Complete the Square to Find Solutions - Algebra 1

Solve by completing the square: $x^2+2x+5=0$ . What are the solutions?
Answer: $x=-1\pm2i$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The form (x - p)² = q tells us immediately: if q > 0, there are two real solutions (x = p ± √q); if q = 0, there's one solution (x = p); if q < 0, there are no real solutions (we'd need to take square root of negative). So completing the square not only solves the equation, it also reveals the nature of the solutions before we calculate them! Solving x² + 2x + 5 = 0 by completing the square: (1) Move constant to right: x² + 2x = -5. (2) Take half of 2 to get 1, square it to get 1, add to both sides: x² + 2x + 1 = -5 + 1. (3) Left side is perfect square (x + 1)², right side simplifies to -4: (x + 1)² = -4. (4) Take square roots: x + 1 = ±√(-4) = ±2i. (5) Solve: x = -1 ± 2i. Done! Choice B correctly completes the square to get x = -1 ± 2i with accurate arithmetic and proper form, recognizing that √(-4) = 2i. Choice A forgets that we need the imaginary unit i when taking the square root of a negative number: √(-4) = 2i, not just 2. When q < 0 in (x - p)² = q, we get complex solutions! The ± is crucial: from (x - 3)² = 16, taking square roots gives x - 3 = ±4 (both +4 and -4), so x = 3 + 4 = 7 OR x = 3 - 4 = -1. Two solutions! Don't forget the ± and don't forget to split it into two separate solutions. When the right side is negative, remember to include i for the imaginary unit!
A teacher asks students to derive the quadratic formula by completing the square on $ax^2 + bx + c = 0$ . After dividing by $a$ and completing the square, a student obtains $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ . What is the next step to arrive at the quadratic formula?
Answer: Take the square root of both sides to get $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$ When you're completing the square to derive the quadratic formula, you're working toward isolating the variable x. At this stage, you have a perfect square on the left side equal to a fraction on the right side, so the logical next step is to eliminate the square by taking the square root of both sides. Taking the square root of both sides gives you $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$ , which is answer choice B. When you take the square root of a fraction, you take the square root of both the numerator and denominator separately. The square root of $4a^2$ is $2|a|$ , but since we're in the context of deriving a general formula, we write this as $2a$ with the understanding that the $\pm$ symbol will account for both positive and negative cases. Choice A incorrectly includes absolute value bars around the $a$ in the denominator. While $\sqrt{4a^2} = 2|a|$ is technically correct, in the context of the quadratic formula derivation, we handle the sign through the $\pm$ symbol rather than absolute values. Choice C suggests multiplying by $4a^2$ , which would eliminate the fraction but move you away from isolating x—the opposite of what you want at this step. Choice D suggests expanding and rearranging, which would completely undo the completing the square work you just did. Remember: when solving equations with perfect squares, always take the square root of both sides as your next move, and don't forget the $\pm$ symbol since squares have two possible roots.
Transform and solve by completing the square: $x^2 - 6x + 2 = 0$ . First write it in the form $(x-p)^2=q$ , then find the solutions.
Answer: $(x-3)^2=7\;\Rightarrow\;x=3\pm\sqrt{7}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The completing-the-square process creates a perfect square trinomial on one side: for x² + bx, we add (b/2)²—half the middle coefficient, squared—to both sides. This turns x² + bx + (b/2)² into (x + b/2)², a perfect square that factors as a binomial squared. Then we solve by taking square roots. It's like setting up a problem in a form where the answer is obvious! Solving x² - 6x + 2 = 0 by completing the square: (1) Move constant to right: x² - 6x = -2. (2) Take half of -6 to get -3, square it to get 9, add to both sides: x² - 6x + 9 = -2 + 9. (3) Left side is perfect square (x - 3)², right side simplifies to 7: (x - 3)² = 7. (4) Take square roots: x - 3 = ±√7. (5) Solve: x = 3 ± √7. Done! Choice A correctly completes the square to get (x - 3)² = 7 and solves to get x = 3 ± √7 with accurate arithmetic and proper form. Choice B has the right perfect square form but makes a sign error: from (x - 3)², the sign inside is negative, but B uses (x + 3)² which would come from a positive coefficient. Watch those signs! When we have (x - 3)², it's equivalent to (x + (-3))², so the sign depends on b. The sign in the factored form is opposite to the p value. The (b/2)² trick: if you have x² + 6x, half of 6 is 3, and 3² = 9, so you add 9 to complete the square, getting (x + 3)². If you have x² - 10x, half of -10 is -5, and (-5)² = 25 (positive!), so add 25 to get (x - 5)². The sign of b affects the sign in the binomial, but (b/2)² is always positive. Practice this on a few examples and it becomes automatic!
