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!

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!

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!

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!

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.

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, take the square root of both sides to get x - p = ±√k, then solve for x: x = p ± √k. The ± symbol means you get two solutions: p + √k and p - √k. Starting with (x - 4)² = 25, we take the square root of both sides: x - 4 = ±5. Now we solve for x by adding 4 to both sides: x = 4 ± 5. This gives us two solutions: x = 4 + 5 = 9 and x = 4 - 5 = -1. The ± splits into two separate calculations! Choice C is correct because it properly expresses the solutions as x = 4 ± 5, which represents both x = 9 and x = -1. Choice A has the right final answers but doesn't show the ± form, while Choice B makes an error by writing ±25 instead of ±5 (we take the square root of 25, which is 5, not 25 itself). Quick check: after solving, substitute your answers back into the original equation. (9-4)² = 5² = 25 ✓ and (-1-4)² = (-5)² = 25 ✓. If you get the original right side, great! This catch-your-own-mistakes habit is one of the best math skills you can develop!

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!

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!

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 $x^2 = k$, 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: $x = \pm \sqrt{k}$. Looking at $x^2 = 64$, we recognize that 64 is a perfect square, $8^2$. Taking the square root of both sides: $x = \pm 8$. This means $x = 8$ or $x = -8$. Both work because $8^2 = 64$ AND $(-8)^2 = 64$. Always remember that $\pm$ symbol—it gives us both solutions! Choice B 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 $\pm$ (both positive and negative), so $x^2 = 64$ gives $x = \pm 8$, which means both $x = 8$ AND $x = -8$. Here's how to choose your method: (1) If it's $x^2 =$ 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!

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!