Last updated .
2.4 — A Square Deal
Read the explanations and worked examples before beginning each part of the packet. The aim is to know exactly why the rewriting steps work, not just to remember a shortcut.
1. In class: A Solidify Understanding Task
Review: One square, then several squares
In Section 2.3, a square whose side is \(x+q\) has area \((x+q)^2=x^2+2qx+q^2\). The four regions have areas \(x^2\), \(qx\), \(qx\), and \(q^2\). A perfect-square trinomial is an expression with exactly this structure.
Take a square with side \(x+4\). Divide each side into a segment of length \(x\) and a segment of length 4. The regions contribute \(x^2\), \(4x\), \(4x\), and 16 square units.
\[(x+4)^2=x^2+8x+16.\]Expanding confirms the geometry: \((x+4)(x+4)=x^2+4x+4x+16\).
Now suppose there are three separate identical squares. Their total area is three times the area of one square. This is why the factor 3 multiplies all parts of the expansion.
Each square has side length \(x+2\), so its area is \((x+2)^2=x^2+4x+4\).
\[3(x+2)^2=3(x^2+4x+4)=3x^2+12x+12.\]The coefficient 3 means three copies; it does not mean the side of each square becomes \(3(x+2)\). Three squares have total area \(3(x+2)^2\), whereas one square of side \(3(x+2)\) has area \(9(x+2)^2\).
Recognizing a multiple of a perfect square
Consider \(2x^2+12x+18\). The leading coefficient 2 tells us to look for two copies of one perfect square. Factor 2 out of every term:
\[2x^2+12x+18=2(x^2+6x+9)=2(x+3)^2.\]The interior square has side \(x+3\), because twice 3 is 6 and \(3^2=9\). The total area is the area of two such squares.
Complete \(4x^2+24x+\square\) so it represents four copies of a perfect square.
Step 1. Factor 4 from the terms containing \(x\): \(4(x^2+6x+\square)\).
Step 2. Half of 6 is 3; square it to get 9 inside the parentheses.
Step 3. Multiply by the outside 4. The missing constant in the original expression is \(4\cdot9=36\).
\[4x^2+24x+36=4(x+3)^2.\]Each square is \((x+3)\) by \((x+3)\); there are four squares.
Complete \(3x^2-30x+\square\).
\[3(x^2-10x+\square),\qquad \left(\frac{-10}{2}\right)^2=25.\]\[3x^2-30x+75=3(x-5)^2.\]The sign of the interior binomial follows the middle term, even though the missing constant is positive. For a physical square, require \(x-5\geq0\).
When the original trinomial is not a perfect square
Sometimes the required constant does not match the constant already present. We can still build a perfect square and then subtract or add a correction to keep the expression unchanged. This process is called completing the square.
Rewrite \(2x^2+12x+13\) without changing its value.
1. Factor the coefficient from the variable terms: \(2(x^2+6x)+13\).
2. Inside the parentheses, half of 6 is 3, and \(3^2=9\).
3. Add 9 inside, but also subtract its full value \(2\cdot9=18\) outside:
\[2(x^2+6x+9)+13-18.\]4. Replace the perfect square and combine constants:
\[\boxed{2x^2+12x+13=2(x+3)^2-5.}\]Check: \(2(x^2+6x+9)-5=2x^2+12x+18-5\), exactly the original expression.
The quantity \(2(x+3)^2\) is the matching completed-square expression; the original is 5 less. “Closest perfect square” in the packet refers to matching the \(x^2\) and \(x\) coefficients, not to the physically closest shape or the smallest numerical distance between functions.
Rewrite \(x^2-8x+21\).
The perfect-square part is \(x^2-8x+16=(x-4)^2\). Since 21 is 5 greater than 16:
\[x^2-8x+21=(x-4)^2+5.\]Rewrite \(2x^2+6x-1\). Factor 2 out of the variable terms:
\[2(x^2+3x)-1.\]Half of 3 is \(\frac32\); its square is \(\frac94\). Multiplying by 2 adds \(\frac92\), which we must subtract:
\[2\left(x^2+3x+\frac94\right)-1-\frac92\]\[\boxed{2\left(x+\frac32\right)^2-\frac{11}{2}.}\]This is valid even if the midway number or the final correction is not an integer.
- Factor \(a\) out of the \(x^2\) and \(x\) terms only.
- Inside parentheses, take half the new coefficient of \(x\), then square it.
- Add that square inside the parentheses and subtract \(a\) times that same square outside, preserving equality.