Use completing the square to find the solutions to $x^2 + 2x + 5 = 0$ . (Your answer may indicate if there are no real solutions.)
Answer: $(x+1)^2=-4\;\Rightarrow\; \text{no real solutions}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The form (x - p)² = q tells us immediately: if q > 0, there are two real solutions (x = p ± √q); if q = 0, there's one solution (x = p); if q < 0, there are no real solutions (we'd need to take square root of negative). So completing the square not only solves the equation, it also reveals the nature of the solutions before we calculate them! Solving x² + 2x + 5 = 0 by completing the square: (1) Move constant to right: x² + 2x = -5. (2) Take half of 2 to get 1, square it to get 1, add to both sides: x² + 2x + 1 = -5 + 1. (3) Left side is perfect square (x + 1)², right side simplifies to -4: (x + 1)² = -4. (4) Since -4 is negative, we cannot take its real square root—there are no real solutions! (We'd get complex solutions x = -1 ± 2i.) Done! Choice C correctly completes the square to get (x + 1)² = -4 and recognizes that since we'd need √(-4), there are no real solutions, with accurate arithmetic and proper interpretation. Choice A makes an arithmetic error when simplifying the right side: -5 + 1 = -4, not 4, so we should get (x + 1)² = -4, not (x + 1)² = 4. This critical error changes the equation from having no real solutions to having two real solutions—a completely different answer! After completing the square to get (x + 1)² = -4, we can see: since -4 is negative, we'd need √(negative), so no real solutions exist—we'd get complex solutions x = -1 ± i√4 = -1 ± 2i. The value of q immediately tells us how many real solutions exist!
Two students complete the square for the same equation $x^2 - 8x + 7 = 0$ . Student A writes $(x - 4)^2 = 9$ and Student B writes $(x - 4)^2 - 9 = 0$ . Which statement about their work is correct?
Answer: Both forms are mathematically equivalent and correct for solving the equation When you're completing the square, there are multiple valid ways to express your final result. The key is understanding that algebraically equivalent forms are all correct as long as they help you solve the equation. Let's verify both students completed the square correctly. Starting with $x^2 - 8x + 7 = 0$ , you take half of the coefficient of $x$ (which is $-8/2 = -4$ ), square it to get $16$ , then add and subtract this value: $x^2 - 8x + 16 - 16 + 7 = 0$ . This gives you $(x - 4)^2 - 9 = 0$ , which can be rearranged to $(x - 4)^2 = 9$ . Both forms are mathematically sound. Student A's form $(x - 4)^2 = 9$ isolates the perfect square, making it easy to take the square root of both sides. Student B's form $(x - 4)^2 - 9 = 0$ keeps everything on one side, which some prefer. Both lead to the same solutions: $x = 7$ or $x = 1$ . Choice A is wrong because there's no mathematical requirement that all terms stay on one side when completing the square. Choice B is incorrect because there's no single "required format" - both approaches work. Choice D contains a sign error; since we have $-8x$ , we need $(x - 4)^2$ , not $(x + 4)^2$ . The correct answer is C because both forms are mathematically equivalent and will yield the same solutions. Study tip: When completing the square, focus on whether the algebra is correct rather than the specific format. Multiple equivalent forms can all be valid.
Solve $x^2+2x+5=0$ by completing the square. What does the completed square form reveal about the solutions?
