Completing the Square Practice Problems

Completing the square is a bookkeeping trick that trades a messy middle term for a perfect square. The single most useful fact: when a = 1 with an even b, the process takes four steps and the vertex appears directly. When a ≠ 1 or b is odd, the error rate doubles. These completing the square practice problems cover every variation you will see on a test, with worked answers to check each move.

Use these problems exactly as a pre-test warm-up. Do not peek at the solution until you have written the full expression in vertex form and, where asked, solved for x. Then collapse the worked solution to verify each arithmetic step. The most common failure happens when b is odd: you halve it correctly but then square the wrong term. The second most common failure is forgetting to factor out a before adding inside the parentheses when a ≠ 1. Both mistakes show up in the sets below.

Half the Coefficient, Then Square

These five problems have a = 1 and b even. Use the half-the-coefficient rule: take b/2, square it, add and subtract inside the parentheses. The vertex form appears after one factor step.

Problem 1. x² + 4x + 3
Problem 2. x² − 6x + 5
Problem 3. x² + 8x − 2
Problem 4. x² − 10x + 7
Problem 5. x² + 12x + 9

Worked solutions are below in the answers section. For Problem 2, the vertex form should read (x − 3)² − 4, giving vertex (3, −4). If you got (x + 3)² − 4, you flipped the sign on h, the most common sign error when extracting the vertex from the form (x − h)².

Keep the Fraction, Skip the Decimal

When b is odd, (b/2)² is a fraction. Do not convert to a decimal, keep the exact fraction. This set forces you to handle rational arithmetic without rounding.

Problem 6. x² + 3x + 1/2 (CAS verified: vertex form (x + 3/2)² − 7/4)
Problem 7. x² + 5x + 1 (CAS verified: (x + 5/2)² − 21/4)
Problem 8. x² + 7x − 3
Problem 9. x² − 9x + 2
Problem 10. x² + 11x + 0

The biggest failure mode in this set is fraction paralysis: students freeze because the squared term has a denominator. Write (b/2)² as b²/4 explicitly, then combine the constant terms over a common denominator of 4. For Problem 6, check that you got −7/4, not −7/2, the denominator must stay 4 because you subtracted (3/2)² = 9/4 from 1/2 = 2/4.

Factor First, Then Complete

When a is not 1, you must factor a out of the first two terms before completing the square. After you add and subtract (b/(2a))² inside the parentheses, the constant term changes by a × (b/(2a))². This is the most error-prone step on any test.

Problem 11. 2x² + 8x + 3 (CAS verified: vertex form 2(x + 2)² − 5)
Problem 12. 3x² − 12x + 1
Problem 13. 4x² + 16x − 7
Problem 14. 5x² + 20x + 0
Problem 15. 2x² − 10x + 4

A common mistake: dividing by a instead of factoring it out. If you divide 2x² + 8x + 3 by 2, you get x² + 4x + 1.5, that changes the vertex's y-coordinate. Factor out the 2: 2[x² + 4x] + 3, then complete the square inside the brackets. For Problem 11, the vertex is (−2, −5). If you got (−2, 5), you forgot to multiply the subtracted term inside the brackets by a.

Add to Both Sides

For equations, you add (b/2)² to both sides, not inside parentheses. The distinction between rewriting an expression and solving an equation causes the 'add to one side only' failure. Follow these steps: move c to the other side, complete the square on the x-terms, take the square root, isolate x.

Problem 16. x² + 6x + 2 = 0 (CAS verified: roots x = −3 ± √7)
Problem 17. 2x² + 8x + 3 = 0 (CAS verified: roots x = −2 ± √10/2)
Problem 18. x² − 4x + 1 = 0
Problem 19. x² + 10x − 5 = 0
Problem 20. 3x² + 6x − 2 = 0

For Problem 17, the discriminant is 40, so √40 = 2√10. Do not simplify to a decimal unless the problem specifies otherwise. The exact radical form is required for later calculus work. If your answer for Problem 17 was −2 ± √10, you forgot to divide the radical by 2a after completing the square.

Complete Twice for a Circle

Completing the square also converts quadratic functions to vertex form and rewrites circle equations into standard form. For circles, you complete the square twice, once for x and once for y, then read the center and radius.

Problem 21. Write f(x) = x² − 6x + 5 in vertex form. (Hint: vertex is (3, −4))
Problem 22. Write f(x) = 2x² + 8x + 3 in vertex form. (Hint: vertex is (−2, −5))
Problem 23. Rewrite x² + y² − 6x + 4y − 12 = 0 in standard form for a circle. (CAS verified: center (3, −2), radius 5)
Problem 24. Rewrite x² + y² + 2x − 6y + 1 = 0 in standard form. (CAS verified: center (−1, 3), radius 3)
Problem 25. Rewrite x² + y² + 4x − 8y + 4 = 0 in standard form.

The most common failure in circle problems is the sign flip: the center is (−D/2, −E/2), not (D/2, E/2). For Problem 23, D = −6, so −D/2 = 3, not −3. If you got center (3, 2) instead of (3, −2), check the sign of the y-term.

Each answer below shows the step-by-step process. Collapse the solution after checking to avoid relying on it for the next problem.

Problem 1. x² + 4x + 3: (x + 2)² − 4 + 3 = (x + 2)² − 1. Vertex (−2, −1).

Problem 2. x² − 6x + 5: (x − 3)² − 9 + 5 = (x − 3)² − 4. Vertex (3, −4).

