Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games


Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← Back to quizzes

Algebra Quiz

Algebra Quiz: Complete The Square To Find Solutions

Practice Complete The Square To Find Solutions in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 (with a≠0a\ne 0a=0). Which expression for xxx results?

Select an answer to continue

What this quiz covers

This quiz focuses on Complete The Square To Find Solutions, 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

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 (with a≠0a\ne 0a=0). Which expression for xxx results?

  1. x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​ (correct answer)
  2. x=−b±b2−2acax=\dfrac{-b\pm\sqrt{b^2-2ac}}{a}x=a−b±b2−2ac​​
  3. x=b±b2−4ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}x=2ab±b2−4ac​​
  4. x=−b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}x=2a−b±b2+4ac​​

Explanation: 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 algebra and proper signs throughout the derivation. Choice A has the wrong sign under the radical: when we move c/a to the right and add b²/(4a²), we get (b² - 4ac)/(4a²), not (b² + 4ac)/(4a²). The -4ac term comes from -c/a = -4ac/(4a²) when we find a common denominator. Sign errors in derivations are common but critical! 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.

Question 2

Solve the quadratic equation x2+8x+7=0x^2 + 8x + 7 = 0x2+8x+7=0 by completing the square. Give both solutions.

  1. x=−4±23x=-4\pm\sqrt{23}x=−4±23​
  2. x=−4±3x=-4\pm3x=−4±3 (correct answer)
  3. x=−8±7x=-8\pm\sqrt{7}x=−8±7​
  4. x=−4+3x=-4+3x=−4+3

Explanation: 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!

Question 3

Solve by completing the square: x2+2x+5=0x^2+2x+5=0x2+2x+5=0. What are the solutions?

  1. x=−1±2x=-1\pm2x=−1±2
  2. x=−1±2ix=-1\pm2ix=−1±2i (correct answer)
  3. x=1±2ix=1\pm2ix=1±2i
  4. x=−1±4x=-1\pm\sqrt{4}x=−1±4​

Explanation: 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!

Question 4

Transform and solve by completing the square: x2−6x+2=0x^2 - 6x + 2 = 0x2−6x+2=0. First write it in the form (x−p)2=q(x-p)^2=q(x−p)2=q, then find the solutions.

  1. (x−6)2=2  ⇒  x=6±2(x-6)^2=2\;\Rightarrow\;x=6\pm\sqrt{2}(x−6)2=2⇒x=6±2​
  2. (x−3)2=7  ⇒  x=3±7(x-3)^2=7\;\Rightarrow\;x=3\pm\sqrt{7}(x−3)2=7⇒x=3±7​ (correct answer)
  3. (x+3)2=7  ⇒  x=−3±7(x+3)^2=7\;\Rightarrow\;x=-3\pm\sqrt{7}(x+3)2=7⇒x=−3±7​
  4. (x−3)2=11  ⇒  x=3±11(x-3)^2=11\;\Rightarrow\;x=3\pm\sqrt{11}(x−3)2=11⇒x=3±11​

Explanation: 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!

Question 5

What is the completed square form of x2+4x−5=0x^2+4x-5=0x2+4x−5=0 written as (x−p)2=q(x-p)^2=q(x−p)2=q?

  1. (x+2)2=1(x+2)^2=1(x+2)2=1
  2. (x+2)2=9(x+2)^2=9(x+2)2=9 (correct answer)
  3. (x−2)2=−9(x-2)^2=-9(x−2)2=−9
  4. (x−2)2=9(x-2)^2=9(x−2)2=9

Explanation: 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 + 4x - 5 = 0 by completing the square: (1) Move constant to right: x² + 4x = 5. (2) Take half of 4 to get 2, square it to get 4, add to both sides: x² + 4x + 4 = 5 + 4. (3) Left side is perfect square (x + 2)², right side simplifies to 9: (x + 2)² = 9. (4) Take square roots: x + 2 = ±3. (5) Solve: x = -2 ± 3. Done! Choice C correctly completes the square to get (x + 2)² = 9 with accurate arithmetic and proper form. Choice D doesn't add (b/2)² to both sides—only adds to one side or miscalculates, leading to negative q incorrectly! Completing the square requires adding the perfect square term to BOTH sides to maintain equality. If you only add to the left, you've changed the equation and will get wrong solutions. Both sides, always! 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!

