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3.6 — Curbside Rivalry
Suppose you know a rectangle’s area but not its side lengths. If one side is a fixed amount longer than the other, a quadratic equation describes the missing dimensions. A different quadratic equation appears when a business changes its prices and the number of customers changes in response. Both problems start with ordinary relationships, not with a formula to memorize.
1. In class: A Solidify Understanding Task
Write a quadratic equation from a rectangular area
For a rectangle, area equals length times width. If its width is \(x\) and its length is \(n\) units greater, its area is \(x(x+n)\). Both lengths must be positive for a physical rectangle.
Width: \(x\)
Length: \(x+n\)
Area: \(A\)
\[\boxed{x(x+n)=A}\]Expanding and moving everything to one side gives \(x^2+nx-A=0\). The geometry requires \(x>0\) and \(x+n>0\).
A square has four equal sides, so \(x^2=48\). Take the nonnegative root because a length cannot be negative.
\[x=\sqrt{48}=\sqrt{16\cdot3}=4\sqrt3\approx6.93\text{ inches}.\]The logo is \(4\sqrt3\) inches by \(4\sqrt3\) inches. Do not use the negative solution for a side length.
The area is 48 square inches. Let width be \(x\) inches; length is \(x+8\).
\[x(x+8)=48\quad\Longrightarrow\quad x^2+8x-48=0.\]Find integers whose product is \(-48\) and whose sum is \(8\): 12 and −4.
\[(x+12)(x-4)=0.\]So \(x=-12\) or \(x=4\). Only \(x=4\) is a possible width, giving a 4-inch by 12-inch rectangle. Check: \(4\cdot12=48\).
Again let width be \(x\):
\[x(x+13)=48\quad\Longrightarrow\quad x^2+13x-48=0.\] \[(x+16)(x-3)=0\quad\Longrightarrow\quad x=-16\text{ or }x=3.\]Discard the negative width; the rectangle is 3 inches by 16 inches.
When factoring is not convenient, use a different strategy
The same quadratic can be viewed as an equation, \(x^2+nx-A=0\), or a function, \(y=x^2+nx-A\). Its algebraic solutions are exactly the x-values of the graph’s x-intercepts. A graph can help estimate solutions; algebra gives exact values.
Find a rectangle of area 48 square inches whose length is 6 inches greater than its width.
\[x^2+6x-48=0\quad\Longrightarrow\quad x^2+6x=48.\]Add \((6/2)^2=9\) to both sides:
\[(x+3)^2=57\quad\Longrightarrow\quad x=-3\pm\sqrt{57}.\]The usable width is \(\sqrt{57}-3\approx4.55\) inches and the length is \(\sqrt{57}+3\approx10.55\) inches. The exact product is \((\sqrt{57}-3)(\sqrt{57}+3)=57-9=48\).
Apply the formula to \(x^2+6x-48=0\) with \(a=1\), \(b=6\), \(c=-48\):
\[x=\frac{-6\pm\sqrt{6^2-4(1)(-48)}}{2(1)}=\frac{-6\pm\sqrt{228}}2=-3\pm\sqrt{57}.\]The formula agrees with completing the square. This gives two algebraic roots, but only one valid width.
Factoring: use when the quadratic has easily recognized factors, as in \(x^2+8x-48\).
Completing the square: use to isolate one squared expression, as in \((x+3)^2=57\).
Quadratic formula: use for any quadratic equation \(ax^2+bx+c=0\), where \(a\ne0\).
\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]Graphing: use the x-intercepts to visualize or check solutions; a sketch usually supplies estimates rather than exact roots.
A rectangle is 6 feet longer than it is wide. Write \(x(x+6)=16\).
\[x^2+6x-16=(x+8)(x-2)=0.\]The valid width is 2 feet and length is 8 feet, since \(2\cdot8=16\).
