Solve Quadratics by Multiple Methods - Algebra 1

Solve by taking square roots: $(x - 4)^2 = 25$
Answer: $x = 9, -1$ This question tests your ability to solve quadratic equations by 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 A is correct because it properly includes both solutions with correct arithmetic. Well done! Choice C has only one solution when there should be two. Quadratic equations (unless they're perfect squares with a repeated root) always have two solutions—don't forget to use the ± in your solving process! The ± symbol is your reminder to find both solutions: x = 4 ± 5 means two answers (4 + 5 = 9 and 4 - 5 = -1). Don't just calculate the plus—do both! This is one of the most common places students lose a solution. 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!
The quadratic equation $ax^2 + bx + c = 0$ has discriminant $\Delta = 0$ . If this equation is written in the form $(px + q)^2 = 0$ by completing the square, what is the relationship between the solution and the coefficients?
Answer: The solution is $x = -\frac{q}{p}$ and equals $-\frac{b}{2a}$ When you see a quadratic with discriminant $\Delta = 0$ , 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 $ax^2 + bx + c = 0$ , we complete the square by first factoring out $a$ : $a(x^2 + \frac{b}{a}x + \frac{c}{a}) = 0$ . To complete the square, we add and subtract $(\frac{b}{2a})^2$ : $a\left[(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right] = 0$ Since $\Delta = b^2 - 4ac = 0$ , the terms $-\frac{b^2}{4a^2} + \frac{c}{a}$ simplify to zero, giving us: $a(x + \frac{b}{2a})^2 = 0$ , or $(x + \frac{b}{2a})^2 = 0$ Comparing this to $(px + q)^2 = 0$ , we see that $p = 1$ and $q = \frac{b}{2a}$ . The solution is $x = -\frac{q}{p} = -\frac{b}{2a}$ . Choice A incorrectly states the solution equals $\frac{b}{2a}$ (missing the negative). Choice B has the wrong sign for $\frac{q}{p}$ 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 $x = -\frac{q}{p}$ and equals $-\frac{b}{2a}$ . Remember: when $\Delta = 0$ , the quadratic formula gives $x = -\frac{b}{2a}$ , which should match your completed square form. This connection helps verify your work.
Solve by factoring: $x^2 + 7x + 12 = 0$
Answer: $x = -3, -4$ 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 identifies both solutions as x = -3 and x = -4. Well done! Choice A makes a sign error: from (x + 3) = 0, we get x = -3 (not x = 3). When you solve x + 3 = 0, you subtract 3 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!
Using the quadratic formula, solve $2x^2 - 4x - 6 = 0$ .
Answer: $x = -1,\ 3$ 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 = 2, b = -4, c = -6: x = (-(-4) ± √((-4)² - 4(2)(-6)))/(2·2) = (4 ± √(16 + 48))/4 = (4 ± √64)/4 = (4 ± 8)/4. This gives x = (4 + 8)/4 = 12/4 = 3 and x = (4 - 8)/4 = -4/4 = -1. 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 B has a sign error in the quadratic formula: with b = -4, -b = 4 (positive), but perhaps confusing the order of solutions. Those signs are tricky! 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!
Using the quadratic formula, solve $x^2+6x+5=0$ .
Answer: $x=-1,\,-5$ 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 = 6, c = 5: x = (-6 ± √(6² - 4(1)(5)))/(2·1) = (-6 ± √(36 - 20))/2 = (-6 ± √16)/2 = (-6 ± 4)/2. This gives us x = (-6 + 4)/2 = -2/2 = -1 and x = (-6 - 4)/2 = -10/2 = -5. The ± in the formula is what gives us two solutions! Choice B is correct because it properly applies the quadratic formula and gives both solutions: x = -1 and x = -5. Choice A has the wrong signs—it gives x = -1, 5 instead of x = -1, -5. When using the quadratic formula, be extra careful with signs, especially the -b at the beginning! 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.
Solve: $x^2 = 64$
Answer: $x = \pm 8$ 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!
Solve by factoring: $x^2 + 9x + 20 = 0$
Answer: $x = -4, -5$ 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!
Solve by factoring: $x^2 + 5x + 6 = 0$
Answer: $x = -2,\ -3$ 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!