Question 6

Solve x2+6x+5=0x^2 + 6x + 5 = 0x2+6x+5=0 by completing the square. Give the solutions for xxx.

  1. x=−3±2x=-3\pm 2x=−3±2, so x=−1x=-1x=−1 or x=−5x=-5x=−5 (correct answer)
  2. x=3±2x=3\pm 2x=3±2, so x=1x=1x=1 or x=5x=5x=5
  3. x=−3±5x=-3\pm \sqrt{5}x=−3±5​
  4. x=−3x=-3x=−3 only

Explanation: 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 + 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, giving x = -3 + 2 = -1 or x = -3 - 2 = -5. Done! Choice A correctly completes the square to get (x + 3)² = 4 and solves to get x = -3 ± 2, which gives x = -1 or x = -5, with accurate arithmetic and proper form. Choice C makes an arithmetic error when simplifying the right side: -5 + 9 = 4, not 5, so we should get (x + 3)² = 4, not (x + 3)² = 5. Completing the square involves several arithmetic steps, so take your time and double-check! 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!

Question 7

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 (assume a≠0a\ne0a=0). Which expression correctly gives the solutions for xxx?

  1. x=−b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}x=2a−b±b2+4ac​​
  2. x=b±b2−4ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}x=2ab±b2−4ac​​
  3. x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​ (correct answer)
  4. x=−b±b2−2acax=\dfrac{-b\pm\sqrt{b^2-2ac}}{a}x=a−b±b2−2ac​​

Explanation: 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.

Question 8

Transform and solve: x2−6x+2=0x^2 - 6x + 2 = 0x2−6x+2=0. Use completing the square to write it as (x−p)2=q(x - p)^2 = q(x−p)2=q and then find xxx.

  1. (x+3)2=7  ⇒  x=−3±7(x+3)^2=7\;\Rightarrow\; x=-3\pm\sqrt{7}(x+3)2=7⇒x=−3±7​
  2. (x−3)2=7  ⇒  x=3±7(x-3)^2=7\;\Rightarrow\; x=3\pm\sqrt{7}(x−3)2=7⇒x=3±7​ (correct answer)
  3. (x−6)2=2  ⇒  x=6±2(x-6)^2=2\;\Rightarrow\; x=6\pm\sqrt{2}(x−6)2=2⇒x=6±2​
  4. (x−3)2=11  ⇒  x=3±11(x-3)^2=11\;\Rightarrow\; x=3\pm\sqrt{11}(x−3)2=11⇒x=3±11​

Explanation: 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.

Question 9

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 (assume a≠0a\ne0a=0). Which expression for xxx results?

  1. x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​ (correct answer)
  2. x=−b±b2−2acax=\dfrac{-b\pm\sqrt{b^2-2ac}}{a}x=a−b±b2−2ac​​
  3. x=b±b2−4ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}x=2ab±b2−4ac​​
  4. x=−b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}x=2a−b±b2+4ac​​

Explanation: 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!

Question 10

Solve 2x2+12x+10=02x^2 + 12x + 10 = 02x2+12x+10=0 by completing the square (divide first if needed).

  1. x=−3±2x=-3\pm\sqrt{2}x=−3±2​
  2. x=−1x=-1x=−1 or x=−5x=-5x=−5 (correct answer)
  3. x=3±2x=3\pm\sqrt{2}x=3±2​
  4. x=−3±2x=-3\pm 2x=−3±2

Explanation: 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 = -3 ± 2, giving x = -1 or x = -5. Dividing by a first makes the completing square much cleaner! Choice C correctly solves to get x = -1 or x = -5 with accurate arithmetic and proper form. Choice A forgets to simplify after taking the square root: from (x + 3)² = 4, we get x + 3 = ±2 (not ±√4), since √4 = 2. Always simplify radicals: √4 = 2, √9 = 3, √16 = 4, etc. Don't leave them as √4 when they simplify to whole numbers! 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!