If the rectangle is instead 10 feet longer than wide, write \(x^2+10x-16=0\). Complete the square:
\[(x+5)^2=41\quad\Longrightarrow\quad x=-5\pm\sqrt{41}.\]The positive width is \(\sqrt{41}-5\approx1.40\) feet; the length is \(\sqrt{41}+5\approx11.40\) feet.
Build a revenue model from price and quantity
Suppose a service can sell 100 driveway mascots at $20 each. For each $5 price increase, it sells 10 fewer. Let \(x\) be the number of $5 increases. In the survey model, both price and quantity are linear functions of \(x\).
Price per mascot:
\[\boxed{P(x)=20+5x}\]Quantity sold:
\[\boxed{Q(x)=100-10x}\]Revenue = price × quantity:
\[\boxed{R(x)=(20+5x)(100-10x)}\]For integer price-increase steps with a nonnegative selling price and a nonnegative number sold, the main task uses \(0\le x\le10\). The polynomial can be graphed beyond that interval algebraically, but those values need not have business meaning.
Compute the three functions at sample increments.
| \(x\): number of $5 increases | \(P(x)\): price ($) | \(Q(x)\): sales | \(R(x)=P(x)Q(x)\): revenue ($) |
|---|---|---|---|
| 0 | 20 | 100 | 2000 |
| 1 | 25 | 90 | 2250 |
| 2 | 30 | 80 | 2400 |
| 3 | 35 | 70 | 2450 |
| 4 | 40 | 60 | 2400 |
| 5 | 45 | 50 | 2250 |
| 10 | 70 | 0 | 0 |
Multiplying and combining like terms gives
\[R(x)=2000+300x-50x^2.\]Revenue is quadratic because it is the product of two linear expressions. It first rises and then falls: eventually the lower sales outweigh the higher price.
Price \(P(x)\) rises in a straight line; sales \(Q(x)\) falls in a straight line. Because their vertical units differ, avoid interpreting their heights as comparable quantities even if both are plotted on a shared coordinate grid, as the packet requests.
Solve a target-revenue equation
The twins want $2,000 profit after $250 of fixed supplies. Since profit = revenue − cost, their revenue target is \(2000+250=2250\) dollars. The problem gives a fixed cost estimate; it does not provide separate per-mascot costs.
Set the revenue function equal to $2,250:
\[(20+5x)(100-10x)=2250.\]Expand and simplify:
\[2000+300x-50x^2=2250\] \[-50x^2+300x-250=0\] \[x^2-6x+5=0\] \[(x-1)(x-5)=0.\]Therefore \(x=1\) or \(x=5\). Translate increments back into prices:
\[P(1)=20+5(1)=\$25,\qquad P(5)=20+5(5)=\$45.\]Check: at $25, 90 mascots give $2,250; at $45, 50 mascots also give $2,250. Both are valid within the model. After subtracting $250, profit is $2,000 either way.
Complete the square in the revenue equation:
\[R(x)=-50(x^2-6x)+2000=-50(x-3)^2+2450.\]The squared term is never negative, so \(-50(x-3)^2\le0\). Therefore maximum revenue is $2,450, at \(x=3\), meaning a price of \(20+5(3)=\$35\). The graph is symmetric around \(x=3\), which explains why \(x=1\) and \(x=5\) yield the same revenue.
2. Ready — x- and y-intercepts of linear equations
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 an x-intercept, set \(y=0\) and solve for \(x\). For a y-intercept, set \(x=0\) and solve for \(y\). Record intercepts as ordered pairs. This works regardless of whether the line appears in standard, slope-intercept, or point-slope form.
Find the intercepts of \(3x+4y=24\).
Set \(y=0\): \(3x=24\), so the x-intercept is \((8,0)\).
Set \(x=0\): \(4y=24\), so the y-intercept is \((0,6)\).
Find the intercepts of \(y=-2x+10\).
The y-intercept \((0,10)\) is immediately visible. For the x-intercept, \(0=-2x+10\) gives \((5,0)\).
Find the intercepts of \(y-3=2(x+4)\).
