ACCUPLACER Advanced Algebra & Functions Quiz: Completing The Square
20 questions · exam conditions
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Completing The SquareQuestion 1 of 20

After completing the square, the expression x2+bx+15x^2 + bx + 15 becomes (x+4)2+k(x + 4)^2 + k. What are the values of bb and kk?

b=4,k=−1b = 4, k = -1
b=8,k=−1b = 8, k = -1
b=8,k=31b = 8, k = 31
b=4,k=31b = 4, k = 31
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Completing The Square

Practice Completing The Square in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Completing The Square, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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

After completing the square, the expression x2+bx+15x^2 + bx + 15 becomes (x+4)2+k(x + 4)^2 + k. What are the values of bb and kk?

  1. b=4,k=−1b = 4, k = -1
  2. b=8,k=−1b = 8, k = -1 (correct answer)
  3. b=8,k=31b = 8, k = 31
  4. b=4,k=31b = 4, k = 31
Explanation: When you encounter completing the square problems, you're working backwards from a completed square form to find the original expression's coefficients. To find bb and kk, expand (x+4)2+k(x + 4)^2 + k and match it to x2+bx+15x^2 + bx + 15. Expanding: (x+4)2+k=x2+8x+16+k(x + 4)^2 + k = x^2 + 8x + 16 + k For this to equal x2+bx+15x^2 + bx + 15, the coefficients must match:
  • The xx coefficient: b=8b = 8
  • The constant term: 16+k=1516 + k = 15, so k=−1k = -1
Let's examine why the other answers are incorrect: Choice A (b=4,k=−1b = 4, k = -1) uses the number inside the parentheses (4) as the coefficient of xx, but when you expand (x+4)2(x + 4)^2, you get 8x8x, not 4x4x. Choice C (b=8,k=31b = 8, k = 31) correctly identifies b=8b = 8 but calculates kk incorrectly. This might come from adding instead of subtracting: 16+15=3116 + 15 = 31 rather than solving 16+k=1516 + k = 15. Choice D (b=4,k=31b = 4, k = 31) combines both errors from choices A and C—using 4 as the xx-coefficient and miscalculating kk. The key strategy here is to always expand the completed square form and match coefficients term by term. Remember that (x+a)2=x2+2ax+a2(x + a)^2 = x^2 + 2ax + a^2, so the coefficient of xx is always twice the number in the parentheses.

Question 2

The expression −x2+8x−10-x^2 + 8x - 10 can be written in the vertex form −(x−h)2+k-(x-h)^2 + k. What is the value of h+kh+k?

  1. 2
  2. 10 (correct answer)
  3. 12
  4. 30
Explanation: First, factor out -1 from the first two terms: −(x2−8x)−10-(x^2 - 8x) - 10. To complete the square, take half of -8 (which is -4) and square it to get 16. Add and subtract 16 inside the parentheses: −(x2−8x+16−16)−10-(x^2 - 8x + 16 - 16) - 10. This becomes −((x−4)2−16)−10-((x-4)^2 - 16) - 10. Distribute the negative sign: −(x−4)2+16−10-(x-4)^2 + 16 - 10, which simplifies to −(x−4)2+6-(x-4)^2 + 6. From this form, h=4h=4 and k=6k=6. The sum h+kh+k is 4+6=104+6 = 10.

Question 3

If x2+bx+25x^2 + bx + 25 is a perfect square trinomial and b>0b > 0, what is the value of bb?

  1. 5
  2. 10 (correct answer)
  3. 25
  4. 50
Explanation: A perfect square trinomial has the form x2+2kx+k2=(x+k)2x^2 + 2kx + k^2 = (x+k)^2. In the expression x2+bx+25x^2 + bx + 25, the constant term is 2525, which is 525^2. So, k2=25k^2=25, which means k=5k=5 or k=−5k=-5. The middle term is b=2kb = 2k. Since we are given that b>0b > 0, we must use k=5k=5. Therefore, b=2(5)=10b = 2(5) = 10.

Question 4

The expression x2+18x+cx^2 + 18x + c is equivalent to (x+h)2(x+h)^2. What is the value of c−hc-h?

  1. 9
  2. 72 (correct answer)
  3. 90
  4. 153
Explanation: For x2+18x+cx^2 + 18x + c to be a perfect square (x+h)2(x+h)^2, we must have 2h=182h = 18, so h=9h=9. The constant term cc must be h2h^2, so c=92=81c = 9^2 = 81. The question asks for the value of c−hc-h, which is 81−9=7281 - 9 = 72.

Question 5

When the expression 4x2+8x−54x^2 + 8x - 5 is written in the form 4(x−h)2+k4(x-h)^2 + k, what is the value of kk?

