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3.11 — Quadratic Quandaries
In previous sections you solved quadratic equations such as \(x^2-5x+6=0\). Today you solve quadratic inequalities, which often have infinitely many solutions. The familiar skills of factoring, graphing, and completing the square will do most of the work.
1. In class: A Develop & Solidify Understanding Task
From a quadratic equation to a quadratic inequality
An equation asks, “Where are two quantities equal?” The equation \(x^2-5x+6=0\) is true only when \(x=2\) or \(x=3\). An inequality asks, “For which inputs is one quantity larger or smaller?” For example, \(x^2-5x+6<0\) asks where the corresponding graph is below the x-axis.
Factor: \(x^2-5x+6=(x-2)(x-3)\). Its zeroes, \(2\) and \(3\), divide the number line into three intervals. On an interval that contains no zeros, neither factor changes sign, so their product keeps the same sign there.
Solve \(x^2-5x+6<0\).
Step 1. Find boundary values by solving \((x-2)(x-3)=0\): \(x=2,3\).
Step 2. Test the three intervals. For \(x=0\), \((-2)(-3)=6>0\). For \(x=2.5\), \((0.5)(-0.5)=-0.25<0\). For \(x=4\), \((2)(1)=2>0\).
Step 3. We need the negative interval, giving \(\boxed{2<x<3}\). The endpoints are excluded because the inequality is strict.
A reliable procedure for quadratic inequalities
The idea is the same whether you use a sign chart, a sketch, or a table. Put everything on one side, find the zeros of the corresponding quadratic, then determine the sign of the quadratic in each interval.
Solve \(x^2+3x-10\ge0\).
Factor: \((x+5)(x-2)\ge0\). The zeros are \(-5\) and \(2\). Outside the roots, both factors are negative (to the far left) or both positive (to the far right), so the product is nonnegative. Between the roots the product is negative.
\(\boxed{x\le-5\quad\text{or}\quad x\ge2}\).
Solve \(2x^2-5x<12\).
First subtract 12: \(2x^2-5x-12<0\). Factor: \((2x+3)(x-4)<0\). The zeros are \(-\frac32\) and \(4\). Test \(x=0\): \((3)(-4)=-12<0\); the graph lies below zero between the roots.
\(\boxed{-\frac32<x<4}\).
Quadratic inequalities in a revenue problem
Carlos and Clarita’s driveway-mascot situation is revisited in the packet. They can sell 100 mascots at $20 each, but sell 10 fewer for every $5 price increase. Let \(x\) be the number of $5 price increases. Thus the price is \(20+5x\) dollars, the quantity is \(100-10x\), and the revenue is
\[R(x)=(100-10x)(20+5x)=-50x^2+300x+2000.\]To model nonnegative prices and quantities, the meaningful interval is \(-4\le x\le10\). If “price increase” is meant literally, add \(x\ge0\); if only whole $5 increments are allowed, \(x\) must also be an integer. We will first treat \(x\) as continuous, as the packet’s graphs suggest, and then explain the discrete interpretation.
The twins want $2,000 in profit after $250 in supplies, so they need $2,250 in revenue. Solve
\[(100-10x)(20+5x)=2250.\]Expand, subtract 2250, and divide by \(-50\):
\[-50x^2+300x-250=0\quad\Longrightarrow\quad x^2-6x+5=0.\]Factor: \((x-1)(x-5)=0\), so \(x=1\) or \(x=5\). The two prices are \(20+5(1)=\$25\) and \(20+5(5)=\$45\). Both produce exactly $2,250.
“At least” means \(R(x)\ge2250\), not equality. Rearranging:
\[-50(x-1)(x-5)\ge0.\]Divide by \(-50\), reversing the inequality:
\[(x-1)(x-5)\le0.\]The product is nonpositive between its roots, inclusive: \(\boxed{1\le x\le5}\). The possible prices therefore run from \(\boxed{\$25\text{ to }\$45}\), inclusive, under the continuous model. For whole $5 increments the allowable prices are $25, $30, $35, $40, and $45.
