Algebra · Eqora guide
Completing the square in homework: which number is missing?
Find the missing term from the middle coefficient, keep the equation balanced, and test your next quadratic without copying an answer.
Completing the square in homework can begin with a line such as x² − 8x = 9, which looks unfinished in a strangely specific way. You may remember that a number must be added, but is it 4, 8, 16, or something taken from the 9 on the right? The useful question is not which number a calculator suggests. It is which number turns x² − 8x into the square of one binomial while preserving the original equation. Here the missing number is 16: x² − 8x + 16 = (x − 4)².
We will derive that choice from expansion, solve the complete equation, check both roots, and handle a leading coefficient other than one. The ceramic tiles pictured here are a physical analogy for a corner that finishes a square; they are not a scale-accurate model of x² − 8x. After the mathematics, a narrow Eqora workflow can help with one uncertain homework step: capture the whole task, compare the recognized symbols, inspect the explanation, ask a focused question, and then solve a similar problem with the phone closed.
Eqora publishes this guide and is the only app recommended here. An app can help you examine a step, but it cannot guarantee that it read a handwritten minus sign, followed your instructor's requested method, or reached a correct final answer. Make a first attempt, observe your school's rules for homework and assessment, and verify the notation, assumptions, algebra, and roots yourself. Eqora is a learning aid, not a substitute for exam work or a promise of grades.
Completing the square in homework starts with a binomial
Expand (x + h)² carefully: it is x² + 2hx + h². The middle coefficient is twice h, not h itself. To turn x² + px into part of a square, choose h = p/2. The last term must then be h² = (p/2)². This is why the instruction says to take half of the coefficient of x and square it. It is a consequence of multiplication, not an isolated recipe to memorize.
In x² − 8x, p is −8 with its sign. Half is −4, and the needed final term is (−4)² = 16. The completed expression is x² − 8x + 16 = (x − 4)². Expanding the right side is a quick proof: x² − 4x − 4x + 16 returns x² − 8x + 16. Writing (x − 8)² would produce −16x in the middle, so it cannot represent the original expression. The photograph's missing tile corner suggests the idea; only expansion establishes the exact coefficient.
Balance the equation before taking a square root
Return to the homework equation x² − 8x = 9. Adding 16 only to the left would change its solutions. Add the same 16 to both sides: x² − 8x + 16 = 9 + 16. Now (x − 4)² = 25. Square-rooting gives x − 4 = 5 or x − 4 = −5. The two answers are x = 9 and x = −1. The ± sign is essential because both 5² and (−5)² equal 25.
Substitute into the original, not just the transformed line. For x = 9, 9² − 8·9 = 81 − 72 = 9. For x = −1, (−1)² − 8(−1) = 1 + 8 = 9. Both pass. If you obtained x = 4 ± 5 but recorded only 9, the missing answer is a square-root branch. If your supposed solution makes the original left side something other than 9, locate the first line where you changed one side without the other. A clean final answer is not a replacement for this check.
The sign of the middle term controls the bracket
For x² + 6x = 7, half of +6 is +3 and its square is 9. Add 9 to both sides: (x + 3)² = 16. Thus x + 3 = ±4, giving x = 1 or x = −7. This time the bracket contains +3 because the middle term is positive. The added term itself is still positive: squaring either +3 or −3 produces 9. Keep the sign in the bracket and the sign of the added constant as separate decisions.
A common error is to take half of the number on the right, or to square the whole coefficient rather than its half. Test the proposed square by expanding it before continuing. (x + 6)² has middle term 12x, not 6x. Another error is to subtract 9 on one side because the bracket contains +3. For an equation, you add 9 to each side to preserve equality; for a single expression, you may add and subtract 9 within the same expression. These are related but not identical tasks.
Normalize a leading coefficient before halving
The simple half-the-middle rule assumes the coefficient of x² is 1. Consider 2x² + 8x − 10 = 0. Since 2 is nonzero, divide every term on both sides by 2. The equation becomes x² + 4x − 5 = 0, or x² + 4x = 5. Half of 4 is 2, so add 2² = 4 to both sides. Now (x + 2)² = 9, leading to x + 2 = ±3 and x = 1 or x = −5.
Check in the original equation: at x = 1, 2 + 8 − 10 = 0; at x = −5, 50 − 40 − 10 = 0. Do not halve 8 before dividing by 2 and then add 16 to the normalized line. That mixes coefficients from two different equations. You could instead factor 2 from the x terms and complete a square inside the parentheses, but the outside 2 must then multiply every term inside. Dividing the complete equation first is often easier to audit on homework paper.
