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3.7 — Perfecting My Quads
You already know several methods for quadratic equations. The new challenge is fluency: understanding why each method works, noticing when it gives an exact result and when it only gives an estimate, and choosing an efficient one for the equation in front of you.
1. In class: A Practice Understanding Task
What does solving a quadratic mean?
To solve \(x^2-2x-15=0\), we need inputs that make the left side zero. If we define \(f(x)=x^2-2x-15\), the solutions are the zeros of the function, or the x-coordinates of its x-intercepts. These are three descriptions of the same condition: \(f(x)=0\).
Try \(x=5\):
\[5^2-2(5)-15=25-10-15=0.\]Therefore 5 is a solution. Try \(x=0\): \(0-0-15=-15\), so 0 is not a solution. Checking confirms a candidate; it does not automatically reveal every solution.
Carlos and Zac: tables and graphs
Carlos’s table method evaluates the related function at chosen x-values. An output of zero identifies an exact solution. If no table entry equals zero, values on either side may help locate an approximate crossing, but a finite table alone does not prove that every zero has been found.
| \(x\) | −4 | −3 | −2 | 0 | 3 | 5 |
|---|---|---|---|---|---|---|
| \(f(x)=x^2-2x-15\) | 9 | 0 | −7 | −15 | −12 | 0 |
Zeros occur at \(x=-3\) and \(x=5\). A complete quadratic graph or an algebraic method can establish that these are the two real solutions.
Zac’s graph method finds where \(y=f(x)\) crosses or touches the x-axis. It provides a visual estimate and shows whether the parabola has two, one, or no x-intercepts. Use algebra if exact roots are required.
Clarita: factor and use the zero-product property
The zero-product property says that if \(AB=0\), then \(A=0\) or \(B=0\). This is why factoring can turn one quadratic equation into two simple equations.
1. Move every term to one side so the other side is zero.
2. Factor completely, taking out a greatest common factor if needed.
3. Set each factor equal to zero.
4. Solve and check the roots.
Factoring is especially efficient here because the trinomial has simple integer factors.
Solve \(3x^2+14x-5=0\). Split the middle term using 15 and −1:
\[3x^2+15x-x-5=0\] \[3x(x+5)-1(x+5)=0\] \[(3x-1)(x+5)=0\] \[\boxed{x=\frac13\quad\text{or}\quad x=-5.}\]Notice that \(3x-1=0\) gives \(x=1/3\), not 1.
Tia: complete the square
Completing the square rewrites part of a quadratic as the square of a binomial. Then the square-root property isolates x. Both the positive and negative roots are necessary whenever the right side is positive.
1. Move the constant: \(x^2+bx=-c\).
2. Add \((b/2)^2\) to both sides.
3. Factor the perfect square on the left.
4. Take both square-root branches and solve for x.
For \(x^2+4x-8=0\):
\[x^2+4x=8\] \[x^2+4x+4=12\] \[(x+2)^2=12\] \[x+2=\pm\sqrt{12}=\pm2\sqrt3\] \[\boxed{x=-2\pm2\sqrt3.}\]Decimal approximations are about 1.464 and −5.464. The radical answers are exact.
Solve \(-2x^2+3x+1=0\). Divide by −2:
\[x^2-\frac32x-\frac12=0\] \[x^2-\frac32x=\frac12\] \[\left(x-\frac34\right)^2=\frac12+\frac9{16}=\frac{17}{16}\] \[x-\frac34=\pm\frac{\sqrt{17}}4\] \[\boxed{x=\frac{3\pm\sqrt{17}}4.}\]Completing the square works even when factors are not integers. Keep fractions exact rather than rounding midway.
Tehani: use the quadratic formula
The quadratic formula is obtained by completing the square in the general quadratic equation. For real-coefficient equations where the expression under the square root is nonnegative, it returns every real root.
First arrange the equation as \(ax^2+bx+c=0\), with \(a\ne0\). Identify the coefficients carefully.