Complete the square:
\[R(x)=-50(x^2-6x)+2000=-50(x-3)^2+2450.\]Since a square is never negative, \(-50(x-3)^2\le0\). Revenue is largest at \(x=3\), for a price of \(20+5(3)=\$35\). Maximum revenue is \(\boxed{\$2450}\); after $250 in supply costs, profit is \(\boxed{\$2200}\).
Quadratic inequalities with geometric constraints
Suppose a rectangular logo has width \(x\) inches and length \(x+8\) inches. Its area is \(A=x(x+8)\). The requirement “less than 48 square inches” translates to \(x(x+8)<48\).
Move all terms left: \(x^2+8x-48<0\). Factor:
\[(x+12)(x-4)<0.\]This is negative between the zeros, giving \(-12<x<4\) as a purely algebraic solution. But a physical width must be positive. Intersect with \(x>0\):
\(\boxed{0<x<4}\) inches; the length \(x+8\) satisfies \(\boxed{8<x+8<12}\) inches. Each pair is linked by \(\text{length}=\text{width}+8\); the width and length cannot be chosen independently.
Problem 5 asks about a physical rectangle, so negative or zero widths must be excluded. Problem 6 explicitly removes the physical context: its answer is \(-12<x<4\), whereas the dimensions in Problem 5 require \(0<x<4\). In both cases “less than” excludes an area exactly equal to 48 square inches.
Compare a parabola to a line in two equivalent ways
Not every inequality begins with zero on one side. For example, \(x^2-4\le4x+1\) asks where the quadratic expression is no greater than the linear expression. Two graphs can be compared directly, or the expressions can be combined to create a single quadratic.
Subtract the right side to create \(x^2-4x-5\le0\). Factor:
\[(x-5)(x+1)\le0.\]The graph \(y=x^2-4x-5\) is on or below the x-axis between the roots: \(\boxed{-1\le x\le5}\).
Graph \(y=x^2-4\) and \(y=4x+1\). The graphs meet where \(x^2-4=4x+1\), or \((x-5)(x+1)=0\), at \(x=-1\) and \(x=5\). Between them, the parabola lies below the line (at \(x=0\), \(-4<1\)); include the intersection inputs because the inequality uses \(\le\). Again, \(\boxed{-1\le x\le5}\).
2. Ready — Factor polynomials completely (Problems 1–10)
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.
Factoring translates a quadratic into a product and makes its zeros visible. For a monic trinomial \(x^2+bx+c\), seek numbers \(m,n\) with \(m+n=b\) and \(mn=c\). For \(ax^2+bx+c\), where \(a\ne1\), use the product \(ac\), split the middle term, and factor by grouping. Check by distributing.
Factor \(x^2-4x-21\). Find numbers whose product is \(-21\) and sum is \(-4\): \(-7,3\). Therefore \(\boxed{(x-7)(x+3)}\).
Factor \(3x^2+7x-6\). Since \(ac=-18\), use \(9\) and \(-2\):
\[3x^2+9x-2x-6=3x(x+3)-2(x+3)=(3x-2)(x+3).\]The product is fully factored. Its zeroes \(\frac23\) and \(-3\) are the graph’s x-intercepts.
Factor \(x^2+10x+25\). Since \(25=5^2\) and \(10x=2(5)x\), \(\boxed{(x+5)^2}\). The corresponding parabola touches the x-axis once at \(-5\), rather than crossing it.
Check your work — Ready 1–10 answer key
- \((x+4)(x-3)\).
- \((x-4)(x+2)\).
- \((x+7)(x-2)\).
- \((x-3)(x+2)\).
- \((x+3)^2\).
- \((x-5)(x-2)\).
- \((2x+1)(x-5)\).
- \(3(x-3)(x+2)\).
- \(2(x+7)(x-3)\).
- Factored form shows the x-intercepts by setting factors to zero. The symmetry axis is halfway between the roots, and evaluating there gives the vertex. The leading coefficient gives opening direction and vertical scale.
3. Set — Solve inequalities and match shaded graphs (Problems 11–23)
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 ideas used in these exercises, revisit the In class explanation.