Question 11

Solve the quadratic equation x2+8x+7=0x^2 + 8x + 7 = 0x2+8x+7=0 by completing the square. Give the solutions for xxx.

  1. x=−1, −7x=-1,\,-7x=−1,−7 (correct answer)
  2. x=4±9x=4\pm\sqrt{9}x=4±9​
  3. x=−4±9x=-4\pm\sqrt{9}x=−4±9​
  4. x=−4±7x=-4\pm\sqrt{7}x=−4±7​

Explanation: 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 = -1 or -7. Done! Choice A correctly solves to get x = -1 and x = -7 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 in (x - p)² is -4, not +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 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!

Question 12

Show that the equations x2+4x−5=0x^2 + 4x - 5 = 0x2+4x−5=0 and (x+2)2=9(x+2)^2 = 9(x+2)2=9 have the same solutions. Which set of solutions do they share?

  1. {−5, 1}\{-5,\,1\}{−5,1} (correct answer)
  2. {−1, 5}\{-1,\,5\}{−1,5}
  3. {−2, 2}\{-2,\,2\}{−2,2}
  4. {1}\{1\}{1}

Explanation: 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! Solving x² + 4x - 5 = 0 by completing the square: (1) Move constant to right: x² + 4x = 5. (2) Take half of 4 to get 2, square it to get 4, add to both sides: x² + 4x + 4 = 5 + 4. (3) Left side is perfect square (x + 2)², right side simplifies to 9: (x + 2)² = 9. (4) Take square roots: x + 2 = ±3. (5) Solve: x = -2 ± 3 = 1 or -5. Done! This matches the given (x + 2)² = 9 directly. Choice A correctly identifies the shared solutions as {-5, 1} with accurate arithmetic and proper form. Choice B has a sign error in the solutions: from x = -2 ± 3, it's -2 + 3 = 1 and -2 - 3 = -5, not {-1, 5}. Watch those signs when solving after taking square roots! 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!

Question 13

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 (assume a≠0a\ne 0a=0). Which expression for xxx results?

  1. x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​ (correct answer)
  2. x=b±b2−4ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}x=2ab±b2−4ac​​
  3. x=−b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}x=2a−b±b2+4ac​​
  4. x=−b±(b−2a)2−4acax=\dfrac{-b\pm\sqrt{(b-2a)^2-4ac}}{a}x=a−b±(b−2a)2−4ac​​

Explanation: 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.

Question 14

Two students complete the square for the same equation x2−8x+7=0x^2 - 8x + 7 = 0x2−8x+7=0. Student A writes (x−4)2=9(x - 4)^2 = 9(x−4)2=9 and Student B writes (x−4)2−9=0(x - 4)^2 - 9 = 0(x−4)2−9=0. Which statement about their work is correct?

  1. Only Student B's form is correct because all terms should be on one side
  2. Only Student A's form is correct because it matches the required format
  3. Both forms are mathematically equivalent and correct for solving the equation (correct answer)
  4. Both students made errors because the correct form should be (x+4)2=9(x + 4)^2 = 9(x+4)2=9

Explanation: 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 x2−8x+7=0x^2 - 8x + 7 = 0x2−8x+7=0, you take half of the coefficient of xxx (which is −8/2=−4-8/2 = -4−8/2=−4), square it to get 161616, then add and subtract this value: x2−8x+16−16+7=0x^2 - 8x + 16 - 16 + 7 = 0x2−8x+16−16+7=0. This gives you (x−4)2−9=0(x - 4)^2 - 9 = 0(x−4)2−9=0, which can be rearranged to (x−4)2=9(x - 4)^2 = 9(x−4)2=9. Both forms are mathematically sound. Student A's form (x−4)2=9(x - 4)^2 = 9(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(x - 4)^2 - 9 = 0(x−4)2−9=0 keeps everything on one side, which some prefer. Both lead to the same solutions: x=7x = 7x=7 or x=1x = 1x=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-8x−8x, we need (x−4)2(x - 4)^2(x−4)2, not (x+4)2(x + 4)^2(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.

