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2.3 — Building the Perfect Square
Start with a square whose side length is made from two pieces. We will count the pieces of its area, connect them to multiplication, and then reverse that reasoning to discover the missing piece of a perfect-square trinomial. Work the examples before attempting the corresponding packet exercises.
1. In class: A Develop Understanding Task
A square built from four areas
The area of a square of side length \(x\) is \(x^2\). If each side becomes \(x+3\) long, the new area is \((x+3)^2\), not \(x^2+9\). Both dimensions change, creating two rectangular strips as well as the corner square.
Add the four pieces:
\[A=x^2+3x+3x+9=x^2+6x+9.\]The outer square has side \(x+3\), so \(A=(x+3)^2\). Thus
\[(x+3)^2=x^2+6x+9.\]The two expressions are equivalent: they give the same area for every allowed \(x\).
Using lengths \(x+2\) and \(x+1\) gives the same four-piece arrangement. The particular numbers change, but the structure does not.
For side \(x+2\), the four pieces have areas \(x^2,2x,2x,4\), giving
\[(x+2)^2=x^2+4x+4.\]For side \(x+1\), they have areas \(x^2,x,x,1\), giving
\[(x+1)^2=x^2+2x+1.\]These model the same skills as the main task's problems 3–5.
The algebraic pattern behind a perfect square
A binomial has two terms. Squaring a binomial means multiplying it by itself; it does not mean squaring each term separately. A trinomial has three terms. A perfect-square trinomial is a trinomial that equals a binomial multiplied by itself.
The middle term is \(10x\) because there are two products of \(5x\). The constant term is \(25=5^2\).
If the number inside the parentheses is negative, the two matching middle products are negative.
The last term is still positive because \((-4)(-4)=16\).
To expand: square the first term, add twice its product with the second term, then square the second term. The rule works for negative, fractional, and other real values of \(q\).
Finding the missing piece of a perfect square
Suppose you already know the \(x^2\) part and the middle term. You want a constant that makes the expression a square of \(x+q\). Since the middle term must be \(2qx\), find \(q\) by halving its coefficient. Then square \(q\) to find the constant.
Complete \(x^2+14x+\square\).
Step 1: The middle coefficient is 14. Half of 14 is 7.
Step 2: Square 7 to get 49.
\[x^2+14x+49=(x+7)^2.\]Check by multiplying: \((x+7)(x+7)=x^2+14x+49\).
Complete \(x^2-9x+\square\). Half of \(-9\) is \(-\frac92\); squaring gives \(\frac{81}{4}\):
\[x^2-9x+\frac{81}{4}=\left(x-\frac92\right)^2.\]An odd middle coefficient is no problem. The missing constant can be a fraction.
2. Ready review — Graph a line from its intercepts
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.
The x-intercept is a point where a graph meets the horizontal x-axis. Its y-coordinate is 0. The y-intercept meets the vertical y-axis. Its x-coordinate is 0.
| Intercept to find | What to substitute | How to record it |
|---|---|---|
| x-intercept | Set \(y=0\); solve for \(x\) | \((x,0)\) |
| y-intercept | Set \(x=0\); solve for \(y\) | \((0,y)\) |
Graph \(4x+3y=24\).
Set \(y=0\): \(4x=24\), so the x-intercept is \((6,0)\).
Set \(x=0\): \(3y=24\), so the y-intercept is \((0,8)\).
Plot the two points and draw the unique straight line through them. Substitute each into the original equation to check.
Find the intercepts of \(5x-2y=10\).
\[y=0:\ 5x=10\Rightarrow(2,0).\]\[x=0:\ -2y=10\Rightarrow(0,-5).\]A negative y-intercept goes below the x-axis.
Sometimes both intercepts are the origin \((0,0)\). Then you need a second point to draw the line: choose another convenient \(x\)-value and solve for \(y\).
Check your work — Ready 1–6 answer key
- (a) x-intercept \((4,0)\). (b) y-intercept \((0,6)\). Plot not shown.