Solve $x^2 + 2x + 5 = 0$ . Express complex solutions in $a \pm bi$ form.
Answer: $x = -1 \pm 2i$ 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!
Using the quadratic formula, solve $x^2 + 6x + 5 = 0$ .
Answer: $x = -1, -5$ This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula $x = (-b \pm \sqrt{b^2 - 4ac})/(2a)$ works for ANY quadratic equation $ax^2 + bx + c = 0$ —it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions. Using the quadratic formula with a = 1, b = 6, c = 5: $x = (-6 \pm \sqrt{6^2 - 4(1)(5)})/(2 \cdot 1) = (-6 \pm \sqrt{36 - 20})/2 = (-6 \pm \sqrt{16})/2 = (-6 \pm 4)/2$ . This gives us $x = (-6 + 4)/2 = -2/2 = -1$ and $x = (-6 - 4)/2 = -10/2 = -5$ . The $\pm$ in the formula is what gives us two solutions! Choice A is correct because it properly applies the quadratic formula and gets both solutions: x = -1 and x = -5. Well done! Choice B has a sign error in the quadratic formula: the formula starts with -b (negative b), so with b = 6, we use -6, not +6. Those signs are tricky! For the quadratic formula, here's a trick: write out 'x = ( $-b \pm \sqrt{b^2 - 4ac}$ )/(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.
Solve: $x^2 = 81$
Answer: $x = 9, -9$ 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!
A ball's height (in feet) is modeled by $h(t)=-16t^2+64t$ . When does it hit the ground (solve $h(t)=0$ )?
Answer: $t=0,\,4$ 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!
Solve: $2x^2-8x=0$
Answer: $x=0,\,4$ 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!
A quadratic equation of the form $x^2 + bx + c = 0$ has solutions $x = 3 \pm 2i$ . What is the value of $b + c$ ?
Answer: $7$ When you see a quadratic equation with complex solutions, you need to work backwards from the solutions to find the original equation. The key insight is that if a quadratic has real coefficients, complex solutions always come in conjugate pairs. Given the solutions $x = 3 \pm 2i$ , you have $x = 3 + 2i$ and $x = 3 - 2i$ . To find the original quadratic, use the fact that if $r$ and $s$ are solutions, then the equation is $(x - r)(x - s) = 0$ . So: $(x - (3 + 2i))(x - (3 - 2i)) = 0$ Expanding this: $((x - 3) - 2i)((x - 3) + 2i) = 0$ This is a difference of squares pattern: $(a - b)(a + b) = a^2 - b^2$ $(x - 3)^2 - (2i)^2 = 0$ $(x - 3)^2 - 4i^2 = 0$ Since $i^2 = -1$ : $(x - 3)^2 + 4 = 0$ $x^2 - 6x + 9 + 4 = 0$ $x^2 - 6x + 13 = 0$ Therefore, $b = -6$ and $c = 13$ , so $b + c = -6 + 13 = 7$ . Looking at the wrong answers: A) $19$ likely comes from adding the coefficients incorrectly. B) $13$ is just the value of $c$ alone, missing the $b$ term. D) $-7$ probably results from sign errors during expansion. Remember: when working backwards from complex solutions, always use the conjugate pair formula and watch your signs carefully during algebraic manipulation.
Which method is most efficient for solving $x^2 - 49 = 0$ ?
Answer: Inspection (mental math) This question tests your ability to solve quadratic equations using multiple methods and choose the most efficient one, 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 $x^2 - 49 = 0$ , we can rewrite it as $x^2 = 49$ . We immediately recognize that 49 is a perfect square, $7^2$ . By inspection, we know $x = \pm 7$ . This means $x = 7$ or $x = -7$ . Both work because $7^2 = 49$ AND $(-7)^2 = 49$ . This is much faster than setting up the quadratic formula or trying to factor! Choice C is correct because inspection (mental math) is the most efficient method for this equation—you can see the answer immediately without any written work. Well done! While factoring ( $x^2 - 49 = (x+7)(x-7) = 0$ ) or the quadratic formula would also work, they require more steps for something you can solve in your head. 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!
Solve $x^2+2x+5=0$ . Express complex solutions in $a\pm bi$ form.