Question 15

A quadratic equation is given as x2−6x+2=0x^2 - 6x + 2 = 0x2−6x+2=0. When completing the square to transform this equation into the form (x−p)2=q(x - p)^2 = q(x−p)2=q, what is the value of qqq?

  1. 777 (correct answer)
  2. 999
  3. −7-7−7
  4. 111111

Explanation: To complete the square for x2−6x+2=0x^2 - 6x + 2 = 0x2−6x+2=0, first move the constant: x2−6x=−2x^2 - 6x = -2x2−6x=−2. Take half of the coefficient of xxx and square it: (−6/2)2=9(-6/2)^2 = 9(−6/2)2=9. Add 9 to both sides: x2−6x+9=−2+9=7x^2 - 6x + 9 = -2 + 9 = 7x2−6x+9=−2+9=7. This gives (x−3)2=7(x - 3)^2 = 7(x−3)2=7, so q=7q = 7q=7. Choice B incorrectly uses the value added to complete the square. Choice C uses −7-7−7 from a sign error. Choice D results from adding instead of subtracting the constant term.

Question 16

The quadratic equation x2−4x+k=0x^2 - 4x + k = 0x2−4x+k=0 is transformed by completing the square to (x−2)2=7(x - 2)^2 = 7(x−2)2=7. What is the value of kkk, and what are the solutions to the original equation?

  1. k=11k = 11k=11; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​
  2. k=3k = 3k=3; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​
  3. k=−3k = -3k=−3; solutions are x=−2±7x = -2 \pm \sqrt{7}x=−2±7​
  4. k=−3k = -3k=−3; solutions are x=2±7x = 2 \pm \sqrt{7}x=2±7​ (correct answer)

Explanation: When you encounter a quadratic equation that's been transformed by completing the square, you need to work backwards to find the original coefficient and then solve using the completed square form. Starting with x2−4x+k=0x^2 - 4x + k = 0x2−4x+k=0, let's complete the square ourselves. Take half of the coefficient of xxx (which is −4-4−4), square it: (−4/2)2=(−2)2=4(-4/2)^2 = (-2)^2 = 4(−4/2)2=(−2)2=4. So we rewrite as x2−4x+4+k−4=0x^2 - 4x + 4 + k - 4 = 0x2−4x+4+k−4=0, which becomes (x−2)2+k−4=0(x - 2)^2 + k - 4 = 0(x−2)2+k−4=0, or (x−2)2=4−k(x - 2)^2 = 4 - k(x−2)2=4−k. Since we're told this equals (x−2)2=7(x - 2)^2 = 7(x−2)2=7, we have 4−k=74 - k = 74−k=7, so k=−3k = -3k=−3. To find the solutions, we solve (x−2)2=7(x - 2)^2 = 7(x−2)2=7 by taking the square root of both sides: x−2=±7x - 2 = \pm\sqrt{7}x−2=±7​, giving us x=2±7x = 2 \pm \sqrt{7}x=2±7​. Looking at the wrong answers: Choice A incorrectly calculates k=11k = 11k=11 (likely from 4+k=74 + k = 74+k=7). Choice B gets k=3k = 3k=3 (perhaps from k−4=7k - 4 = 7k−4=7) but has the right solutions. Choice C has the correct k=−3k = -3k=−3 but wrong solutions x=−2±7x = -2 \pm \sqrt{7}x=−2±7​, missing that the vertex form shows (x−2)(x - 2)(x−2), not (x+2)(x + 2)(x+2). Study tip: When completing the square, always double-check your work by expanding the completed square form back to standard form. This catches sign errors and helps you verify both the constant term and the solutions.

