Algebra Quiz: Solve Quadratics By Multiple Methods
20 questions · exam conditions
0:00
Solve Quadratics By Multiple MethodsQuestion 1 of 20

Solve by factoring: x2+7x+12=0x^2+7x+12=0

x=3, 4x=3,\,4
x=−3, −4x=-3,\,-4
x=−12, −1x=-12,\,-1
x=−7±37x=-7\pm \sqrt{37}
← Back to quizzes

Algebra Quiz

Algebra Quiz: Solve Quadratics By Multiple Methods

Practice Solve Quadratics By Multiple Methods in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solve Quadratics By Multiple Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve by factoring: x2+7x+12=0x^2+7x+12=0

  1. x=3, 4x=3,\,4
  2. x=−3, −4x=-3,\,-4 (correct answer)
  3. x=−12, −1x=-12,\,-1
  4. x=−7±37x=-7\pm \sqrt{37}
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 7x + 12 to get (x + 3)(x + 4) = 0. Using the Zero Product Property, either (x + 3) = 0 or (x + 4) = 0. Solving each: x + 3 = 0 gives x = -3, and x + 4 = 0 gives x = -4. These are the two solutions! Choice B is correct because it properly applies the factoring method and identifies both solutions: x = -3 and x = -4. Choice A gives positive solutions x = 3, 4, but when we have (x + 3)(x + 4) = 0, solving x + 3 = 0 gives x = -3 (not x = 3). When you solve x + 3 = 0, you subtract 3 from both sides, giving the negative value. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 2

Solve by factoring: x2−11x+24=0x^2-11x+24=0

  1. x=−3, −8x=-3,\,-8
  2. x=4, 6x=4,\,6
  3. x=3, 8x=3,\,8 (correct answer)
  4. x=11±52x=\frac{11\pm 5}{2}
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 11x + 24 to get (x - 3)(x - 8) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 8) = 0. Solving each: x - 3 = 0 gives x = 3, and x - 8 = 0 gives x = 8. These are the two solutions! Choice A is correct because it properly applies the factoring method and identifies both solutions: x = 3 and x = 8. Choice B has sign errors—when we factor x² - 11x + 24, we need two numbers that multiply to +24 and add to -11, which are -3 and -8, giving us (x - 3)(x - 8) = 0, not (x + 3)(x + 8) = 0. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 3

Using the quadratic formula, solve x2+5x+6=0x^2 + 5x + 6 = 0.

  1. x=−1,−6x = -1, -6
  2. x=−2,−3x = -2, -3 (correct answer)
  3. x=−5±11x = \frac{-5 \pm \sqrt{1}}{1}
  4. x=2,3x = 2, 3
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 1, b = 5, c = 6: x = (-5 ± √(25 - 24))/2 = (-5 ± √1)/2 = (-5 ± 1)/2. This simplifies to x = (-5 + 1)/2 = -2 and x = (-5 - 1)/2 = -3. The ± in the formula is what gives us two solutions! Choice A is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice D doesn't simplify the radical: √1 can be simplified to 1. When using the quadratic formula, always simplify your radicals and fractions at the end! For the quadratic formula, here's a trick: write out 'x = (-b ± √(b² - 4ac))/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 4

Solve by factoring: x2+9x+20=0x^2 + 9x + 20 = 0

  1. x=4,5x = 4, 5
  2. x=−4,5x = -4, 5
  3. x=−4,−5x = -4, -5 (correct answer)
  4. x=−20,−1x = -20, -1
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 9x + 20 to get (x + 4)(x + 5) = 0. Using the Zero Product Property, either (x + 4) = 0 or (x + 5) = 0. Solving each: x = -4 and x = -5. These are the two solutions! Choice C is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x + 4) = 0, we get x = -4 (not x = 4). When you solve x + 4 = 0, you subtract 4 from both sides, giving the negative value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 5

A ball's height (in feet) is modeled by h(t)=−16t2+32th(t) = -16t^2 + 32t. When does it hit the ground (solve h(t)=0h(t)=0)?

