Quadratics Error Analysis Worksheet: Factoring and the Quadratic Formula
Quadratics error analysis, twenty times over: somebody already did the work — badly. Your job is to find the slip, name it, and fix it. Full answer key included.
Published · Last updated · Every answer checked for accuracy before publishing · Free to print for classroom & home use
- Level
- Algebra I and Algebra II, grades 8–11
- Standards
- CCSS.MATH.CONTENT.HSA.REI.B.4 and HSA.SSE.B.3.a
- Includes
- 20 quadratics error analysis problems in two parts, three color-coded figures, and a full answer key that names each mistake
- Time
- About 35–45 minutes for both parts
- Format
- Read on this page or print the PDF — no sign-up
Quadratics error analysis hands you a worked solution that already contains a mistake. You find the first line that goes wrong, name the error in plain words, and rewrite the problem correctly from that line down. The twenty problems below cover factoring and the quadratic formula, and every answer is explained.
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Quadratics error analysis flips the usual assignment on its head. Solving a quadratic correctly proves you can do it once; catching somebody else's wrong solution proves you actually know what the steps mean. The same three mistakes show up over and over, so here they are, dressed up in their usual disguises.
The roots are −2 and −3, not 2 and 3.
Flip the sign on the way out.
Zero is the only number with veto power.
Expand, move everything over, then factor.
−b, not b. 2a, not 2.
And the whole top gets divided.
How Quadratics Error Analysis Works
Three steps, and skipping the middle one is cheating.
- Find it. Which line is the first one that's wrong? Not the last one — the first.
- Name it. In words. "Sign error." "Divided by 2 instead of 2a." If you can't name it, you haven't found it.
- Fix it. Redo the problem from that line and give the correct solution.
Step 2 is where the learning happens. A student who says "it should be −2" has spotted a typo. A student who says "they forgot to flip the sign when they set the factor to zero" has spotted a habit — theirs, probably, next Tuesday.
Quadratics Error Analysis for Factoring: Read the Signs First
Before you hunt for factor pairs, read the two signs in x² + bx + c. They narrow the field before you've written anything down.
| Color | Case | What to write |
|---|---|---|
| Green | c > 0, b > 0 | (x + m)(x + n) |
| Red | c > 0, b < 0 | (x − m)(x − n) |
| Purple | c < 0 | (x + m)(x − n), bigger one follows b |
For x² − 5x + 6, the pair −1 and −6 multiplies to 6 perfectly well. It also adds to −7, which is not −5. Checking the product and calling it done is the single most common factoring error I mark. Both conditions, every time.
The Zero Product Trap
The zero product property says: if two things multiply to zero, one of them is zero. That's it. It does not say anything about two things multiplying to 6, or 12, or your locker number.
(x − 3)(x + 2) = 6
x − 3 = 6 → x = 9
x + 2 = 6 → x = 4
x = 4 happens to work. x = 9 gives 66. Being right by accident is still wrong.
x² − x − 6 = 6
x² − x − 12 = 0
(x − 4)(x + 3) = 0
x = 4, x = −3
Expand, move everything to one side, then factor.
Same equation, two roads. Only one of them is legal.
Error analysis only works once the correct method is already familiar, so if the factoring itself is shaky, run the drills on solving quadratics by factoring first. Once the roots are in hand, vertex and axis of symmetry and graphing quadratic functions turn them into a picture — and a picture is the fastest lie detector there is.
Quadratic Formula: Four Places It Goes Wrong
The formula itself never makes a mistake. The four spots below are where fingers do.
| Color | Piece | The classic slip |
|---|---|---|
| Red | −b | b = −6 makes −b = +6 |
| Purple | ± | Writing only one root |
| Green | b² − 4ac | (−3)² = 9, never −9 |
| Blue | 2a | Dividing by 2 when a ≠ 1 |
One more, free of charge: the discriminant is a traffic light, not a decoration. Positive means two real roots, zero means one, negative means none that live on the number line. A student who reports "no solutions" for a discriminant of zero has stopped one thought too early — the discriminant worksheet drills exactly that.
| Color | Equation | Discriminant & roots |
|---|---|---|
| Green | y = x² − 4 | b² − 4ac = 16 > 0 → crosses twice, two real roots |
| Gold | y = x² | b² − 4ac = 0 → touches once, one double root |
| Red | y = x² + 2 | b² − 4ac = −8 < 0 → never touches, no real roots |
Need the underlying skills first? Start with solving quadratics by factoring or the quadratic formula worksheet, and if the equation won't factor cleanly, completing the square is the way through. For the plain-English version of the formula, Math is Fun is hard to beat, and the standards these problems align to sit at Common Core HSA-REI with matching classroom tasks at Illustrative Mathematics.
Quadratics Error Analysis Practice Problems
For each one: find the first wrong line, name the error in words, then give the correct solution.
Part A — factoring errors
- Solve x² + 5x + 6 = 0.
Student's work(x + 2)(x + 3) = 0, so x = 2 and x = 3.
- Solve x² − 7x = 0.