Answer: $(x+1)^2=-4$ , so there are no real solutions This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) actually comes FROM completing the square on the general equation ax² + bx + c = 0! If you complete the square on this general form (treating a, b, c as constants), you derive the quadratic formula. So completing the square isn't just one method among many—it's the fundamental method that gives us the formula itself! After completing the square to get (x - p)² = q, we can see: if q > 0, since q is positive, we can take its square root, giving two real solutions x = p ± √q; if q = 0, we have a perfect square equal to zero, giving one solution x = p; if q < 0, we'd need √(negative), so no real solutions exist—we'd get complex solutions x = p ± i√|q|. The value of q immediately tells us how many real solutions exist! Choice A correctly completes the square to get (x + 1)² = -4, revealing no real solutions with accurate arithmetic and proper form. Choice B makes an arithmetic error when simplifying the right side: -5 + 1 = -4, not +4; completing the square involves several arithmetic steps, so take your time and double-check! To verify your completing-the-square work: expand your (x - p)² + k form back out using FOIL, and you should get back to your original quadratic. If you don't, there's an error. Also, solve the original equation using the quadratic formula—you should get the same solutions. These double-checks catch mistakes and build confidence!
Complete the square to solve $x^2-8x+12=0$ .
Answer: $(x-4)^2=4\;\Rightarrow\;x=4\pm2$ , so $x=2$ or $x=6$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 8x + 12 = 0 by completing the square: (1) Move constant to right: x² - 8x = -12. (2) Take half of -8 to get -4, square it to get 16, add to both sides: x² - 8x + 16 = -12 + 16. (3) Left side is perfect square (x - 4)², right side simplifies to 4: (x - 4)² = 4. (4) Take square roots: x - 4 = ±2. (5) Solve: x = 4 ± 2, so x = 2 or x = 6. Done! Choice A correctly completes the square to get (x - 4)² = 4 and solves to get x = 4 ± 2 with accurate arithmetic and proper form. Choice B makes an arithmetic error when simplifying the right side: -12 + 16 = 4, not -4; this leads to incorrect no real solutions conclusion. The ± is crucial: from (x - 3)² = 16, taking square roots gives x - 3 = ±4 (both +4 and -4), so x = 3 + 4 = 7 OR x = 3 - 4 = -1. Two solutions! Don't forget the ± and don't forget to split it into two separate solutions. Check both: 7 and -1 both satisfy the original equation? Yes!
Solve the quadratic equation $x^2 + 8x + 7 = 0$ by completing the square. Give both solutions.
Answer: $x=-4\pm3$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² + 8x + 7 = 0 by completing the square: (1) Move constant to right: x² + 8x = -7. (2) Take half of 8 to get 4, square it to get 16, add to both sides: x² + 8x + 16 = -7 + 16. (3) Left side is perfect square (x + 4)², right side simplifies to 9: (x + 4)² = 9. (4) Take square roots: x + 4 = ±√9. (5) Solve: x = -4 ± 3. Done! Choice B correctly completes the square to get (x + 4)² = 9 and solves to get x = -4 ± 3 with accurate arithmetic and proper form. Choice D forgets the ± when taking square roots: from (x + 4)² = 9, we get x + 4 = ±3 (both positive and negative square root), giving TWO solutions x = -4 + 3 and x = -4 - 3. Missing the ± means missing a solution! This is one of the most common mistakes in completing the square. The complete completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move the constant term to the right side, (3) Take half the x-coefficient, square it: (b/2)², (4) Add this to BOTH sides, (5) Factor left side as (x + b/2)² (it's a perfect square now!), (6) Simplify right side, (7) Take ± square root of both sides, (8) Solve for x. Follow these 8 steps systematically and you'll get the right answer every time!
Derive the quadratic formula by completing the square on the general equation $ax^2+bx+c=0$ (assume $a\ne 0$ ). Which expression for $x$ results?
