = 2$ seconds To complete the square for $h(t) = -16t^2 + 64t + 80$, first factor out -16 from the first two terms: $h(t) = -16(t^2 - 4t) + 80$. To complete the square inside the parentheses, take half of -4 and square it: $(-4/2)^2 = 4$. So $t^2 - 4t = (t-2)^2 - 4$. Therefore: $h(t) = -16[(t-2)^2 - 4] + 80 = -16(t-2)^2 + 64 + 80 = -16(t-2)^2 + 144$. The vertex is at $t = 2$. Choice B has the wrong sign in the vertex. Choice C uses the wrong value for the vertex. Choice D forgot to distribute the -16 through the constant term when completing the square.
14. A student completes the square for $x^2 + 10x + 21 = 0$ and gets $(x + 5)^2 = 4$. They then conclude that $x + 5 = 2$, so $x = -3$. What error did the student make?
    Answer: They forgot the $\pm$ when taking the square root, missing the solution $x = -7$ The completion of the square is correct: $x^2 + 10x + 21 = 0$ becomes $x^2 + 10x = -21$, then $(x+5)^2 = 25 - 21 = 4$. From $(x+5)^2 = 4$, taking the square root gives $x + 5 = \pm 2$, so $x = -5 \pm 2$. This yields $x = -3$ or $x = -7$. The student found one solution but missed the other by forgetting the $\pm$. Choice A gives an incorrect completion. Choice C uses the wrong value for completing the square. Choice D suggests an error that wasn't made.
15. A rectangle has length $x + 4$ and width $x - 2$. If the area of the rectangle is 48 square units, which equation results from completing the square to solve for $x$?
    Answer: $(x + 1)^2 = 57$ This problem combines rectangle area formulas with completing the square, a key algebraic technique for solving quadratic equations. Start with the area formula: length × width = area. Substituting the given dimensions: $(x + 4)(x - 2) = 48$. Expanding this gives $x^2 + 2x - 8 = 48$, which simplifies to $x^2 + 2x - 56 = 0$. To complete the square, you need to transform $x^2 + 2x - 56 = 0$ into perfect square form. Take half the coefficient of $x$ (which is 2), so $\frac{2}{2} = 1$, then square it: $1^2 = 1$. Add and subtract this inside the equation: $x^2 + 2x + 1 - 1 - 56 = 0$. The first three terms form a perfect square: $(x + 1)^2 - 57 = 0$, giving you $(x + 1)^2 = 57$. Looking at the wrong answers: Choice A has $(x - 1)^2$ instead of $(x + 1)^2$, which occurs if you incorrectly handle the sign when completing the square. Choice B gives $(x + 1)^2 = 49$, suggesting an arithmetic error when combining the constants (-1 and -56). Choice D shows $(x + 2)^2 = 60$, which happens if you mistakenly use the full coefficient (2) instead of half of it when completing the square. When completing the square on $x^2 + bx$, always remember to add and subtract $\left(\frac{b}{2}\right)^2$. This technique transforms any quadratic into a perfect square, making it easier to solve by taking square roots.
16. A student is solving $3x^2 + 18x - 5 = 0$ by completing the square. After factoring out the coefficient of $x^2$ and moving the constant term, they have $3(x^2 + 6x) = 5$. What should be their next step?
    Answer: Add 9 inside the parentheses and add $3 \cdot 9 = 27$ to the right side To complete the square for $x^2 + 6x$, we need to add $(\frac{6}{2})^2 = 9$. Since this 9 is inside parentheses that are multiplied by 3, adding 9 inside actually adds $3 \cdot 9 = 27$ to the left side. To maintain equality, we must add 27 to the right side as well. Choice B forgets to account for the factor of 3. Choice C incorrectly squares 6 instead of half of 6. Choice D uses the coefficient instead of half the linear coefficient.