Student's workDivide both sides by x: x − 7 = 0, so x = 7.
- Solve x² + 2x = 15.
Student's workx(x + 2) = 15, so x = 15 or x + 2 = 15, giving x = 15 and x = 13.
- Factor 2x² + 10x + 12 completely.
Student's work(x + 2)(x + 3)
- Factor x² − 9.
Student's work(x − 3)(x − 3)
- Factor x² + 16.
Student's work(x + 4)(x + 4)
- Solve x² − 4x + 4 = 0.
Student's work(x − 2)(x − 2) = 0, so x = 2 and x = −2.
- Factor 3x² − 5x − 2.
Student's work(3x − 1)(x + 2)
- Solve x² = 36.
Student's workx = 6
- Factor x² − 5x + 6.
Student's work−1 × −6 = 6, so (x − 1)(x − 6).
Part B — quadratic formula errors
- Solve x² − 6x + 5 = 0.
Student's workx = (−6 ± √(36 − 20)) / 2
- Solve 2x² + 3x − 5 = 0.
Student's workx = (−3 ± √(9 + 40)) / 2 = (−3 ± 7) / 2
- Solve x² + 4x + 1 = 0.
Student's workx = −4 ± √12 / 2
- Solve x² − 3x − 4 = 0.
Student's workb² = (−3)² = −9, so the discriminant is −9 + 16 = 7.
- Solve 3x² + 2x = 8.
Student's worka = 3, b = 2, c = 8
- Solve x² + 2x + 5 = 0.
Student's workDiscriminant = 4 − 20 = −16, so x = (−2 ± 4) / 2 = 1 and −3.
- Solve x² − 5x + 6 = 0.
Student's workx = (5 + 1) / 2 = 3
- Solve 4x² − 12x + 9 = 0.
Student's workDiscriminant = 144 − 144 = 0, so there are no solutions.
- Solve −x² + 4x − 3 = 0.
Student's worka = 1, b = 4, c = −3
- Solve 2x² − 4x − 6 = 0.
Student's workx = (4 ± 8) / 4, and cancelling the 4s gives x = 1 ± 8, so x = 9 and x = −7.
Quadratics Error Analysis: Answer Key
Part A
1. x² + 5x + 6 = 0
Error: signs not flipped when solving the factors.
The factoring is fine. But x + 2 = 0 gives x = −2, not 2.
x = −2 and x = −3
2. x² − 7x = 0
Error: divided by x and threw away a root.
Dividing by a variable deletes the case x = 0. Factor instead: x(x − 7) = 0.
x = 0 and x = 7
3. x² + 2x = 15
Error: zero product property used against 15.
Set it to zero first: x² + 2x − 15 = 0, then (x + 5)(x − 3) = 0.
x = −5 and x = 3
4. 2x² + 10x + 12
Error: the GCF vanished.
Pull out the 2 first, then factor what's left. Multiply the student's answer back out and you get x² + 5x + 6 — exactly half of what you started with.
2(x + 2)(x + 3)
5. x² − 9
Error: difference of squares written with matching signs.
(x − 3)(x − 3) expands to x² − 6x + 9. The middle terms have to cancel, so the signs must be opposite.
(x + 3)(x − 3)
6. x² + 16
Error: treated a sum of squares as if it factors.
(x + 4)(x + 4) is x² + 8x + 16, which has an extra 8x nobody asked for. Only a difference of squares factors; a sum has no real factorization.
Prime over the real numbers
7. x² − 4x + 4 = 0
Error: invented a ± that isn't there.
Both factors are (x − 2), so both give the same root. This is a double root, and the parabola just kisses the x-axis.
x = 2 only
8. 3x² − 5x − 2
Error: both signs are backwards.
(3x − 1)(x + 2) expands to 3x² + 5x − 2 — right numbers, wrong middle sign. Flip the sign inside each bracket and the middle term flips with it.
(3x + 1)(x − 2)
9. x² = 36
Error: forgot the negative root.
x² − 36 = 0 factors as (x − 6)(x + 6) = 0. Square roots in an equation come in pairs.
x = 6 and x = −6
10. x² − 5x + 6
Error: checked the product, skipped the sum.
−1 and −6 do multiply to 6, but they add to −7. You need a pair that does both jobs: −2 and −3.
(x − 2)(x − 3)
Part B
11. x² − 6x + 5 = 0
Error: used b instead of −b.
Here b = −6, so −b = +6. The numerator starts with 6, not −6. Then √16 = 4.
x = (6 ± 4) / 2 → x = 5 and x = 1
12. 2x² + 3x − 5 = 0
Error: divided by 2 instead of 2a.
a = 2, so the denominator is 4. The discriminant work was correct.
x = (−3 ± 7) / 4 → x = 1 and x = −5/2
13. x² + 4x + 1 = 0
Error: only the radical got divided.
The fraction bar sits under the entire numerator. Write it as (−4 ± √12) / 2, then simplify √12 = 2√3.
x = −2 ± √3
14. x² − 3x − 4 = 0
Error: squared a negative and kept it negative.