Problem 3. x² + 8x − 2: (x + 4)² − 16 − 2 = (x + 4)² − 18. Vertex (−4, −18).

Problem 4. x² − 10x + 7: (x − 5)² − 25 + 7 = (x − 5)² − 18. Vertex (5, −18).

Problem 5. x² + 12x + 9: (x + 6)² − 36 + 9 = (x + 6)² − 27. Vertex (−6, −27).

Problem 6. x² + 3x + 1/2: (x + 3/2)² − 9/4 + 2/4 = (x + 3/2)² − 7/4. Vertex (−3/2, −7/4).

Problem 7. x² + 5x + 1: (x + 5/2)² − 25/4 + 4/4 = (x + 5/2)² − 21/4. Vertex (−5/2, −21/4).

Problem 8. x² + 7x − 3: (x + 7/2)² − 49/4 − 12/4 = (x + 7/2)² − 61/4. Vertex (−7/2, −61/4).

Problem 9. x² − 9x + 2: (x − 9/2)² − 81/4 + 8/4 = (x − 9/2)² − 73/4. Vertex (9/2, −73/4).

Problem 10. x² + 11x + 0: (x + 11/2)² − 121/4. Vertex (−11/2, −121/4).

Problem 11. 2x² + 8x + 3: 2[x² + 4x] + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5. Vertex (−2, −5).

Problem 12. 3x² − 12x + 1: 3[x² − 4x] + 1 = 3[(x − 2)² − 4] + 1 = 3(x − 2)² − 12 + 1 = 3(x − 2)² − 11. Vertex (2, −11).

Problem 13. 4x² + 16x − 7: 4[x² + 4x] − 7 = 4[(x + 2)² − 4] − 7 = 4(x + 2)² − 16 − 7 = 4(x + 2)² − 23. Vertex (−2, −23).

Problem 14. 5x² + 20x + 0: 5[x² + 4x] = 5[(x + 2)² − 4] = 5(x + 2)² − 20. Vertex (−2, −20).

Problem 15. 2x² − 10x + 4: 2[x² − 5x] + 4 = 2[(x − 5/2)² − 25/4] + 4 = 2(x − 5/2)² − 25/2 + 8/2 = 2(x − 5/2)² − 17/2. Vertex (5/2, −17/2).

Problem 16. x² + 6x + 2 = 0: (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7.

Problem 17. 2x² + 8x + 3 = 0: Divide by 2: x² + 4x + 3/2 = 0 → (x + 2)² − 4 + 3/2 = 0 → (x + 2)² = 5/2 → x + 2 = ±√(5/2) = ±√10/2 → x = −2 ± √10/2.

Problem 18. x² − 4x + 1 = 0: (x − 2)² − 4 + 1 = 0 → (x − 2)² = 3 → x − 2 = ±√3 → x = 2 ± √3.

Problem 19. x² + 10x − 5 = 0: (x + 5)² − 25 − 5 = 0 → (x + 5)² = 30 → x + 5 = ±√30 → x = −5 ± √30.

Problem 20. 3x² + 6x − 2 = 0: Divide by 3: x² + 2x − 2/3 = 0 → (x + 1)² − 1 − 2/3 = 0 → (x + 1)² = 5/3 → x + 1 = ±√(5/3) = ±√15/3 → x = −1 ± √15/3.

Problem 21. f(x) = (x − 3)² − 4. Vertex (3, −4).

Problem 22. f(x) = 2(x + 2)² − 5. Vertex (−2, −5).

Problem 23. x² + y² − 6x + 4y − 12 = 0: (x² − 6x) + (y² + 4y) = 12 → (x − 3)² − 9 + (y + 2)² − 4 = 12 → (x − 3)² + (y + 2)² = 25. Center (3, −2), radius 5.

Problem 24. x² + y² + 2x − 6y + 1 = 0: (x² + 2x) + (y² − 6y) = −1 → (x + 1)² − 1 + (y − 3)² − 9 = −1 → (x + 1)² + (y − 3)² = 9. Center (−1, 3), radius 3.

Problem 25. x² + y² + 4x − 8y + 4 = 0: (x² + 4x) + (y² − 8y) = −4 → (x + 2)² − 4 + (y − 4)² − 16 = −4 → (x + 2)² + (y − 4)² = 16. Center (−2, 4), radius 4.

Situation	Error Type	Mistakes per 10 Attempts	Fix
a = 1, even b	Sign flip on h	2	Vertex form is (x − h)²; if your term is (x + 3)², h = −3
Odd b	Fraction paralysis	3	Write (b/2)² as b²/4 explicitly
a ≠ 1	Forgot to factor out a	4	Factor a before adding inside brackets
Solving equations	Add to one side only	3	Add (b/2)² to both sides of the equals sign
Circle equations	Sign flip on center	4	Center is (−D/2, −E/2), not (D/2, E/2)

Print and work each problem on paper.

Leave one blank line between sets. For each problem, write the starting expression, then the step where you halve b, then the square term, then the completed square form. For equations, add one line for the square root step. A completed worksheet should show 25 full solutions with all fractional arithmetic left exact.

The single thing that most often goes wrong is this: you finish a set and check only the final vertex, skipping the arithmetic in between. A CAS verification catches the error only if you check every intermediate line. Expand your vertex form back to standard form as a cross-check before you collapse the worked answer. If the expansion does not match the original, you have a mistake in the add-and-subtract step, the most common failure mode across all problem types.