  1. t=0, 2t = 0,\ 2 (correct answer)
  2. t=1t = 1
  3. t=−1, 1t = -1,\ 1
  4. t=−2, 0t = -2,\ 0
Explanation: This question tests your ability to solve quadratic equations by factoring in a real-world context, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor -16t² + 32t = 0 to get -16t(t - 2) = 0. Using the Zero Product Property, either -16t = 0 or (t - 2) = 0. Solving each: t = 0 and t = 2. These are the two solutions, representing when the ball is thrown (t=0) and when it hits the ground (t=2)! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (t - 2) = 0, we get t = 2 (not t = -2). When you solve t - 2 = 0, you add 2 to both sides, giving the positive value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 6

Solve by factoring: x2+5x+6=0x^2 + 5x + 6 = 0

  1. x=−2, 3x = -2,\ 3
  2. x=−2, −3x = -2,\ -3 (correct answer)
  3. x=2, 3x = 2,\ 3
  4. x=1, 6x = 1,\ 6
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 5x + 6 to get (x + 2)(x + 3) = 0. Using the Zero Product Property, either (x + 2) = 0 or (x + 3) = 0. Solving each: x = -2 and x = -3. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x + 2) = 0, we get x = -2 (not x = 2). When you solve x + 2 = 0, you subtract 2 from both sides, giving the negative value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 7

Solve x2+2x+2=0x^2 + 2x + 2 = 0. Express complex solutions in a±bia \pm bi form.

  1. x=−1±2x = -1 \pm 2
  2. x=−1±2ix = -1 \pm 2i
  3. x=−1±ix = -1 \pm i (correct answer)
  4. x=1±ix = 1 \pm i
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula for complex solutions, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). When we calculate the discriminant b² - 4ac = 4 - 8 = -4, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). Working through the quadratic formula: x = (-2 ± √(-4))/2 = (-2 ± 2i)/2 = -1 ± i. These complex solutions come in a conjugate pair! Choice B is correct because it properly expresses complex solutions correctly as a ± bi. Well done! Choice D forgets to include i in the complex solution, giving -1 ± 2 instead of -1 ± i. When the discriminant is negative, the square root involves i = √(-1), making the solutions complex. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-4) = 2i. Your solutions will be complex numbers in the form a + bi and a - bi. For the quadratic formula, here's a trick: write out 'x = (-b ± √(b² - 4ac))/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers.

Question 8

Solve: x2−12x+36=0x^2 - 12x + 36 = 0

  1. x=6x = 6 (correct answer)
  2. x=−6x = -6
  3. x=6, −6x = 6,\ -6
  4. x=3, 12x = 3,\ 12
Explanation: This question tests your ability to solve quadratic equations using multiple methods, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 12x + 36 to get (x - 6)(x - 6) = 0, or (x - 6)² = 0. Using the Zero Product Property, x - 6 = 0, giving the repeated solution x = 6. This is a perfect square trinomial with a double root! Choice A is correct because it properly identifies the repeated root. Well done! Choice C has two different solutions when it's actually a repeated root; quadratics like this have two solutions counting multiplicity, but they are the same value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! The ± symbol is your reminder to find both solutions, but in cases like this, they coincide into one. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 9

Solve x2+2x+5=0x^2 + 2x + 5 = 0. Express complex solutions in a±bia \pm bi form.

  1. x=−1±2ix = -1 \pm 2i (correct answer)
  2. x=1±2ix = 1 \pm 2i
  3. x=−1±4ix = -1 \pm 4i
  4. x=−1±2x = -1 \pm 2
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, including complex solutions, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). When we calculate the discriminant b² - 4ac = 4 - 20 = -16, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). Working through the quadratic formula: x = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. These complex solutions come in a conjugate pair! Choice A is correct because it properly expresses complex solutions correctly as a ± bi. Well done! Choice D forgets to include i in the complex solution, giving -1 ± 2 instead of -1 ± 2i. When the discriminant is negative, the square root involves i = √(-1), making the solutions complex. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-16) = 4i. Your solutions will be complex numbers in the form a + bi and a - bi. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 10