  1. -9 (correct answer)
  2. -6
  3. -4
  4. -1
Explanation: First, factor out 4 from the first two terms: 4(x2+2x)−54(x^2 + 2x) - 5. To complete the square for x2+2xx^2 + 2x, we add (2/2)2=12=1(2/2)^2 = 1^2 = 1 inside the parentheses. This gives 4(x2+2x+1)−54(x^2 + 2x + 1) - 5. However, by adding 1 inside the parentheses, we have effectively added 4×1=44 \times 1 = 4 to the expression. To keep it equivalent, we must subtract 4: 4(x2+2x+1)−5−44(x^2 + 2x + 1) - 5 - 4. This simplifies to 4(x+1)2−94(x+1)^2 - 9. Thus, the value of kk is -9.

Question 6

The expression x2+kx+12x^2 + kx + 12 is rewritten in the form (x+h)2+3(x+h)^2 + 3. If k>0k > 0, what is the value of kk?

  1. 3
  2. 6 (correct answer)
  3. 9
  4. 12
Explanation: Expand the vertex form: (x+h)2+3=x2+2hx+h2+3(x+h)^2 + 3 = x^2 + 2hx + h^2 + 3. Compare the coefficients of this expanded form to the original expression x2+kx+12x^2 + kx + 12. The constant terms must be equal: h2+3=12h^2 + 3 = 12. Solving for h2h^2 gives h2=9h^2 = 9, so h=3h = 3 or h=−3h = -3. The coefficients of the xx-term must also be equal: k=2hk = 2h. Since the problem states that k>0k > 0, we must have h>0h > 0, so we choose h=3h=3. Therefore, k=2(3)=6k = 2(3) = 6.

Question 7

If y=x2+5x−2y = x^2 + 5x - 2 is rewritten in the form y=(x+h)2+ky = (x+h)^2 + k, what is the value of kk?

  1. -33/4 (correct answer)
  2. -25/4
  3. 3/2
  4. 17/4
Explanation: To complete the square for x2+5x−2x^2 + 5x - 2, take half of the xx-coefficient (5), which is 5/25/2, and square it: (5/2)2=25/4(5/2)^2 = 25/4. Add and subtract this value: (x2+5x+25/4)−25/4−2(x^2 + 5x + 25/4) - 25/4 - 2. The expression in parentheses is (x+5/2)2(x + 5/2)^2. The constants are −25/4−2=−25/4−8/4=−33/4-25/4 - 2 = -25/4 - 8/4 = -33/4. So, the vertex form is y=(x+5/2)2−33/4y = (x+5/2)^2 - 33/4. The value of kk is −33/4-33/4.

Question 8

Which of the following expressions is equivalent to x2−x+1x^2 - x + 1 after completing the square?

  1. (x−1/2)2+3/4(x - 1/2)^2 + 3/4 (correct answer)
  2. (x−1/2)2+5/4(x - 1/2)^2 + 5/4
  3. (x−1)2+0(x - 1)^2 + 0
  4. (x+1/2)2+3/4(x + 1/2)^2 + 3/4
Explanation: For the expression x2−x+1x^2 - x + 1, the coefficient of the xx-term is -1. Half of -1 is −1/2-1/2, and squaring this gives (−1/2)2=1/4(-1/2)^2 = 1/4. Rewrite the expression by adding and subtracting 1/4: (x2−x+1/4)−1/4+1(x^2 - x + 1/4) - 1/4 + 1. The terms in the parentheses form a perfect square: (x−1/2)2(x - 1/2)^2. The constants combine to −1/4+1=−1/4+4/4=3/4-1/4 + 1 = -1/4 + 4/4 = 3/4. The equivalent expression is (x−1/2)2+3/4(x - 1/2)^2 + 3/4.

Question 9

When 5x2−10x+125x^2 - 10x + 12 is written in vertex form a(x−h)2+ka(x-h)^2+k, what is the product hkhk?

  1. -7
  2. 5
  3. 7 (correct answer)
  4. 12
Explanation: First, factor out the leading coefficient, 5: 5(x2−2x)+125(x^2 - 2x) + 12. Complete the square inside the parentheses by adding (−2/2)2=1(-2/2)^2 = 1. To balance the expression, subtract 5×1=55 \times 1 = 5 from the constant term: 5(x2−2x+1)+12−55(x^2 - 2x + 1) + 12 - 5. This simplifies to 5(x−1)2+75(x-1)^2 + 7. From this form, we identify h=1h=1 and k=7k=7. The product hkhk is 1×7=71 \times 7 = 7.

Question 10

The function f(x)=2x2+8x−1f(x) = 2x^2 + 8x - 1 is rewritten in vertex form f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k. What is the value of the sum a+h+ka+h+k?