- (a) x-intercept \((-3,0)\). (b) y-intercept \((0,2)\). Plot not shown.
- (a) x-intercept \((7,0)\). (b) y-intercept \((0,-3)\). Plot not shown.
- (a) x-intercept \((4,0)\). (b) y-intercept \((0,-2)\). Plot not shown.
- (a) x-intercept \((3,0)\). (b) y-intercept \((0,-9)\). Plot not shown.
- (a) x-intercept \((1,0)\). (b) y-intercept \((0,6)\). Plot not shown.
3. Set — Expand and complete perfect squares (7–26)
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.
For the shared perfect-square pattern and missing-constant method, revisit the algebraic pattern and finding the missing piece in the In class section.
Square a binomial with a multiple of x
The Set homework also squares expressions beginning with a multiple of \(x\). Here the two rectangular products are \(ax\cdot q\) rather than \(x\cdot q\):
In general, \((ax+q)^2=a^2x^2+2aqx+q^2\). The coefficient of the middle \(x\)-term is \(B=2aq\).
Find a missing middle coefficient
You may instead know the final constant but not the middle coefficient. A perfect square ending in \(q^2\) has middle term \(2qx\) if the binomial is \(x+q\), or \(-2qx\) if it is \(x-q\).
Complete \(x^2+Bx+64\).
Since \(64=8^2\), there are two possible perfect squares:
\[(x+8)^2=x^2+16x+64,\]\[(x-8)^2=x^2-16x+64.\]Therefore \(B=16\) or \(B=-16\). Both work unless the question places a restriction on the sign.
Complete \(x^2+Bx+\frac9{16}\). The square root of \(\frac9{16}\) is \(\frac34\). Double it to obtain \(\frac32\).
\[B=\pm\frac32,\qquad x^2\pm\frac32x+\frac9{16}=\left(x\pm\frac34\right)^2.\]These problems ask for the value of \(B\) when the constant term is given. Each positive constant shown in these exercises allows two values, positive and negative, because \((x+q)^2\) and \((x-q)^2\) have the same final term. Unless the instructor specifies \(B>0\), report both possibilities with their corresponding factorizations.
Check your work — Set 7–26 answer key
- \(x^2+5x+5x+25=x^2+10x+25\). The two middle terms combine.
- \(9x^2+21x+21x+49=9x^2+42x+49\). The two middle terms combine.
- \(81x^2+9x+9x+1=81x^2+18x+1\). The two middle terms combine.
- \(16x^2+44x+44x+121=16x^2+88x+121\). The two middle terms combine.
- \(B=2aq\), because the cross-products are \(aqx+aqx\).
- (a) \(16\). (b) \((x+4)(x+4)\).
- (a) \(25\). (b) \((x+5)(x+5)\).
- (a) \(64\). (b) \((x+8)(x+8)\).
- (a) \(9\). (b) \((x+3)(x+3)\).
- (a) \(121\). (b) \((x+11)(x+11)\).
- (a) \(81\). (b) \((x+9)(x+9)\).
- (a) \(B=\pm 8\). (b) \((x+4)^2\) for positive B, or \((x-4)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 22\). (b) \((x+11)^2\) for positive B, or \((x-11)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 50\). (b) \((x+25)^2\) for positive B, or \((x-25)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 30\). (b) \((x+15)^2\) for positive B, or \((x-15)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 14\). (b) \((x+7)^2\) for positive B, or \((x-7)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 26\). (b) \((x+13)^2\) for positive B, or \((x-13)^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 5\). (b) \((x+\frac{5}{2})^2\) for positive B, or \((x-\frac{5}{2})^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 3\). (b) \((x+\frac{3}{2})^2\) for positive B, or \((x-\frac{3}{2})^2\) for negative B (each square is two equal factors).
- (a) \(B=\pm 7\). (b) \((x+\frac{7}{2})^2\) for positive B, or \((x-\frac{7}{2})^2\) for negative B (each square is two equal factors).
4. Go review — Horizontal and vertical lines
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.