Question 17

When completing the square for x2+bx+c=0x^2 + bx + c = 0x2+bx+c=0, a student obtains (x+3)2=11(x + 3)^2 = 11(x+3)2=11. If the original equation had integer coefficients, which of the following could be the original equation?

  1. x2+3x−11=0x^2 + 3x - 11 = 0x2+3x−11=0
  2. x2−6x−2=0x^2 - 6x - 2 = 0x2−6x−2=0
  3. x2+6x+2=0x^2 + 6x + 2 = 0x2+6x+2=0
  4. x2+6x−2=0x^2 + 6x - 2 = 0x2+6x−2=0 (correct answer)

Explanation: When you see a question about completing the square, you need to work backwards from the completed square form to find the original equation. This tests your understanding of how the completing the square process transforms equations. Starting with (x+3)2=11(x + 3)^2 = 11(x+3)2=11, you can expand the left side to find the original form. Remember that (x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9(x+3)2=x2+6x+9. So the equation becomes: x2+6x+9=11x^2 + 6x + 9 = 11x2+6x+9=11 Subtracting 11 from both sides gives you the standard form: x2+6x+9−11=0x^2 + 6x + 9 - 11 = 0x2+6x+9−11=0 x2+6x−2=0x^2 + 6x - 2 = 0x2+6x−2=0 This matches choice D exactly. Let's check why the other options don't work. Choice A gives x2+3x−11=0x^2 + 3x - 11 = 0x2+3x−11=0. When completing the square, you'd get (x+32)2=534(x + \frac{3}{2})^2 = \frac{53}{4}(x+23​)2=453​, not (x+3)2=11(x + 3)^2 = 11(x+3)2=11. Choice B gives x2−6x−2=0x^2 - 6x - 2 = 0x2−6x−2=0, which completes to (x−3)2=11(x - 3)^2 = 11(x−3)2=11 – close, but the sign inside the parentheses is wrong. Choice C gives x2+6x+2=0x^2 + 6x + 2 = 0x2+6x+2=0, which completes to (x+3)2=7(x + 3)^2 = 7(x+3)2=7, not 11. Study tip: When working with completing the square problems, always verify your answer by expanding the completed square form back to standard form. The coefficient of xxx in the original equation should be twice the number inside the parentheses of the completed square form.

Question 18

The equation x2+kx+12=0x^2 + kx + 12 = 0x2+kx+12=0 is completed to the form (x+k2)2=q(x + \frac{k}{2})^2 = q(x+2k​)2=q. If this quadratic equation has exactly one solution, what is the value of kkk and what is that solution?

  1. k=0k = 0k=0; solution is x=±23x = \pm 2\sqrt{3}x=±23​
  2. k=±43k = \pm 4\sqrt{3}k=±43​; solution is x=∓23x = \mp 2\sqrt{3}x=∓23​ (correct answer)
  3. k=±43k = \pm 4\sqrt{3}k=±43​; solution is x=0x = 0x=0
  4. k=12k = 12k=12; solution is x=−6x = -6x=−6