Set \(x=0\): \(y-3=8\), giving \((0,11)\). Set \(y=0\): \(-3=2(x+4)\); divide by 2 and subtract 4 to get \(x=-11/2\). The x-intercept is \((-11/2,0)\).
\(y=mx+b\): the y-intercept \((0,b)\) is already shown.
\(Ax+By=C\): set one variable to zero to find either intercept directly.
\(y-y_1=m(x-x_1)\): substitute zero for one coordinate and solve; this form makes the given point \((x_1,y_1)\) visible.
Check your work — Ready 1–4 answer key
- a. \((4,0)\); b. \((\frac35,0)\); c. \((\frac14,0)\); d. \((\frac14,0)\); e. \((-10,0)\); f. \((2,0)\).
- Standard form \(Ax+By=C\) is convenient: setting \(y=0\) immediately gives \(Ax=C\). Slope-intercept form also works directly by setting its output to zero.
- a. \((0,3)\); b. \((0,-3)\); c. \((0,1)\); d. \((0,1)\); e. \((0,20)\); f. \((0,-5)\).
- Slope-intercept form \(y=mx+b\) displays the y-intercept \((0,b)\) directly. Standard form is also convenient: setting \(x=0\) gives \(By=C\).
3. Set — turn quadratic equations back into rectangles
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.
An equation of the form \(x^2+nx-A=0\) can be rearranged as \(x(x+n)=A\). The rectangle has a width \(x\), length \(x+n\), and area \(A\) square units. The interpretation requires both dimensions to be positive; an equation with a positive area can still have a negative algebraic solution.
Interpret and solve \(x^2+9x-36=0\).
Write \(x(x+9)=36\). A width of \(x\) and length \(x+9\) give area 36. Factoring:
\[(x+12)(x-3)=0\quad\Longrightarrow\quad x=-12\text{ or }x=3.\]Only \(x=3\) is a valid width. The rectangle measures 3 by 12.
Interpret \(x^2+4x-18=0\), or \(x(x+4)=18\). The length is 4 units greater than the width and the area is 18.
\[(x+2)^2=22\quad\Longrightarrow\quad x=-2\pm\sqrt{22}.\]Only \(x=\sqrt{22}-2\approx2.69\) is positive. The length is \(\sqrt{22}+2\approx6.69\). Their exact product is 18.
“What constraints have been placed on the dimensions?” refers to the physical restrictions on the rectangle: both lengths must be positive, and the longer side is the fixed amount greater than the shorter side. The quadratic may have a negative solution, but that solution does not represent a possible width.
Check your work — Set 5–12 answer key
- a. Area 170 square units, sides \(x,x+7\). b. Length is 7 units longer than width; \(x>0\).
- a. Area 16 square units, sides \(x,x+15\). b. Length is 15 units longer than width; \(x>0\).
- a. Area 35 square units, sides \(x,x+2\). b. Length is 2 units longer than width; \(x>0\).
- a. Area 80 square units, sides \(x,x+10\). b. Length is 10 units longer than width; \(x>0\).
- \(x=10,-17\).
- \(x=1,-16\).
- \(x=5,-7\).
- \(x=-5\pm\sqrt{105}\).
4. Go — factor quadratic expressions thoroughly
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 Go homework asks for factored expressions, not necessarily solutions to equations. In particular, \(x^2-5x+6=(x-2)(x-3)\) is a factorization, but writing \(x=2\) or \(x=3\) would only be appropriate if an equation equal to zero were given.
1. Greatest common factor (GCF): remove any factor shared by every term.
2. Trinomial \(x^2+bx+c\): find \(m,n\) with \(m+n=b\) and \(mn=c\), then write \((x+m)(x+n)\).
3. Difference of squares: \(u^2-v^2=(u-v)(u+v)\).
4. Perfect-square trinomial: \(u^2\pm2uv+v^2=(u\pm v)^2\).
Factor \(x^2+9x+20\).
Two numbers must multiply to 20 and add to 9: 4 and 5.
\[\boxed{x^2+9x+20=(x+4)(x+5).}\]Check by expansion: \(x^2+5x+4x+20=x^2+9x+20\).