- Rewrite the trinomial as a binomial squared, then combine the remaining constants.
Using the completed square to understand a graph
From Section 2.2, \(y=a(x-h)^2+k\) has vertex \((h,k)\). So \(2(x+3)^2-5\) has vertex \((-3,-5)\), opens upward, and is vertically stretched by 2. The new algebra does not change the graph; it reveals features already present in the original expression.
For \(y=3x^2-18x+20\), factor and complete the square:
\[y=3(x^2-6x)+20=3(x^2-6x+9)+20-27\]\[\boxed{y=3(x-3)^2-7.}\]The vertex is \((3,-7)\). At \(x=3\), the original gives \(27-54+20=-7\), which verifies the minimum.
2. Ready — Matching a parabola to its equation
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Ready 1–6 asks for each graph’s y-intercept and then a match to one of six quadratic equations. A y-intercept is where the graph crosses the vertical axis, so \(x=0\). In \(f(x)=ax^2+bx+c\), this gives \(f(0)=c\): the y-intercept is \((0,c)\).
When matching, check the y-intercept and opening direction (positive \(a\) opens upward; negative \(a\) opens downward). Finally use the vertex or another point if needed. A graph crossing at \((0,-5)\) must correspond to an equation with constant term \(-5\), but that alone would not identify it if multiple choices had the same constant.
Compare \(f(x)=x^2-4x+4\) and \(g(x)=-x^2+2x+4\). Both cross the y-axis at \((0,4)\), but \(f\) opens up and \(g\) opens down. Also \(f(x)=(x-2)^2\) has vertex \((2,0)\), whereas \(g(x)=-(x-1)^2+5\) has vertex \((1,5)\). These checks distinguish the graphs decisively.
The printed graphs are small. Treat the grid marks as units and use the equations as an independent check. Read the point where each curve meets the y-axis (the line \(x=0\)), not the parabola’s lowest or highest point, which may be elsewhere.
Check your work — Ready 1–6 answer key
- y-intercept \((0,-5)\); equation c.
- y-intercept \((0,4)\); equation f.
- y-intercept \((0,5)\); equation b.
- y-intercept \((0,1)\); equation e.
- y-intercept \((0,-1)\); equation a.
- y-intercept \((0,-7)\); equation d.
3. Set — Five copies, scrambled tiles, and deciding if a square is complete
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Set 7–9: The packet says each expression represents five perfect squares. Write \(5(x+q)^2\), expand to \(5x^2+10qx+5q^2\), and match the middle term. The constant is \(5q^2\), not merely \(q^2\). Each of the five square pieces has side \(x+q\).
Complete \(5x^2-40x+\square\).
Match \(10q=-40\), so \(q=-4\). The required constant is \(5(-4)^2=80\).
\[\boxed{5x^2-40x+80=5(x-4)^2.}\]There are five identical \((x-4)\)-by-\((x-4)\) squares (when \(x\ge4\)).
Set 10: Count how many \(x^2\)-tiles, \(x\)-rectangles, and unit squares appear in the drawing. Combine their areas to make the expanded expression; group them into whole enlarged squares where possible for the equivalent squared expression. A tile diagram of disjoint pieces need not already be arranged as squares.
Set 11–13: Find the unique scaled perfect square that has the same quadratic and linear terms as the original; compare constants to determine the adjustment. For example:
For \(3x^2+12x-7\), half of \(12/3=4\) is 2. The matching scaled square is \(3(x+2)^2=3x^2+12x+12\). The desired constant is \(-7\), which is 19 below 12:
\[\boxed{3x^2+12x-7=3(x+2)^2-19.}\]So the original expression is not itself a scaled perfect square; it is the completed square minus 19.
“Closest perfect square” is not a distance measurement. For these problems, it means the expression of the form \(a(x+q)^2\) that matches the given \(x^2\) and \(x\) coefficients, before adding the constant correction. Even for a negative or noninteger correction, the algebra is valid.
Check your work — Set 7–13 answer key
- Missing constant \(45\); each of the five squares has dimensions \((x+3)\times(x+3)\).
- Missing constant \(125\); each of the five squares has dimensions \((x-5)\times(x-5)\).
- Missing constant \(5\); each of the five squares has dimensions \((x+1)\times(x+1)\).
- With the usual algebra-tile convention (large square area \(x^2\), strip area \(x\), small square area 1), \(A(x)=2x^2+4x+5=2(x+1)^2+3\). The diagram does not label these dimensions.