Answer: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) actually comes FROM completing the square on the general equation ax² + bx + c = 0! If you complete the square on this general form (treating a, b, c as constants), you derive the quadratic formula. So completing the square isn't just one method among many—it's the fundamental method that gives us the formula itself! Starting with ax² + bx + c = 0: (1) Divide by a: x² + (b/a)x + (c/a) = 0. (2) Move constant: x² + (b/a)x = -c/a. (3) Complete square, half of b/a is b/(2a), squared is b²/(4a²): x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²). (4) Left side: (x + b/(2a))². Right side: (b² - 4ac)/(4a²). (5) So (x + b/(2a))² = (b² - 4ac)/(4a²). (6) Square root: x + b/(2a) = ±√(b² - 4ac)/(2a). (7) Solve: x = -b/(2a) ± √(b² - 4ac)/(2a) = (-b ± √(b² - 4ac))/(2a). That's the quadratic formula! Choice A correctly derives the formula with accurate arithmetic and proper form. Choice B makes a sign error when solving: the term is -b/(2a), not +b/(2a), because we subtract b/(2a) after the square root step. Completing the square involves several arithmetic steps, especially when deriving the formula with a, b, c. Take your time and double-check fraction arithmetic! Why this works: when you add (b/2)² to x² + bx, you're creating the expansion of (x + b/2)²: remember (a + b)² = a² + 2ab + b², so (x + b/2)² = x² + 2(x)(b/2) + (b/2)² = x² + bx + (b/2)². That's why adding (b/2)² makes a perfect square! It's not random—it completes the perfect square trinomial pattern.
Complete the square to solve $x^2+6x+5=0$ . Write an equivalent equation in the form $(x-p)^2=q$ , then find the solutions.
Answer: $(x+3)^2=4$ , so $x=-3\pm2$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The completing-the-square process creates a perfect square trinomial on one side: for x² + bx, we add (b/2)²—half the middle coefficient, squared—to both sides. This turns x² + bx + (b/2)² into (x + b/2)², a perfect square that factors as a binomial squared. Then we solve by taking square roots. It's like setting up a problem in a form where the answer is obvious! Solving x² + 6x + 5 = 0 by completing the square: (1) Move constant to right: x² + 6x = -5. (2) Take half of 6 to get 3, square it to get 9, add to both sides: x² + 6x + 9 = -5 + 9. (3) Left side is perfect square (x + 3)², right side simplifies to 4: (x + 3)² = 4. (4) Take square roots: x + 3 = ±2. (5) Solve: x = -3 ± 2, so x = -1 or x = -5. Done! Choice B correctly completes the square to get (x + 3)² = 4 and solves to get x = -3 ± 2 with accurate arithmetic and proper form. Choice A makes an arithmetic error when simplifying the right side: -5 + 9 = 4, not 14. Completing the square involves several arithmetic steps. Take your time and double-check fraction arithmetic! Why this works: when you add (b/2)² to x² + bx, you're creating the expansion of (x + b/2)²: remember (a + b)² = a² + 2ab + b², so (x + b/2)² = x² + 2(x)(b/2) + (b/2)² = x² + bx + (b/2)². That's why adding (b/2)² makes a perfect square! It's not random—it completes the perfect square trinomial pattern.
Derive the quadratic formula by completing the square on the general equation $ax^2+bx+c=0$ (assume $a\ne0$ ). Which expression for $x$ results?
Answer: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) actually comes FROM completing the square on the general equation ax² + bx + c = 0! If you complete the square on this general form (treating a, b, c as constants), you derive the quadratic formula. So completing the square isn't just one method among many—it's the fundamental method that gives us the formula itself! Starting with ax² + bx + c = 0: (1) Divide by a: x² + (b/a)x + (c/a) = 0. (2) Move constant: x² + (b/a)x = -c/a. (3) Complete square, half of b/a is b/(2a), squared is b²/(4a²): x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²). (4) Left side: (x + b/(2a))². Right side: (b² - 4ac)/(4a²). (5) So (x + b/(2a))² = (b² - 4ac)/(4a²). (6) Square root: x + b/(2a) = ±√(b² - 4ac)/(2a). (7) Solve: x = -b/(2a) ± √(b² - 4ac)/(2a) = (-b ± √(b² - 4ac))/(2a). That's the quadratic formula! Choice C correctly derives the formula with accurate arithmetic and proper form. Choice A makes an arithmetic error when combining fractions in derivation: the right side is (b² - 4ac)/(4a²), not +4ac; completing the square involves several arithmetic steps, especially when deriving the formula with a, b, c. Take your time and double-check fraction arithmetic! To verify your completing-the-square work: expand your (x - p)² + k form back out using FOIL, and you should get back to your original quadratic. If you don't, there's an error. Also, solve the original equation using the quadratic formula—you should get the same solutions. These double-checks catch mistakes and build confidence!