No real root is a valid conclusion
Completing the square also tells you when an equation has no real solution. For x² + 4x + 7 = 0, move 7 to the right and add 4 to both sides: (x + 2)² = −3. No real number has a negative square, so this equation has no real roots. Do not force x + 2 = ±√3; that would square to +3 and solve a different equation. If your course includes complex numbers, the solutions are −2 ± i√3, but say explicitly that these are not real roots.
This interpretation depends on the number system requested by the assignment. A graph of y = x² + 4x + 7 lies above the horizontal axis because it can be written as (x + 2)² + 3. The smallest real value is 3 at x = −2. A rough graph supports the conclusion, while the square identity proves it. Make the assumption about real or complex solutions visible rather than letting a solver quietly choose one for you.
An expression and an equation require different balancing moves
If the prompt says rewrite x² − 8x + 3 in completed-square or vertex form, there is no equals sign with a second side to modify. Insert +16 and −16 inside the same expression: x² − 8x + 3 = (x² − 8x + 16) − 16 + 3 = (x − 4)² − 13. Expanding (x − 4)² − 13 returns x² − 8x + 3. You have rewritten one expression, not solved for x. Its minimum value is −13 when x = 4.
If the prompt instead says solve x² − 8x + 3 = 0, use the same identity and set (x − 4)² − 13 = 0. Then (x − 4)² = 13 and x = 4 ± √13. The operations look similar, but the requested output differs: a form, a vertex, or a set of roots. Before asking for app help, underline the verb in the assignment. A correct factorization does not answer a question asking for the vertex unless you interpret it, and a numerical root list does not show the requested completed-square form.
Capture the complete homework instruction in Eqora
Make your own first line before opening the app: copy the coefficient of x², the signed coefficient of x, and the equality or expression exactly. Then capture the entire problem with Eqora in steady light. Include an instruction such as “complete the square,” “solve over the reals,” or “give exact answers.” Keep superscripts, parentheses, minus signs, and fraction bars in frame. Remove unnecessary personal information, but not the mathematical context that changes the task.
Compare the expression recognized by the app with your paper before reading its steps. Is −8x still negative? Is x² clearly squared? Is 9 on the right, not a coefficient attached to x? Did the app retain the requested number system and format? If any symbol differs, correct the input or retake the photo. The image here deliberately has a blank phone display; it depicts a capture habit, not an actual Eqora output or a guarantee that recognition will be right. More detailed photo advice is available in the related guide.
Inspect one transition and ask a precise follow-up
Now compare your working with the displayed solution line by line. The important transition for x² − 8x = 9 is the move to x² − 8x + 16 = 25. If Eqora shows it, verify why 16 was chosen by expanding (x − 4)² and why the right side became 25. If it shows a different path, check whether the new line is algebraically equivalent instead of assuming the presentation must match your teacher's exact notation.
A focused question is more useful than “solve it again”: “In x² − 8x = 9, why is the added term 16 rather than 4, and why must the right side become 25?” A helpful explanation should connect −8 to 2h, identify h = −4, square it, and preserve both sides. If the answer skips that connection, do the expansion yourself. Another useful question is why x − 4 = ±5 yields two roots. Verify each candidate in the original equation before deciding the uncertainty is resolved.
Close the explanation and solve a transfer problem
Put the phone face down and try x² − 10x = 11. Half of −10 is −5; square it to get 25. Add 25 to both sides: (x − 5)² = 36. Therefore x − 5 = ±6 and x = 11 or x = −1. Check directly: 121 − 110 = 11, and 1 + 10 = 11. If you wrote (x − 10)², expand it and look at the middle term. The independent check of a new coefficient is the part that shows the method transferred.
Change one feature again: x² − 10x + 30 = 0. Rearranging and completing gives (x − 5)² = −5, so there are no real roots. You should be able to say why this conclusion differs from the previous exercise even though both share −10x. Only then reopen Eqora to compare, if your class rules allow it. Record any discrepancy at the first differing line, not as a vague judgment that the final answers disagree. The app remains a study aid, not a substitute for work you must submit independently.
Put it into practice now
Choose one problem from your current homework or review set. Attempt it before opening Eqora, then use the app only at the point where your own reasoning stops.
- State what the problem is asking before you solve it
- Identify the first step you cannot justify
- Ask Eqora one focused follow-up about that step
- Finish with a similar problem and no solution in view
The session is complete when the method is clearer, not simply when the worksheet has one more answer.