Answer: $x=-1\pm 2i$ 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 = 2, c = 5: x = (-2 ± √(2² - 4(1)(5)))/(2·1) = (-2 ± √(4 - 20))/2 = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. 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)). These complex solutions come in a conjugate pair! Choice C is correct because it properly expresses the complex solutions as x = -1 ± 2i, which means x = -1 + 2i and x = -1 - 2i. Choice A 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.
Solve $x^2 + 4x + 8 = 0$ . Express complex solutions in $a \pm bi$ form.
Answer: $x = -2 \pm 2i$ This question tests your ability to solve quadratic equations using the quadratic formula and express 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, 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: first calculate the discriminant b² - 4ac = 4² - 4(1)(8) = 16 - 32 = -16. Since this is negative, we'll have complex solutions. Working through: x = (-4 ± √(-16))/(2·1) = (-4 ± 4i)/2 = -2 ± 2i. These complex solutions come in a conjugate pair! Choice A is correct because it properly expresses the complex solutions as x = -2 ± 2i. Well done! Choice D gives real solutions 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.
Solve by factoring: $x^2 - x - 12 = 0$
Answer: $x = -3, 4$ 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 need to factor x² - x - 12. Looking for two numbers that multiply to -12 and add to -1, we find -4 and 3. So x² - x - 12 = (x - 4)(x + 3) = 0. Using the Zero Product Property, either (x - 4) = 0 or (x + 3) = 0. Solving each: x - 4 = 0 gives x = 4, and x + 3 = 0 gives x = -3. These are the two solutions! Choice A is correct because it properly identifies both solutions as x = -3 and x = 4. Well done! Choice B reverses the signs: from (x + 3) = 0, we get x = -3 (not x = 3), and from (x - 4) = 0, we get x = 4 (not x = -4). When solving these simple equations, move the constant to the other side with the opposite sign. Quick check: after solving, substitute your answers back into the original equation. For x = -3: (-3)² - (-3) - 12 = 9 + 3 - 12 = 0 ✓. For x = 4: 4² - 4 - 12 = 16 - 4 - 12 = 0 ✓. This catch-your-own-mistakes habit is one of the best math skills you can develop!
A student attempts to solve $x^2 + 10x = 39$ by completing the square and writes: $x^2 + 10x + 25 = 39 + 25$ , so $(x + 5)^2 = 64$ . What should the student do next to find the correct solutions?
Answer: Take the square root of both sides, remembering to include both positive and negative roots When you're completing the square and reach an equation like $(x + 5)^2 = 64$ , you're at a crucial step where you need to "undo" the square to solve for $x$ . The key insight is that when you take the square root of a perfect square, you must consider both the positive and negative possibilities. Taking the square root of both sides gives you $x + 5 = \pm\sqrt{64} = \pm 8$ . This means $x + 5 = 8$ or $x + 5 = -8$ , leading to $x = 3$ or $x = -13$ . You can verify: $3^2 + 10(3) = 9 + 30 = 39$ ✓ and $(-13)^2 + 10(-13) = 169 - 130 = 39$ ✓. Choice A makes a critical error by only considering the positive square root. While $\sqrt{64} = 8$ , forgetting the negative case means missing half the solutions. Choice B suggests abandoning a perfectly good method when you're already at the final step—the quadratic formula would give the same answers with more work. Choice D wants you to subtract 25 from both sides, but this would undo the completing-the-square work you just did correctly, taking you backward instead of forward. Remember: whenever you have a squared expression equal to a positive number, taking the square root always produces both positive and negative solutions. This $\pm$ step is essential—quadratic equations typically have two solutions, and forgetting the negative case is one of the most common mistakes in algebra.
The equation $4x^2 - 12x + 9 = 25$ can be solved most efficiently by which method?
Answer: Taking square roots after recognizing the left side as a perfect square trinomial The left side $4x^2 - 12x + 9$ is a perfect square trinomial: $(2x - 3)^2$ . So the equation becomes $(2x - 3)^2 = 25$ . Taking square roots: $2x - 3 = \pm 5$ , giving $x = 4$ or $x = -1$ . Choice A would require rearranging to $4x^2 - 12x - 16 = 0$ , 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.

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