  1. -9 (correct answer)
  2. -5
  3. 1
  4. 9
Explanation: The value of aa is the leading coefficient, so a=2a=2. Factor 2 from the first two terms: 2(x2+4x)−12(x^2+4x) - 1. Complete the square for x2+4xx^2+4x by adding (4/2)2=4(4/2)^2 = 4. We get 2(x2+4x+4)−1−2(4)2(x^2+4x+4) - 1 - 2(4). This simplifies to 2(x+2)2−1−8=2(x+2)2−92(x+2)^2 - 1 - 8 = 2(x+2)^2 - 9. From this form, h=−2h=-2 and k=−9k=-9. The sum is a+h+k=2+(−2)+(−9)=−9a+h+k = 2 + (-2) + (-9) = -9.

Question 11

What constant must be added to the expression x2−7xx^2 - 7x to create a perfect square trinomial?

  1. -49/4
  2. -7/2
  3. 7/2
  4. 49/4 (correct answer)
Explanation: To complete the square for an expression x2+bxx^2 + bx, we must add the square of half the coefficient of xx. In this case, b=−7b = -7. Half of -7 is −7/2-7/2. Squaring this value gives (−7/2)2=49/4(-7/2)^2 = 49/4. Adding this constant creates the perfect square trinomial x2−7x+49/4x^2 - 7x + 49/4, which is equal to (x−7/2)2(x - 7/2)^2.

Question 12

To convert the quadratic function y=2x2−20x+15y = 2x^2 - 20x + 15 to vertex form, a student's first step is to factor out 2, resulting in y=2(x2−10x)+15y = 2(x^2 - 10x) + 15. What value must be added and subtracted inside the parentheses to complete the square?

  1. 5
  2. 10
  3. 25 (correct answer)
  4. 100
Explanation: The expression inside the parentheses is x2−10xx^2 - 10x. To complete the square, take half of the coefficient of the xx-term, which is −10/2=−5-10/2 = -5. Then, square this result: (−5)2=25(-5)^2 = 25. This is the value that must be added inside the parentheses to form a perfect square trinomial. The student would write y=2(x2−10x+25−25)+15y = 2(x^2 - 10x + 25 - 25) + 15.

Question 13

Which expression represents the correct first step in completing the square for 4x2−20x+94x^2 - 20x + 9?

  1. 4(x2−20x)+94(x^2 - 20x) + 9
  2. 4(x2−5x+9)4(x^2 - 5x + 9)
  3. (4x)2−20x+9(4x)^2 - 20x + 9
  4. 4(x2−5x)+94(x^2 - 5x) + 9 (correct answer)
Explanation: When completing the square for a quadratic expression, your first step is always to factor out the coefficient of the squared term from the first two terms only. This isolates the variable terms so you can work with a simpler expression inside the parentheses. For 4x2−20x+94x^2 - 20x + 9, you need to factor out the 4 from just the x2x^2 and xx terms. When you factor 4 out of 4x2−20x4x^2 - 20x, you get 4(x2−5x)4(x^2 - 5x). The constant term 9 stays separate, giving you 4(x2−5x)+94(x^2 - 5x) + 9. This matches answer choice D. Let's see why the other options are incorrect. Choice A factors out 4 from 4x2−20x4x^2 - 20x, but incorrectly writes 4(x2−20x)4(x^2 - 20x) instead of 4(x2−5x)4(x^2 - 5x). When you factor 4 from −20x-20x, you get −5x-5x, not −20x-20x. Choice B attempts to factor the entire expression at once as 4(x2−5x+9)4(x^2 - 5x + 9), but this doesn't equal the original expression since 4×9=364 \times 9 = 36, not 9. Choice C rewrites 4x24x^2 as (4x)2(4x)^2, which is mathematically correct but doesn't help with completing the square—you need the coefficient outside the parentheses. Remember: when completing the square, always factor out the leading coefficient from only the first two terms. Keep the constant term separate until the final step. This sets you up to easily complete the square inside the parentheses.

Question 14

When rewriting the equation 3x2−6x=123x^2 - 6x = 12 in the form (x−h)2=k(x-h)^2 = k, what is the value of kk?

  1. 1
  2. 3
  3. 5 (correct answer)
  4. 13
Explanation: First, divide the entire equation by the leading coefficient, 3, to get x2−2x=4x^2 - 2x = 4. Now, complete the square on the left side. Half of the xx-coefficient (-2) is -1, and (−1)2=1(-1)^2 = 1. Add 1 to both sides of the equation: x2−2x+1=4+1x^2 - 2x + 1 = 4 + 1. This simplifies to (x−1)2=5(x-1)^2 = 5. Therefore, the value of kk is 5.

Question 15

A quadratic function is rewritten as y=(x+3.5)2−1.25y = (x+3.5)^2 - 1.25. Which of the following was the original function in standard form?