Problems 11–19 ask for intervals of inputs. They include both product signs and inequalities such as \(x^2\le25\), which can be factored by difference of squares.
Solve \(x^2\le36\):
\[x^2-36\le0\quad\Longrightarrow\quad(x-6)(x+6)\le0.\]The product is nonpositive between the roots: \(\boxed{-6\le x\le6}\).
Solve \(3x^2-12x-15<0\). Factor \(3(x-5)(x+1)<0\). Since 3 is positive, the product is negative between \(-1\) and \(5\): \(\boxed{-1<x<5}\).
Problems 20–23 ask you to match two-variable inequalities to graphs. The boundary is the parabola \(y=f(x)\); the solution consists of all points above it for \(y>f(x)\) or below it for \(y<f(x)\). A strict inequality uses a dashed boundary.
Sketch \(y>x^2-4\). Draw the upward-opening parabola with vertex \((0,-4)\), passing through \((-2,0)\) and \((2,0)\). Make the curve dashed, because equality is excluded. Shade the region above the curve. To check, \((0,0)\) is a solution because \(0>-4\).
The matching graphs show shaded regions in two dimensions, not solution intervals on the x-axis. The boundaries are dashed. Compare (1) whether the parabola opens up or down, (2) its vertex and intercepts, and (3) whether the shading is above or below the boundary. In the provided graphs the matches are 20 → b, 21 → c, 22 → d, and 23 → a.
Check your work — Set 11–23 answer key
- \(x<-4\text{ or }x>3\).
- \(-2\le x\le4\).
- \(x\le-7\text{ or }x\ge2\).
- \(x\le-\frac12\text{ or }x\ge5\).
- \(-2<x<3\).
- \(-7<x<3\).
- \(0\le x\le4\).
- \(-5\le x\le5\).
- \(-1\le x\le5\).
- Graph b: dashed upward boundary, shaded above.
- Graph c: dashed upward boundary, shaded below.
- Graph d: dashed downward boundary \(y=-x^2-x+2\), shaded below.
- Graph a: dashed downward boundary \(y=-(x+2)^2+9\), shaded above.
4. Go — Write quadratic functions in vertex form (Problems 24–29)
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 homework finishes with a familiar skill from earlier modules: turning standard or factored form into vertex form, \(f(x)=a(x-h)^2+k\). The vertex is \((h,k)\). To rewrite an expression, complete the square carefully; remember to factor out the leading coefficient before completing the square when it is not 1.
Rewrite \(f(x)=x^2+8x+11\). Half of 8 is 4; its square is 16:
\[f(x)=(x^2+8x+16)-16+11=\boxed{(x+4)^2-5}.\]The vertex is \((-4,-5)\).
Rewrite \(g(x)=(x-4)(x+2)\). Expand, then complete the square:
\[g(x)=x^2-2x-8=(x^2-2x+1)-9=\boxed{(x-1)^2-9}.\]Check: the vertex x-coordinate is also the midpoint of the roots 4 and −2, namely 1.
Rewrite \(h(x)=3x^2+18x+14\). Factor out 3 from the variable terms:
\[h(x)=3(x^2+6x)+14.\]Complete the square inside: \(x^2+6x=(x+3)^2-9\).
\[h(x)=3[(x+3)^2-9]+14=\boxed{3(x+3)^2-13}.\]The vertex is \((-3,-13)\). Subtracting 9 outside the parentheses would give the wrong constant because the bracket is multiplied by 3.
New, fully worked Go practice
Convert \(p(x)=x^2-10x+21\) to vertex form and identify its vertex.
Solution: Half of \(-10\) is \(-5\); square it to get 25. Add and subtract 25:
\[p(x)=x^2-10x+25-25+21=\boxed{(x-5)^2-4}.\]Vertex: \(\boxed{(5,-4)}\). Verification: \((x-5)^2-4=x^2-10x+25-4\).
Convert \(q(x)=-2x^2+12x-7\) into vertex form. State whether the vertex gives a maximum or a minimum.