- Not a perfect square; \(x^2+10x+14=(x+5)^2-11\).
- Not a perfect square; \(2x^2+16x+6=2(x+4)^2-26\).
- Not a perfect square; \(3x^2+18x-12=3(x+3)^2-39\).
4. Go — Evaluate quadratic functions
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
To evaluate \(f(t)\), substitute \(t\) for every occurrence of \(x\) in the function rule, then compute exponents before multiplication and addition. For a negative input, always use parentheses around the substituted value.
Given \(f(x)=2x^2-5x+1\), calculate \(f(3)\).
\[f(3)=2(3)^2-5(3)+1=18-15+1=\boxed4.\]We calculate the square first, then multiply. \(f(3)\) means an output, not \(f\) multiplied by 3.
Given \(g(x)=-3x^2+2x-4\), calculate \(g(-2)\).
\[g(-2)=-3(-2)^2+2(-2)-4=-3(4)-4-4=\boxed{-20}.\]Important: \(-3(-2)^2=-12\), because the square applies to \((-2)\), not to the coefficient \(-3\).
For \(f(x)=x^2-4x+3\) and \(g(x)=2x^2+x-6\), find \(f(0)+g(-1)\).
\[f(0)=0-0+3=3.\]\[g(-1)=2(-1)^2+(-1)-6=2-1-6=-5.\]\[\boxed{f(0)+g(-1)=3+(-5)=-2.}\]Evaluate each function independently before adding the outputs.
If \(h(x)=4x^2-3x+2\), find \(h(\tfrac12)\).
\[h\left(\frac12\right)=4\left(\frac12\right)^2-3\left(\frac12\right)+2=1-\frac32+2=\boxed{\frac32}.\]Check your work — Go 14–18 answer key
- \(f(1)=-24\).
- \(f(5)=60\).
- \(g(10)=-195\).
- \(g(-5)=-60\).
- \(f(0)+g(0)=-30\).
Additional Learning Resources
These resources were selected for Section 2.4. Use the scope guidance so that later algebra topics do not become accidental prerequisites.
| Resource | What to use | Scope / what to skip |
|---|---|---|
| OpenStax — 9.7 Graph Quadratic Functions Using Transformations | Read the part on completing the square for a quadratic function with leading coefficient different from 1; closely matches the algebra in task 8–9. | Some extra material. Focus on the rewriting procedure, not additional graphing exercises unless desired. |
| Khan Academy — Worked example: completing the square (leading coefficient ≠ 1) | Shows factoring out a leading coefficient and completing the square with that coefficient. | Some extra material. The video goes on to solve an equation; stop after the completed-square rewriting if only studying 2.4. |
| OpenStax Algebra 1 — 9.4, Perfect Squares with Coefficients Other Than 1 | Start with section 9.4.1 for squares with a nonunit leading coefficient. | Significantly broader. Skip the later sections on solving quadratic equations by completing the square. |
Teacher / parent review of the source packet
Section 2.4, Task 8; Set 11–13. The phrase “closest perfect square” lacks a defined metric and can suggest several unrelated interpretations. A precise version is: Find the scaled binomial square with the same coefficients of \(x^2\) and \(x\), and then give the additive constant required to recover the original expression. This interpretation also permits negative constants and avoids treating a symbolic expression as a literal positive area.
Section 2.4, Task 5 and Set 7–9. The wording refers to multiple physical squares, but a binomial side such as \(x-5\) only represents a physical length when nonnegative. If using the quilts as literal geometry, give the domain restriction. The polynomial identities themselves are valid for every real \(x\).
No other material error requiring correction identified.
Packet-to-lesson coverage check
| Packet portion | Required skills | Taught above |
|---|---|---|
| Main task 1–4 | Represent multiple squares in diagrams; expand and rewrite scaled squares. | In class; examples 1–3 |
| Main task 5 | Find missing constants with nonunit leading coefficients. | In class; examples 3–4 |
| Main task 6–9 | Match incomplete area models, complete squares with corrections, generalize a procedure. | In class; examples 5–8 |
| Ready 1–6 | Read y-intercepts and match quadratic graphs to equations. | Section 2; example 9 and comparison graph |
| Set 7–9 | Five equal square areas, missing constants and dimensions. | Section 3; example 10 |
| Set 10–13 | Count algebra tiles, equivalent forms, completed-square adjustments. | Section 3; example 11 |
| Go 14–18 | Substitute positive, negative, zero, or fractional inputs and combine function outputs. | Section 4; four new worked problems |
Last updated .