Solve the quadratic equation $x^2-4x+1=0$ by completing the square. First rewrite it in the form $(x-p)^2=q$ , then find the solutions.
Answer: $(x-2)^2=3$ , so $x=2\pm\sqrt{3}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 4x + 1 = 0 by completing the square: (1) Move constant to right: x² - 4x = -1. (2) Take half of -4 to get -2, square it to get 4, add to both sides: x² - 4x + 4 = -1 + 4. (3) Left side is perfect square (x - 2)², right side simplifies to 3: (x - 2)² = 3. (4) Take square roots: x - 2 = ±√3. (5) Solve: x = 2 ± √3. Done! Choice A correctly completes the square to get (x - 2)² = 3 and solves to get x = 2 ± √3 with accurate arithmetic and proper form. Choice B has the wrong sign in the binomial: from x² - 4x, we get (x - 2)², not (x + 2)². When the middle term is -4x, half is -2, so we have (x - 2)². The sign in the factored form matches the sign of half the middle coefficient. The complete completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move the constant term to the right side, (3) Take half the x-coefficient, square it: (b/2)², (4) Add this to BOTH sides, (5) Factor left side as (x + b/2)² (it's a perfect square now!), (6) Simplify right side, (7) Take ± square root of both sides, (8) Solve for x. Follow these 8 steps systematically and you'll get the right answer every time!
Complete the square to solve $x^2 + 2x + 5 = 0$ . What are the solutions?
Answer: $x=-1\pm 2i$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The form (x - p)² = q tells us immediately: if q > 0, there are two real solutions (x = p ± √q); if q = 0, there's one solution (x = p); if q < 0, there are no real solutions (we'd need to take square root of negative). So completing the square not only solves the equation, it also reveals the nature of the solutions before we calculate them! After completing the square to get (x + 1)² = -4 for x² + 2x + 5 = 0, we can see: since q = -4 < 0, we'd need √(negative), so no real solutions exist—we'd get complex solutions x = -1 ± 2i. The value of q immediately tells us how many real solutions exist! Choice B correctly solves to get x = -1 ± 2i with accurate arithmetic and proper form. Choice A forgets that q is negative and treats it as real: from (x + 1)² = -4, we must use i for √(-4) = 2i, not just ±2 which would be for positive q. This leads to complex solutions, not real! Always check the sign of q. To verify your completing-the-square work: expand your (x - p)² + k form back out using FOIL, and you should get back to your original quadratic. If you don't, there's an error. Also, solve the original equation using the quadratic formula—you should get the same solutions. These double-checks catch mistakes and build confidence!
Use completing the square to find the solutions to $x^2 - 4x + 1 = 0$ .
Answer: $x=2\pm\sqrt{3}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 4x + 1 = 0 by completing the square: (1) Move constant to right: x² - 4x = -1. (2) Take half of -4 to get -2, square it to get 4, add to both sides: x² - 4x + 4 = -1 + 4. (3) Left side is perfect square (x - 2)², right side simplifies to 3: (x - 2)² = 3. (4) Take square roots: x - 2 = ±√3. (5) Solve: x = 2 ± √3. Done! Choice B correctly solves to get x = 2 ± √3 with accurate arithmetic and proper form. Choice C has the right perfect square form but makes a sign error: from (x - 2)², the p is 2, but C uses -2 which would come from a positive coefficient. Watch those signs! When we have (x - 2)², it's for negative b. The sign in the factored form is opposite to the p value. The (b/2)² trick: if you have x² + 6x, half of 6 is 3, and 3² = 9, so you add 9 to complete the square, getting (x + 3)². If you have x² - 10x, half of -10 is -5, and (-5)² = 25 (positive!), so add 25 to get (x - 5)². The sign of b affects the sign in the binomial, but (b/2)² is always positive. Practice this on a few examples and it becomes automatic!
A student is completing the square for $x^2 + 5x - 3 = 0$ and writes the intermediate step: $x^2 + 5x + \frac{25}{4} = 3 + \frac{25}{4}$ . What should be the next step in their work?