Solution: Factor out \(-2\) from the quadratic and linear terms:
\[q(x)=-2(x^2-6x)-7.\]Now \(x^2-6x=(x-3)^2-9\), so
\[q(x)=-2[(x-3)^2-9]-7=\boxed{-2(x-3)^2+11}.\]The vertex is \((3,11)\); it is a maximum because the coefficient of the square is negative.
Convert \(r(x)=2(x-1)(x-7)\) to vertex form.
Solution: The symmetry line lies halfway between roots 1 and 7, so \(x=4\). Evaluate \(r(4)=2(3)(-3)=-18\). Therefore, with leading coefficient 2,
\[\boxed{r(x)=2(x-4)^2-18}.\]Check by expanding: \(2(x^2-8x+16)-18=2x^2-16x+14=2(x-1)(x-7)\).
Convert \(s(x)=\frac12x^2-3x+8\) to vertex form.
Solution: Factor \(\frac12\): \(s(x)=\frac12(x^2-6x)+8\). Complete the square with 9:
\[s(x)=\frac12[(x-3)^2-9]+8=\boxed{\frac12(x-3)^2+\frac72}.\]The vertex is \(\left(3,\frac72\right)\).
Check your work — Go 24–29 answer key
- \(f(x)=(x+3)^2-4\).
- \(f(x)=(x-1)^2-16\).
- \(f(x)=(x+2)^2-16\).
- \(f(x)=(x-6)^2-16\).
- \(f(x)=2(x+4)^2-24\).
- \(f(x)=(x-1)^2-9\).
5. Additional learning resources
| Resource | Why use it | Scope / what to skip |
|---|---|---|
| Khan Academy — Quadratic inequalities | Short, direct example of factoring and determining a solution interval. | Close match. Focus on intervals and endpoints. |
| Khan Academy — Quadratic inequalities, example 2 | Another worked sign-analysis problem, including a negative leading coefficient. | Close match. Useful for Set problems 11–19. |
| Professor Leonard — Solving Quadratic Inequalities (Precalculus / College Algebra 25) | Optional extended classroom-style lecture on quadratic inequalities. | Some extra material. Concentrate on inequalities of a single quadratic expression and their intervals. Timestamps not verified; skip examples requiring concepts not used in the assigned packet. |
| Khan Academy — Quadratic inequalities within quadratic equations and functions | Provides related graph-based explanations; includes a link to the graphical approach. | Significantly broader page. Use only the “Quadratic inequalities” segment; other topics are unrelated to today’s assignment. |
6. Teacher’s review (for teachers and parents)
No serious mathematical errors in the assigned Section 3.11 problems were identified. Two scope distinctions are worth making explicit for students:
Main-task Problems 5–7: Problem 5’s dimensions require \(x>0\); Problem 6 deliberately drops this physical restriction. Students should distinguish the algebraic and contextual solution sets.
Main-task Problems 1–4: Treating the count of “additional $5” increases as continuous gives price intervals and a vertex; if sales price may change only in increments of $5, use discrete admissible prices. A clearer prompt would state explicitly whether fractional values of \(x\) are allowed.
7. Curriculum coverage and study checklist
| Packet section / problems | Skills and where taught |
|---|---|
| Main task 1–4 | Equation versus inequality, model revenue, exact target, minimum target, maximum revenue: In class and 3. |
| Main task 5–7 | Rectangle area inequalities with and without contextual dimension restrictions: Section 4. |
| Main task 8–10 | Factor and solve quadratic inequalities with strict and inclusive signs: Section 2. |
| Main task 11–12 | Single difference graph versus intersecting line and parabola: Section 5. |
| Ready 1–10 | Factoring, perfect squares, leading coefficients, roots and graph intercepts: Section 6. |
| Set 11–19 | Solution intervals and endpoints: In class and 7. |
| Set 20–23 | Graph boundaries, dashed lines, shading above or below: Section 7. |
| Go 24–29 | Complete the square and rewrite from standard / factored into vertex form; additional original Go examples A–D: Section 8. |
Source: Mathematics Vision Project, Secondary Mathematics II, Module 3, Section 3.11, “Quadratic Quandaries,” main task and Ready, Set, Go (packet pp. 70–76). This is independent supplementary instruction, not an answer-key replacement.
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