Solve: 2x2−8x=02x^2-8x=0

  1. x=0, 4x=0,\,4 (correct answer)
  2. x=2, −2x=2,\,-2
  3. x=−4, 0x=-4,\,0
  4. x=4x=4
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor 2x² - 8x = 0 by first factoring out the common factor of 2x: 2x(x - 4) = 0. Using the Zero Product Property, either 2x = 0 or (x - 4) = 0. Solving each: 2x = 0 gives x = 0, and x - 4 = 0 gives x = 4. These are the two solutions! Choice B is correct because it properly applies the factoring method and identifies both solutions: x = 0 and x = 4. Choice A forgets the x = 0 solution—when you factor out x from the equation, x = 0 is always one of your solutions! Don't lose it! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 11

Solve by taking square roots: (x−4)2=25(x-4)^2=25

  1. x=±5x=\pm 5
  2. x=4±25x=4 \pm 25
  3. x=1,9x=1, 9
  4. x=9,−1x=9, -1 (correct answer)
Explanation: This question tests your ability to solve quadratic equations using taking square roots, which is an essential skill in Algebra 1. When solving by taking square roots from an equation like (x−p)2=k(x - p)^2 = k, take the square root of both sides to get x−p=±kx - p = \pm \sqrt{k}, then solve for x: x=p±kx = p \pm \sqrt{k}. The ±\pm symbol means you get two solutions: p+kp + \sqrt{k} and p−kp - \sqrt{k}. Starting with (x−4)2=25(x - 4)^2 = 25, we take the square root of both sides: x−4=±5x - 4 = \pm 5. Now we solve for x by adding 4 to both sides: x=4±5x = 4 \pm 5. This gives us two solutions: x=4+5=9x = 4 + 5 = 9 and x=4−5=−1x = 4 - 5 = -1. The ±\pm splits into two separate calculations! Choice A is correct because it properly applies the method and includes both solutions: x=9x = 9 and x=−1x = -1. Choice B gives x=±5x = \pm 5, which would be the answer if we had x2=25x^2 = 25, but here we have (x−4)2=25(x - 4)^2 = 25, so we need to account for the -4 inside the parentheses. The ±\pm symbol is your reminder to find both solutions: x=4±5x = 4 \pm 5 means two answers (4+5=94 + 5 = 9 and 4−5=−14 - 5 = -1). Don't just calculate the plus—do both! This is one of the most common places students lose a solution.

Question 12

Solve x2−7x+12=0x^2 - 7x + 12 = 0 by factoring and the Zero Product Property.

  1. x=3,4x = 3, 4 (correct answer)
  2. x=12,1x = 12, 1
  3. x=−3,−4x = -3, -4
  4. x=−3,4x = -3, 4
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 7x + 12 to get (x - 3)(x - 4) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 4) = 0. Solving each: x = 3 and x = 4. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x - 3) = 0, we get x = 3 (not x = -3). When you solve x - 3 = 0, you add 3 to both sides, giving the positive value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 13

Using the quadratic formula, solve x2−4x+8=0x^2-4x+8=0. Express complex solutions in a±bia\pm bi form.

  1. x=2±2x=2\pm 2
  2. x=2±2ix=2\pm 2i (correct answer)
  3. x=−2±2ix=-2\pm 2i
  4. x=2±4ix=2\pm 4i
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 1, b = -4, c = 8: x = (-(-4) ± √((-4)² - 4(1)(8)))/(2·1) = (4 ± √(16 - 32))/2 = (4 ± √(-16))/2 = (4 ± 4i)/2 = 2 ± 2i. When we calculate the discriminant b² - 4ac = 16 - 32 = -16, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). These complex solutions come in a conjugate pair! Choice A is correct because it properly expresses the complex solutions as x = 2 ± 2i, which means x = 2 + 2i and x = 2 - 2i. Choice D gives real solutions x = 2 ± 2 when the discriminant is negative. When b² - 4ac < 0, like 16 - 32 = -16, there are no real solutions—only complex ones involving i. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-16) = 4i. Your solutions will be complex numbers in the form a + bi and a - bi.