  1. y=x2+7x+11y = x^2 + 7x + 11 (correct answer)
  2. y=x2+3.5x−1.25y = x^2 + 3.5x - 1.25
  3. y=x2+7x+13.5y = x^2 + 7x + 13.5
  4. y=x2−7x+11y = x^2 - 7x + 11
Explanation: Expand the vertex form y=(x+3.5)2−1.25y = (x+3.5)^2 - 1.25. The term (x+3.5)2(x+3.5)^2 expands to x2+2(3.5)x+(3.5)2=x2+7x+12.25x^2 + 2(3.5)x + (3.5)^2 = x^2 + 7x + 12.25. Now, subtract the constant: y=(x2+7x+12.25)−1.25y = (x^2 + 7x + 12.25) - 1.25. This simplifies to y=x2+7x+11y = x^2 + 7x + 11.

Question 16

The first step in solving the equation x2+20x=4x^2+20x=4 by completing the square is to add a constant to both sides. What is that constant?

  1. 10
  2. 20
  3. 100 (correct answer)
  4. 400
Explanation: To complete the square for the expression x2+20xx^2+20x, we take half of the coefficient of the xx-term (20), which is 10, and square it. 102=10010^2 = 100. This is the constant that must be added to both sides of the equation to create a perfect square trinomial on the left side.

Question 17

A student is rewriting the expression x2−12x+5x^2 - 12x + 5 by completing the square. The result is in the form (x−h)2+k(x-h)^2 + k. What is the value of kk?

  1. -31 (correct answer)
  2. -11
  3. 5
  4. 41
Explanation: To complete the square for x2−12x+5x^2 - 12x + 5, take half of the coefficient of the xx-term (which is -12) to get -6, and then square it: (−6)2=36(-6)^2 = 36. Add and subtract 36 to the expression: (x2−12x+36)−36+5(x^2 - 12x + 36) - 36 + 5. The part in the parentheses is a perfect square: (x−6)2(x-6)^2. Combining the constants gives −36+5=−31-36 + 5 = -31. So the expression is (x−6)2−31(x-6)^2 - 31. The value of kk is -31.

Question 18

By completing the square, the equation x2−10x=−1x^2 - 10x = -1 can be rewritten as (x−p)2=q(x-p)^2 = q. What is the value of qq?

  1. 9
  2. 11
  3. 24 (correct answer)
  4. 26
Explanation: The equation is x2−10x=−1x^2 - 10x = -1. To complete the square on the left side, take half of the xx-coefficient (-10), which is -5, and square it: (−5)2=25(-5)^2 = 25. Add 25 to both sides of the equation: x2−10x+25=−1+25x^2 - 10x + 25 = -1 + 25. The left side becomes a perfect square: (x−5)2=24(x-5)^2 = 24. In the form (x−p)2=q(x-p)^2 = q, we have p=5p=5 and q=24q=24. The value of qq is 24.

Question 19

If the expression 2x2+12x+82x^2 + 12x + 8 is written in the form a(x−h)2+ka(x-h)^2 + k, what is the value of hh?

  1. -6
  2. -3 (correct answer)
  3. 3
  4. 6
Explanation: First, factor out the leading coefficient, 2, from the first two terms: 2(x2+6x)+82(x^2 + 6x) + 8. To complete the square inside the parentheses, take half of the xx-coefficient (6), which is 3, and square it to get 9. The expression becomes 2((x+3)2)−18+8=2(x+3)2−102((x+3)^2) - 18 + 8 = 2(x+3)^2 - 10. The general vertex form is a(x−h)2+ka(x-h)^2 + k. Comparing (x−h)2(x-h)^2 to (x+3)2(x+3)^2, we see that x−h=x+3x-h = x+3, which means −h=3-h = 3, so h=−3h = -3.

Question 20

A quadratic equation is manipulated by completing the square, resulting in the equation (x+6)2=10(x+6)^2 = 10. What was the original equation in the form x2+bx+c=0x^2 + bx + c = 0?

  1. x2+6x−4=0x^2 + 6x - 4 = 0
  2. x2+12x−10=0x^2 + 12x - 10 = 0
  3. x2+12x+26=0x^2 + 12x + 26 = 0 (correct answer)
  4. x2+6x+26=0x^2 + 6x + 26 = 0
Explanation: To find the original equation, expand the completed square form and set it equal to zero. Expand (x+6)2(x+6)^2 to get x2+12x+36x^2 + 12x + 36. So the equation is x2+12x+36=10x^2 + 12x + 36 = 10. To put it in the form x2+bx+c=0x^2 + bx + c = 0, subtract 10 from both sides: x2+12x+36−10=0x^2 + 12x + 36 - 10 = 0, which simplifies to x2+12x+26=0x^2 + 12x + 26 = 0.