Answer: Write $(x + \frac{5}{2})^2 = \frac{37}{4}$ and solve for $x$ The student correctly added $\frac{25}{4}$ to both sides. The left side factors as $(x + \frac{5}{2})^2$ since we added $(\frac{5}{2})^2 = \frac{25}{4}$ . The right side simplifies to $3 + \frac{25}{4} = \frac{12}{4} + \frac{25}{4} = \frac{37}{4}$ . The next step is to write the factored form $(x + \frac{5}{2})^2 = \frac{37}{4}$ . Choice B has the wrong sign. Choice C focuses on arithmetic rather than the algebraic transformation. Choice D leaves the work incomplete.
Transform and solve: $x^2-4x+1=0$ . Use completing the square to write it as $(x-p)^2=q$ and find the solutions.
Answer: $(x-2)^2=3\;\Rightarrow\;x=2\pm\sqrt3$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 4x + 1 = 0 by completing the square: (1) Move constant to right: x² - 4x = -1. (2) Take half of -4 to get -2, square it to get 4, add to both sides: x² - 4x + 4 = -1 + 4. (3) Left side is perfect square (x - 2)², right side simplifies to 3: (x - 2)² = 3. (4) Take square roots: x - 2 = ±√3. (5) Solve: x = 2 ± √3. Done! Choice B correctly completes the square to get (x - 2)² = 3 and solves to get x = 2 ± √3 with accurate arithmetic and proper form. Choice A has the process right but doesn't simplify the radical correctly or mismatches q: √3 is already simplified, but A has √5 instead; always reduce radicals to simplest form like √12 = 2√3. The complete completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move the constant term to the right side, (3) Take half the x-coefficient, square it: (b/2)², (4) Add this to BOTH sides, (5) Factor left side as (x + b/2)² (it's a perfect square now!), (6) Simplify right side, (7) Take ± square root of both sides, (8) Solve for x. Follow these 8 steps systematically and you'll get the right answer every time!
Complete the square to solve $2x^2+12x+10=0$ . (Do not use factoring or the quadratic formula.)
Answer: $x=-3\pm2$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! When a ≠ 1, like 2x² + 12x + 10 = 0: (1) First divide everything by 2: x² + 6x + 5 = 0. (2) Now complete the square on this simpler equation: x² + 6x = -5, add (6/2)² = 9 to both sides: x² + 6x + 9 = -5 + 9 = 4. (3) Factor left: (x + 3)² = 4. (4) Solve: x = -3 ± 2, so x = -1 or x = -5. Dividing by a first makes the completing square much cleaner! Choice B correctly completes the square to get (x + 3)² = 4 and solves to get x = -3 ± 2 with accurate arithmetic and proper form. Choice A calculates (b/2)² incorrectly: half of 6 is 3, and squaring gives 9, not something leading to ±√2; this perfect square term is the heart of completing the square—getting it right is crucial! An error here throws off everything that follows. Why this works: when you add (b/2)² to x² + bx, you're creating the expansion of (x + b/2)²: remember (a + b)² = a² + 2ab + b², so (x + b/2)² = x² + 2(x)(b/2) + (b/2)² = x² + bx + (b/2)². That's why adding (b/2)² makes a perfect square! It's not random—it completes the perfect square trinomial pattern.
Derive the quadratic formula by completing the square on the general equation $ax^2+bx+c=0$ (assume $a\ne0$ ). Which expression correctly gives the solutions for $x$ ?