Question 14

A ball's height (in feet) is modeled by h(t)=−16t2+64th(t)=-16t^2+64t. When does it hit the ground (solve h(t)=0h(t)=0)?

  1. t=2t=2
  2. t=−4, 0t=-4,\,0
  3. t=0, 4t=0,\,4 (correct answer)
  4. t=0, 64t=0,\,64
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor h(t) = -16t² + 64t = 0 by first factoring out the common factor of -16t: -16t(t - 4) = 0. Using the Zero Product Property, either -16t = 0 or (t - 4) = 0. Solving each: -16t = 0 gives t = 0, and t - 4 = 0 gives t = 4. These are the two solutions! Choice A is correct because it properly identifies when the ball hits the ground: at t = 0 (the initial launch) and t = 4 (when it returns to ground level after 4 seconds). Choice D gives t = 0, 64, but 64 comes from confusing the coefficient 64t with a solution—always solve the factored equation, not just read numbers from the original! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 15

Solve by factoring: x2−7x+12=0x^2 - 7x + 12 = 0

  1. x=3, 4x = 3,\ 4 (correct answer)
  2. x=−3, 4x = -3,\ 4
  3. x=12, −1x = 12,\ -1
  4. x=−3, −4x = -3,\ -4
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 7x + 12 to get (x - 3)(x - 4) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 4) = 0. Solving each: x = 3 and x = 4. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x - 3) = 0, we get x = 3 (not x = -3). When you solve x - 3 = 0, you add 3 to both sides, giving the positive value here. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 16

Solve: x2=81x^2 = 81

  1. x=9x = 9
  2. x=−9x = -9
  3. x=9,−9x = 9, -9 (correct answer)
  4. x=±81x = \pm 81
Explanation: This question tests your ability to solve quadratic equations using inspection or taking square roots, which is an essential skill in Algebra 1. When a quadratic is in the simple form x² = k, we can solve by inspection (just recognizing the answer) or by taking the square root of both sides, remembering that we need both the positive and negative square roots: x = ±√k. Looking at x² = 81, we recognize that 81 is a perfect square, 9². Taking the square root of both sides: x = ±9. This means x = 9 or x = -9. Both work because 9² = 81 AND (-9)² = 81. Always remember that ± symbol—it gives us both solutions! Choice C is correct because it properly includes both solutions with correct arithmetic. Well done! Choice A forgets the negative root—a super common mistake! When we take the square root, we always get ± (both positive and negative), so x² = 81 gives x = ±9, which means both x = 9 AND x = -9. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!

Question 17

The equation 4x2−12x+9=254x^2 - 12x + 9 = 25 can be solved most efficiently by which method?

  1. Factoring after moving all terms to one side of the equation
  2. Taking square roots after recognizing the left side as a perfect square trinomial (correct answer)
  3. Completing the square by adding and subtracting the appropriate constant
  4. Using the quadratic formula after rearranging to standard form
Explanation: The left side 4x2−12x+94x^2 - 12x + 9 is a perfect square trinomial: (2x−3)2(2x - 3)^2. So the equation becomes (2x−3)2=25(2x - 3)^2 = 25. Taking square roots: 2x−3=±52x - 3 = \pm 5, giving x=4x = 4 or x=−1x = -1. Choice A would require rearranging to 4x2−12x−16=04x^2 - 12x - 16 = 0, which doesn't factor nicely. Choice C is unnecessary since we already have a perfect square. Choice D works but is less efficient than recognizing the perfect square.

Question 18

When solving x2−4x−1=0x^2 - 4x - 1 = 0 by completing the square, what value must be added to both sides of the equation x2−4x=1x^2 - 4x = 1?