17. Which of the following represents the correct process for solving $2x^2 + 8x - 10 = 0$ by completing the square?
    Answer: $2(x^2 + 4x) = 10 \rightarrow 2(x + 2)^2 - 8 = 10 \rightarrow 2(x + 2)^2 = 18$ Completing the square is a systematic method for rewriting quadratic equations in a form that makes them easier to solve. The key is to transform one side into a perfect square trinomial while carefully tracking what happens to both sides of the equation. Starting with $2x^2 + 8x - 10 = 0$, you first move the constant to the right side: $2x^2 + 8x = 10$. Next, factor out the coefficient of $x^2$ from the left side: $2(x^2 + 4x) = 10$. Now comes the crucial step: to complete the square inside the parentheses, you take half of the coefficient of $x$ (which is 4), square it to get 4, then add and subtract this value. This gives you $2(x^2 + 4x + 4 - 4) = 10$, which becomes $2((x + 2)^2 - 4) = 10$. Distributing the 2: $2(x + 2)^2 - 8 = 10$, so $2(x + 2)^2 = 18$. This matches option B exactly. Option A skips the critical step of adding and subtracting 4 inside the parentheses, jumping directly to the perfect square form without showing the algebraic justification. Option C divides by 2 too early in the process, which works but doesn't match the step-by-step format requested. Option D also divides by 2 early and then jumps to the final answer without showing the intermediate completing-the-square steps. When completing the square, always show the "add and subtract" step explicitly—it demonstrates you understand why the method works, not just the mechanical steps.
18. A quadratic equation $ax^2 + bx + c = 0$ with $a > 0$ is solved by completing the square to get $(x + p)^2 = q$. If $q < 0$, what can be concluded about the original equation?
    Answer: The equation has no real solutions because we cannot take the square root of a negative number When you encounter a quadratic equation that's been completed to the form $(x + p)^2 = q$, the sign of $q$ tells you everything about the nature of the solutions. This is because solving requires taking the square root of both sides. If $q < 0$, then you need $x + p = \pm\sqrt{q}$ where $q$ is negative. Since you cannot take the square root of a negative number and get a real result, the equation has no real solutions. The solutions would be complex numbers of the form $x = -p \pm i\sqrt{|q|}$. Answer D correctly identifies this: when $q < 0$, there are no real solutions because taking the square root of a negative number is undefined in the real number system. Answer A contains a partial truth but is incorrect as stated. While the solutions would be complex with real parts equal to $-p$, the question asks what can be concluded about the original equation, and the most fundamental conclusion is that it has no real solutions. Answer B is wrong because a repeated real solution would occur when $q = 0$, not when $q < 0$. Answer C misunderstands the relationship between $a > 0$ and the solutions. While $a > 0$ means the parabola opens upward, this doesn't determine the sign of the solutions—it's the value of $q$ that determines whether real solutions exist at all. Remember: In completed square form $(x + p)^2 = q$, the sign of $q$ determines solution type—positive means two real solutions, zero means one repeated real solution, and negative means no real solutions.
19. The equation $x^2 + 6x + k = 0$ is solved by completing the square, yielding $(x + 3)^2 = 9 - k$. For what values of $k$ does this equation have exactly two distinct real solutions?
    Answer: $k < 9$ When you encounter quadratic equations and need to determine the number of real solutions, focus on what makes the solutions real and distinct. The completed square form $(x + 3)^2 = 9 - k$ is perfect for this analysis. From $(x + 3)^2 = 9 - k$, you solve by taking the square root of both sides: $x + 3 = \pm\sqrt{9 - k}$, so $x = -3 \pm \sqrt{9 - k}$. For real solutions to exist, the expression under the square root must be non-negative: $9 - k \geq 0$. For exactly two distinct real solutions, you need $9 - k > 0$, which means $k < 9$. Here's why: when $k < 9$, the value $9 - k$ is positive, so $\sqrt{9 - k}$ gives you two different real numbers ($-3 + \sqrt{9 - k}$ and $-3 - \sqrt{9 - k}$). When $k = 9$, you get $9 - k = 0$, producing only one solution: $x = -3$. When $k > 9$, you get $9 - k < 0$, making the square root undefined in the real numbers. Looking at the choices: (A) $k = 9$ gives exactly one solution, not two distinct ones. (B) $k > 9$ produces no real solutions. (D) $k \leq 9$ includes $k = 9$, which gives one solution, not two distinct solutions. Only (C) $k < 9$ ensures exactly two distinct real solutions. Study tip: When analyzing quadratic solutions from completed square form $(x + a)^2 = b$, remember that $b > 0$ gives two distinct real solutions, $b = 0$ gives one repeated solution, and $b < 0$ gives no real solutions.
20. A quadratic equation $ax^2 + bx + c = 0$ has solutions that can be found by completing the square. If the completed square form is $3(x - 4)^2 = 27$, what is the value of $b$ in the original equation?
    Answer: $-24$ From $3(x - 4)^2 = 27$, we can expand to get $3(x^2 - 8x + 16) = 27$, which gives us $3x^2 - 24x + 48 = 27$, or $3x^2 - 24x + 21 = 0$. Therefore, $b = -24$. Choice B results from forgetting to distribute the 3. Choice C gives the wrong sign. Choice D forgets the negative sign from the binomial expansion.

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