In an equation \(y=c\), every point has the same y-coordinate; its graph is a horizontal line. If \(c\ne0\), it has the y-intercept \((0,c)\) but no x-intercept. In \(x=c\), every point has the same x-coordinate; its graph is vertical. If \(c\ne0\), it has the x-intercept \((c,0)\) but no y-intercept.
Identify and classify all intercepts of \(y=7\).
All points have y-coordinate 7, so the line is horizontal and crosses the y-axis at \((0,7)\). Because \(y\) never equals 0, there is no x-intercept.
Find and classify the intercepts of \(x=-\frac73\). Then compare with \(y=-\frac73\).
The equation \(x=-\frac73\) describes a vertical line, crossing the x-axis at \((-\frac73,0)\); it has no y-intercept. The equation \(y=-\frac73\) describes a horizontal line, crossing the y-axis at \((0,-\frac73)\); it has no x-intercept. Swapping \(x\) and \(y\) changes both line orientation and intercept type.
Find the intercepts of \(x=0\) and \(y=0\).
\(x=0\) is the entire y-axis, so every point on it is a y-intercept in the geometric sense and the origin is its only x-intercept. Likewise, \(y=0\) is the entire x-axis, so every point on it lies on the x-axis and the origin is its only y-intercept. They do not have one unique intercept of the coincident-axis type. The packet's problems 27–30 avoid these zero cases.
Check your work — Go 27–30 answer key
- y-intercept \((0,-4.5)\); no x-intercept.
- x-intercept \((9.5,0)\); no y-intercept.
- x-intercept \((-8.2,0)\); no y-intercept.
- y-intercept \((0,112)\); no x-intercept.
Additional Learning Resources
These links are supplementary, not needed to follow this lesson. The table separates exact matches from lessons that continue further than today's packet.
| Resource and link | What it covers / why use it | Scope / what to skip |
|---|---|---|
| Khan Academy — Factoring perfect square trinomials | Square-of-a-binomial patterns, their reverse form, and examples with \(x^2\) and \(9x^2\) leading terms. | Some extra material: Study the perfect-square examples; the more general factoring methods are not needed today. |
| Khan Academy — Completing the square (intro) practice | Find the missing constant that turns a trinomial into a perfect square. | Close match: Practice the missing-term items; save quadratic-equation solving for later. |
| OpenStax — Graph with Intercepts | Explicit explanation and examples of finding and graphing intercepts of straight lines; supports Ready and Go. | Close match: Use the sections on finding and graphing intercepts. Other linear graphing methods are optional review. |
Teacher / Parent Review of Source Packet
Section 2.3, Set problems 18–26: Asking for “the value” of \(B\) from a positive constant gives two valid results, \(B=\pm2\sqrt c\), not a unique value. Correct the directions to “Find all values of \(B\)” or explicitly stipulate \(B>0\). This is important because a student providing both signs is mathematically correct.
No other serious ambiguity is required to teach Section 2.3. The square diagrams and intercept review are consistent with the intended mathematical goals.
Section 2.3 Packet-to-Lesson Coverage
| Source material | Required skills | Where addressed |
|---|---|---|
| Main task 1–5 | Area models for \((x+n)^2\), equivalent forms and patterns | In class, area diagram, worked examples 1–4 |
| Main task 6–7 | Missing constants; general relationship between \(b\) and \(c\), including odd/negative coefficients | In class, examples 6–7 |
| Ready 1–6 | x- and y-intercepts of general lines, graph using two points | Section 2, examples 10–11 |
| Set 7–11 | Expand squared binomials with and without a coefficient on \(x\); middle coefficient \(B=2aq\) | In class, examples 3–4; Set, example 5 |
| Set 12–17 | Fill in constants and write squared factors | In class (linked from Set), examples 6–7 |
| Set 18–26 | Find missing middle coefficients from square constants, including fractions; both signs | Set, examples 8–9 and packet clarification |
| Go 27–30 | Horizontal/vertical lines, classify intercepts | Section 4, three original Go examples |
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