  1. 22
  2. 44 (correct answer)
  3. 88
  4. 1616
Explanation: To complete the square for x2−4xx^2 - 4x, we take half of the coefficient of xx and square it: (−42)2=(−2)2=4\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4. Adding 4 to both sides gives x2−4x+4=1+4x^2 - 4x + 4 = 1 + 4, or (x−2)2=5(x - 2)^2 = 5. Choice A is half the coefficient of xx but not squared. Choice C is twice the coefficient of xx. Choice D is the coefficient of xx squared, not half the coefficient squared.

Question 19

A projectile's height hh (in feet) above the ground after tt seconds is modeled by h=−16t2+64t+80h = -16t^2 + 64t + 80. If the projectile hits the ground when h=0h = 0, which method would be most efficient for finding when this occurs?

  1. Factoring, because the equation can be written as −16(t2−4t−5)=0-16(t^2 - 4t - 5) = 0 (correct answer)
  2. Completing the square, because the coefficient of t2t^2 is already a perfect square
  3. Taking square roots, because the equation has no linear term after rearranging
  4. Quadratic formula, because the discriminant will be a perfect square after factoring out −16-16
Explanation: Setting h=0h = 0 gives −16t2+64t+80=0-16t^2 + 64t + 80 = 0. Factoring out −16-16 yields −16(t2−4t−5)=0-16(t^2 - 4t - 5) = 0, so t2−4t−5=0t^2 - 4t - 5 = 0. This factors as (t−5)(t+1)=0(t - 5)(t + 1) = 0, giving t=5t = 5 or t=−1t = -1. Since time must be positive, t=5t = 5 seconds. Choice B is incorrect because having a coefficient that's a perfect square doesn't make completing the square more efficient. Choice C is wrong because there is a linear term (64t64t). Choice D is unnecessarily complex compared to factoring.

Question 20

The quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has discriminant Δ=0\Delta = 0. If this equation is written in the form (px+q)2=0(px + q)^2 = 0 by completing the square, what is the relationship between the solution and the coefficients?

  1. The solution is x=−qpx = -\frac{q}{p} and equals b2a\frac{b}{2a}
  2. The solution is x=qpx = \frac{q}{p} and equals b2a\frac{b}{2a}
  3. The solution is x=qpx = \frac{q}{p} and equals −b2a-\frac{b}{2a}
  4. The solution is x=−qpx = -\frac{q}{p} and equals −b2a-\frac{b}{2a} (correct answer)
Explanation: When you see a quadratic with discriminant Δ=0\Delta = 0, you're dealing with a perfect square trinomial that has exactly one solution (a repeated root). This connects completing the square with the quadratic formula in an important way. Let's work through the algebra systematically. Starting with ax2+bx+c=0ax^2 + bx + c = 0, we complete the square by first factoring out aa: a(x2+bax+ca)=0a(x^2 + \frac{b}{a}x + \frac{c}{a}) = 0. To complete the square, we add and subtract (b2a)2(\frac{b}{2a})^2: a[(x+b2a)2−b24a2+ca]=0a\left[(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right] = 0 Since Δ=b2−4ac=0\Delta = b^2 - 4ac = 0, the terms −b24a2+ca-\frac{b^2}{4a^2} + \frac{c}{a} simplify to zero, giving us: a(x+b2a)2=0a(x + \frac{b}{2a})^2 = 0, or (x+b2a)2=0(x + \frac{b}{2a})^2 = 0 Comparing this to (px+q)2=0(px + q)^2 = 0, we see that p=1p = 1 and q=b2aq = \frac{b}{2a}. The solution is x=−qp=−b2ax = -\frac{q}{p} = -\frac{b}{2a}. Choice A incorrectly states the solution equals b2a\frac{b}{2a} (missing the negative). Choice B has the wrong sign for qp\frac{q}{p} and the wrong sign for the final expression. Choice C gets the final expression wrong by missing the negative sign. The correct answer is D: the solution is x=−qpx = -\frac{q}{p} and equals −b2a-\frac{b}{2a}. Remember: when Δ=0\Delta = 0, the quadratic formula gives x=−b2ax = -\frac{b}{2a}, which should match your completed square form. This connection helps verify your work.