Explanation: When you see a quadratic that "has exactly one solution," this means the discriminant equals zero, creating a perfect square trinomial. The key insight is understanding what happens when you complete the square and how this connects to the number of solutions. Starting with x2+kx+12=0x^2 + kx + 12 = 0x2+kx+12=0, let's complete the square. Moving the constant term: x2+kx=−12x^2 + kx = -12x2+kx=−12. Adding (k2)2(\frac{k}{2})^2(2k​)2 to both sides: (x+k2)2=−12+k24(x + \frac{k}{2})^2 = -12 + \frac{k^2}{4}(x+2k​)2=−12+4k2​. This gives us q=k24−12q = \frac{k^2}{4} - 12q=4k2​−12. For exactly one solution, the right side must equal zero: k24−12=0\frac{k^2}{4} - 12 = 04k2​−12=0. Solving: k24=12\frac{k^2}{4} = 124k2​=12, so k2=48k^2 = 48k2=48, which means k=±43k = \pm 4\sqrt{3}k=±43​. When q=0q = 0q=0, we have (x+k2)2=0(x + \frac{k}{2})^2 = 0(x+2k​)2=0, giving us x=−k2x = -\frac{k}{2}x=−2k​. Since k=±43k = \pm 4\sqrt{3}k=±43​, the solution is x=∓23x = \mp 2\sqrt{3}x=∓23​ (note the sign change). Choice A incorrectly assumes k=0k = 0k=0, which would give q=−12q = -12q=−12, resulting in no real solutions. Choice C correctly finds k=±43k = \pm 4\sqrt{3}k=±43​ but claims the solution is x=0x = 0x=0, which ignores that x=−k2≠0x = -\frac{k}{2} \neq 0x=−2k​=0. Choice D uses k=12k = 12k=12, but this gives q=24q = 24q=24, resulting in two solutions, not one. Remember: "exactly one solution" means the discriminant is zero, which creates a perfect square. Always check that your final answer produces the required number of solutions.

Question 19

A student is completing the square for x2+5x−3=0x^2 + 5x - 3 = 0x2+5x−3=0 and writes the intermediate step: x2+5x+254=3+254x^2 + 5x + \frac{25}{4} = 3 + \frac{25}{4}x2+5x+425​=3+425​. What should be the next step in their work?

  1. Write (x+52)2=374(x + \frac{5}{2})^2 = \frac{37}{4}(x+25​)2=437​ and solve for xxx (correct answer)
  2. Write (x−52)2=374(x - \frac{5}{2})^2 = \frac{37}{4}(x−25​)2=437​ and solve for xxx
  3. Simplify the right side to get 124+254=374\frac{12}{4} + \frac{25}{4} = \frac{37}{4}412​+425​=437​
  4. Write (x+52)2=124+254(x + \frac{5}{2})^2 = \frac{12}{4} + \frac{25}{4}(x+25​)2=412​+425​ without simplifying

Explanation: The student correctly added 254\frac{25}{4}425​ to both sides. The left side factors as (x+52)2(x + \frac{5}{2})^2(x+25​)2 since we added (52)2=254(\frac{5}{2})^2 = \frac{25}{4}(25​)2=425​. The right side simplifies to 3+254=124+254=3743 + \frac{25}{4} = \frac{12}{4} + \frac{25}{4} = \frac{37}{4}3+425​=412​+425​=437​. The next step is to write the factored form (x+52)2=374(x + \frac{5}{2})^2 = \frac{37}{4}(x+25​)2=437​. Choice B has the wrong sign. Choice C focuses on arithmetic rather than the algebraic transformation. Choice D leaves the work incomplete.

Question 20

A teacher asks students to derive the quadratic formula by completing the square on ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0. After dividing by aaa and completing the square, a student obtains (x+b2a)2=b2−4ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}(x+2ab​)2=4a2b2−4ac​. What is the next step to arrive at the quadratic formula?

  1. Take the square root of both sides to get x+b2a=±b2−4ac2∣a∣x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2|a|}x+2ab​=±2∣a∣b2−4ac​​
  2. Take the square root of both sides to get x+b2a=±b2−4ac2ax + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}x+2ab​=±2ab2−4ac​​ (correct answer)
  3. Multiply both sides by 4a24a^24a2 to eliminate the denominator on the right side
  4. Expand the left side and rearrange to standard form before taking the square root

Explanation: 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+b2a=±b2−4ac2ax + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}x+2ab​=±2ab2−4ac​​, 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 4a24a^24a2 is 2∣a∣2|a|2∣a∣, but since we're in the context of deriving a general formula, we write this as 2a2a2a with the understanding that the ±\pm± symbol will account for both positive and negative cases. Choice A incorrectly includes absolute value bars around the aaa in the denominator. While 4a2=2∣a∣\sqrt{4a^2} = 2|a|4a2​=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 4a24a^24a2, 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.