Answer: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) actually comes FROM completing the square on the general equation ax² + bx + c = 0! If you complete the square on this general form (treating a, b, c as constants), you derive the quadratic formula. So completing the square isn't just one method among many—it's the fundamental method that gives us the formula itself! Starting with ax² + bx + c = 0: (1) Divide by a: x² + (b/a)x + (c/a) = 0. (2) Move constant: x² + (b/a)x = -c/a. (3) Complete square, half of b/a is b/(2a), squared is b²/(4a²): x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²). (4) Left side: (x + b/(2a))². Right side: (b² - 4ac)/(4a²). (5) So (x + b/(2a))² = (b² - 4ac)/(4a²). (6) Square root: x + b/(2a) = ±√(b² - 4ac)/(2a). (7) Solve: x = -b/(2a) ± √(b² - 4ac)/(2a) = (-b ± √(b² - 4ac))/(2a). That's the quadratic formula! Choice C correctly derives the formula x = (-b ± √(b² - 4ac))/(2a) with accurate arithmetic and proper form. Choice A has the wrong sign under the radical: it should be b² - 4ac, not b² + 4ac. The discriminant b² - 4ac comes from combining the fractions on the right side when completing the square. This sign is crucial for determining the nature of solutions! Why this works: when you add (b/2)² to x² + bx, you're creating the expansion of (x + b/2)²: remember (a + b)² = a² + 2ab + b², so (x + b/2)² = x² + 2(x)(b/2) + (b/2)² = x² + bx + (b/2)². That's why adding (b/2)² makes a perfect square! It's not random—it completes the perfect square trinomial pattern.
Transform and solve: $x^2 - 6x + 2 = 0$ . Use completing the square to write it as $(x - p)^2 = q$ and then find $x$ .
Answer: $(x-3)^2=7\;\Rightarrow\; x=3\pm\sqrt{7}$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 6x + 2 = 0 by completing the square: (1) Move constant to right: x² - 6x = -2. (2) Take half of -6 to get -3, square it to get 9, add to both sides: x² - 6x + 9 = -2 + 9. (3) Left side is perfect square (x - 3)², right side simplifies to 7: (x - 3)² = 7. (4) Take square roots: x - 3 = ±√7. (5) Solve: x = 3 ± √7. Done! Choice A correctly completes the square to get (x - 3)² = 7 and solves to get x = 3 ± √7 with accurate arithmetic and proper form. Choice C makes an arithmetic error when simplifying the right side: -2 + 9 = 7, not 11, so we should get (x - 3)² = 7, not (x - 3)² = 11. Always double-check your arithmetic when combining terms on the right side after adding (b/2)² to both sides! Why this works: when you add (b/2)² to x² + bx, you're creating the expansion of (x + b/2)²: remember (a + b)² = a² + 2ab + b², so (x + b/2)² = x² + 2(x)(b/2) + (b/2)² = x² + bx + (b/2)². That's why adding (b/2)² makes a perfect square! It's not random—it completes the perfect square trinomial pattern.
Solve using the method of completing the square: $x^2 - 8x + 12 = 0$ .
Answer: $x=4\pm\sqrt{4}$ , so $x=2$ or $x=6$ This question tests your ability to use completing the square to solve quadratic equations by transforming them into the form (x - p)² = q, which makes finding solutions straightforward by taking square roots. Completing the square is a method that transforms any quadratic equation ax² + bx + c = 0 into an equivalent equation (x - p)² = q that has the same solutions but is much easier to solve: once in this form, we just take the square root of both sides (remembering ±!), giving x - p = ±√q, then solve for x = p ± √q. This method works for ALL quadratics, even ones that don't factor nicely! Solving x² - 8x + 12 = 0 by completing the square: (1) Move constant to right: x² - 8x = -12. (2) Take half of -8 to get -4, square it to get 16, add to both sides: x² - 8x + 16 = -12 + 16. (3) Left side is perfect square (x - 4)², right side simplifies to 4: (x - 4)² = 4. (4) Take square roots: x - 4 = ±√4 = ±2. (5) Solve: x = 4 ± 2, so x = 6 or 2. Done! Choice A correctly solves to get x = 4 ± √4, so x = 2 or 6 with accurate arithmetic and proper form. Choice C has the right perfect square form but makes a sign error: from (x - 4)², the p is 4, but C uses -4. Watch those signs! When we have (x - 4)², we can rewrite as (x + (-4))², so p = 4. The sign in the factored form is opposite to the p value. The ± is crucial: from (x - 3)² = 16, taking square roots gives x - 3 = ±4 (both +4 and -4), so x = 3 + 4 = 7 OR x = 3 - 4 = -1. Two solutions! Don't forget the ± and don't forget to split it into two separate solutions. Check both: 7 and -1 both satisfy the original equation? Yes!

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