(−3)² = 9. The discriminant is 9 + 16 = 25, which is a perfect square — a good sign this one also factors.
x = (3 ± 5) / 2 → x = 4 and x = −1
15. 3x² + 2x = 8
Error: read off a, b, c before reaching standard form.
Move the 8 first: 3x² + 2x − 8 = 0, so c = −8. Discriminant = 4 + 96 = 100.
x = (−2 ± 10) / 6 → x = 4/3 and x = −2
16. x² + 2x + 5 = 0
Error: took the square root of a negative and kept walking.
√(−16) is not 4. A negative discriminant means the parabola never crosses the x-axis.
No real solutions (x = −1 ± 2i if complex roots are allowed)
17. x² − 5x + 6 = 0
Error: the ± got left behind.
Discriminant = 25 − 24 = 1, so you need both (5 + 1)/2 and (5 − 1)/2.
x = 3 and x = 2
18. 4x² − 12x + 9 = 0
Error: read a zero discriminant as "none."
Zero means exactly one repeated root: x = 12 / 8. This is the perfect square (2x − 3)² = 0.
x = 3/2 (double root)
19. −x² + 4x − 3 = 0
Error: ignored the minus sign on a.
a = −1, so −b = −4 and 2a = −2. Discriminant = 16 − 12 = 4. Multiplying the whole equation by −1 first is the easier route.
x = (−4 ± 2) / (−2) → x = 1 and x = 3
20. 2x² − 4x − 6 = 0
Error: cancelled one term out of a sum.
You can't cancel the 4 in (4 ± 8)/4 — the whole numerator is one package. Divide after you add or subtract.
x = 12/4 = 3 and x = −4/4 = −1
Every answer above was checked for accuracy before publishing, most recently on . If you spot an error, email burketutoringinfremont@outlook.com or text (510) 453-0350 and it will be corrected.
Common Questions
What is quadratics error analysis?
You're given a worked solution to a quadratic that contains a mistake. You find it, explain it in words, and correct it. Quadratics error analysis targets understanding rather than speed, which is why teachers reach for it right before a test.
Why is finding errors harder than just solving?
Solving lets you follow your own habits. Error analysis makes you follow somebody else's, which means you have to actually read each line instead of running on autopilot.
What's the most common quadratic mistake?
Two, tied. Setting factors equal to something other than zero, and using b where the formula says −b. Both come from copying the shape of a step without reading it.
How do I check a quadratic answer quickly?
Substitute it back. Ten seconds, no ambiguity. For a second opinion on the graph, drop the equation into Desmos or GeoGebra and see where the curve crosses; WolframAlpha will confirm the exact roots.
What grade level is this worksheet?
Algebra I for Part A, Algebra II for Part B, though plenty of Algebra I classes cover the formula too. Part A works as bellwork; Part B is closer to test review.
Can I use this in my classroom?
Yes — print it, copy it, project it. No sign-up and no attribution required, though a link back is always appreciated.
Keep Practicing
Errors get easier to spot once the underlying moves are automatic: factoring, the quadratic formula, completing the square, and the discriminant. On the graphing side, try graphing quadratic functions, vertex and axis of symmetry, vertex and intercept form, reading an equation from a graph, quadratic transformations, and focus and directrix — whose Part B word problems are a good next step once this quadratics error analysis set feels easy. Not sure the function is even quadratic? Compare the function families first.
Standards and References
These problems were written against the following standards and checked against the following references.
- Part A aligns to CCSS.MATH.CONTENT.HSA.SSE.B.3.a — factor a quadratic expression to reveal the zeros of the function it defines.
- Part B aligns to CCSS.MATH.CONTENT.HSA.REI.B.4 — solve quadratic equations in one variable, including by the quadratic formula. Standard text: Common Core, High School Algebra: Reasoning with Equations & Inequalities.
- Matched classroom tasks: Illustrative Mathematics, HSA.REI.B.4.
- Answer verification: each solution was substituted back into the original equation, and the roots and discriminants were cross-checked in Desmos and WolframAlpha.
About the Author
Berke has taught K–12 math and science for more than 10 years, logging over 6,000 tutoring hours with 200+ students. He holds a BS in Biochemistry from Washington University in St. Louis and an MS in Bioinformatics from UMGC.
The twenty mistakes in this quadratics error analysis set are not hypothetical. They are the ones that come up session after session, which is why the zero product trap gets its own figure.
How these are made: problems are prepared from previous class notes and assigned homework, then written up as worked solutions with a single deliberate error. Corrections come in by email or text and get fixed the same week.
More: tutor profile · LinkedIn · burketutoringinfremont@outlook.com
Page Updates
- — page published with 20 error-analysis problems and the printable PDF.
- — added the discriminant graph, per-problem anchor links, and a standards and references section; corrected the wording of the answer explanations for problems 6 and 8.
- — added link to quadratics hub page.
If the same mistake keeps coming back, it usually takes about one session to break the habit. Burke Tutoring offers in-home algebra tutoring in Fremont, Newark, and Union City. Call (510) 453-0350.
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