Factor \(x^2-7x-60\).
We need a negative product and a negative sum. The pair is −12 and 5:
\[\boxed{x^2-7x-60=(x-12)(x+5).}\]Verify: the outer and inner terms add to \(-12x+5x=-7x\).
Factor \(6x^2+18x\).
Both terms contain \(6x\); pull it out rather than trying to form two binomials:
\[\boxed{6x^2+18x=6x(x+3).}\]Factor \(x^2-14x+49\).
Since \(49=7^2\) and \(-14x=-2\cdot7\cdot x\), this is a perfect-square trinomial:
\[\boxed{x^2-14x+49=(x-7)^2.}\]Factor \(-2x^2+18x\).
Factor out \(-2x\) so the binomial begins with a positive \(x\):
\[\boxed{-2x^2+18x=-2x(x-9).}\]Expanding \(-2x\cdot(-9)\) gives the positive \(18x\) term.
Factor \(3x^2-75\).
First remove 3: \(3(x^2-25)\). Now \(x^2-25=x^2-5^2\) is a difference of squares:
\[\boxed{3x^2-75=3(x-5)(x+5).}\]Check your work — Go 13–21 answer key
- \((x-12)(x+11)\).
- \((x-9)(x+4)\).
- \((x+2)(x+3)\).
- \((x+6)(x+7)\).
- \((x+8)(x-7)\).
- \(x(x-1)\).
- \((x-2)(x-6)\).
- \((x-5)^2\).
- \(x(x+5)\).
Additional learning resources
Use these targeted resources for extra instruction and practice. The packet's next section compares more solving approaches, so there is no need to work through unrelated advanced topics today.
| Specific resource | Best use | Scope / What to skip |
|---|---|---|
| Khan Academy — Solving quadratic equations by factoring | Review the zero-product property and using factors to solve the rectangle and revenue equations. | Close match. This covers the core algebra in the main task and Set. |
| Khan Academy — Quadratics by factoring (intro practice) | Short interactive factoring practice, useful before the Go homework. | Close match. Limit practice to quadratics that factor over integers. |
| Professor Leonard — Using Factoring to Solve Quadratics (Precalculus / College Algebra 18) | Optional traditional lecture on factoring quadratic expressions and solving equations. | Some extra material. Focus on monic quadratic factoring and solving; skip more complicated cases not needed for Section 3.6. No verified timestamps available. |
| Khan Academy — Writing quadratic equations from word problems | Practice translating verbal conditions into an algebraic equation. | Some extra material. Choose rectangle/product questions; skip unrelated settings if they require unintroduced modeling. |
Teacher’s review — serious packet issues
No major mathematical error was identified in Section 3.6. The main task's implicit modeling assumption deserves attention: the survey's linear relationship between price and quantity sold is treated as reliable at $5 steps in the modeled interval. It is a model based on survey estimates, not a guaranteed count of customers at every price.
Section 3.6, Set, Problems 5–8: “What constraints have been placed on the dimensions?” can be read as asking only for an algebraic equation rather than real geometric constraints. A clearer version would be: “Identify the two side-length expressions and state which values of \(x\) make both side lengths positive.”
Coverage table — Section 3.6 only
| Packet portion | Skill addressed | In this lesson |
|---|---|---|
| Main task 1–3 | Square and rectangle dimensions; area equations | In class |
| Main task 4–8 | Two solution strategies; x-intercepts; nonfactorable length equations; valid physical roots | In class |
| Main task 9–11 | Price, sales, revenue models, tables and graphs | In class |
| Main task 12 | Profit-to-revenue target; solving quadratic pricing equation | In class |
| Ready 1–4 | Linear x- and y-intercepts in different equation formats | Section 2 |
| Set 5–12 | Rectangle interpretation, dimension constraints, solving quadratics | Section 3 |
| Go 13–21 | Trinomial factoring, perfect squares, GCF | Section 4, six new fully worked problems |
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