\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]The expression \(b^2-4ac\) is called the discriminant. If it is positive, there are two distinct real roots; if it is zero, there is one repeated real root. If it is negative, there are no real roots; the module addresses the broader number system later.
In \(-2x^2+3x+1=0\), \(a=-2\), \(b=3\), \(c=1\):
\[x=\frac{-3\pm\sqrt{3^2-4(-2)(1)}}{2(-2)}\] \[=\frac{-3\pm\sqrt{17}}{-4}=\boxed{\frac{3\pm\sqrt{17}}4}.\]The two signs represent the same pair of solutions after reversing the sign of the denominator. This agrees with completing the square.
Check agreement with Worked Example 5: methods change the work, not the solution set.
Which method is most useful?
| Strategy | Most useful when | Limitation to remember |
|---|---|---|
| Table (Carlos) | Testing candidate whole-number inputs or building intuition | A finite set of inputs may miss roots |
| Graph (Zac) | Estimating and visualizing zero locations and the number of intercepts | Usually approximate, not exact |
| Factor (Clarita) | Factors are apparent or easy to find | Not every quadratic factors nicely over integers |
| Complete square (Tia) | Connecting a quadratic to vertex form, or isolating a square | Fractions can create extra arithmetic |
| Quadratic formula (Tehani) | An exact general-purpose method is needed | Carefully substitute negative coefficients and simplify radicals |
Equation 1: \(x^2-2x-15=0\). Factoring gives \(x=-3,5\).
Equation 2: \(-2x^2+3x+1=0\). Formula or completing the square gives \(x=(3\pm\sqrt{17})/4\).
Equation 3: \(x^2+4x-8=0\). Completing the square or formula gives \(x=-2\pm2\sqrt3\).
In the packet, students are asked to demonstrate all five methods for each equation. Use the table or graph to estimate the irrational roots; use algebra to obtain the exact answers.
The main-task challenge: where two graphs meet
A solution of a system is an ordered pair that satisfies both equations. For a line and parabola, equate their y-expressions. Their common inputs solve a quadratic equation, and substitution gives the corresponding outputs.
Find the intersections:
\[y=x^2-4x+1\qquad y=x-3.\]Set outputs equal:
\[x^2-4x+1=x-3\] \[x^2-5x+4=0\] \[(x-1)(x-4)=0\quad\Longrightarrow\quad x=1,4.\]Use the line equation \(y=x-3\): at \(x=1\), \(y=-2\); at \(x=4\), \(y=1\).
\[\boxed{(1,-2)\ \text{and}\ (4,1).}\]Check in the parabola: \(1-4+1=-2\) and \(16-16+1=1\). The table and graph methods can confirm these points; factoring, completing the square, or the formula solves the resulting quadratic.
1. Isolate the same variable (usually y) in both equations.
2. Equate expressions and solve the resulting quadratic.
3. Substitute each x-value into either original equation to find y.
4. Check each ordered pair in both original equations.
2. Ready — symmetry and the maximum area of 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.
If fencing encloses a rectangle of fixed perimeter, increasing one side forces the other to decrease. The area functions in Ready have the form \(A(x)=x(S-x)\), with \(0\le x\le S\). This is a downward-opening quadratic.
Suppose \(A(x)=x(18-x)\). Then
\[A(5)=5(18-5)=65,\qquad A(13)=13(18-13)=65.\]The two equal areas occur at x-values equally far from 9. At \(x=0\) or \(x=18\), the area is zero.
Since \((x-9)^2\ge0\), the area is at most 81. The maximum is \(81\) square units when \(x=9\), a 9-by-9 square. The symmetry line is \(x=9\). For the packet’s four Ready problems, apply the same method with \(S=10,50,75,48\) respectively.
The two sides sum to S, so the vertex is at \(x=S/2\), where the sides are equal.
\[\boxed{x=\frac S2}\] \[\boxed{A_{\max}=\frac{S^2}{4}}\]The function’s x-intercepts are \(0\) and \(S\). This result comes from completing the square or the symmetry of the parabola; it is not a rule for arbitrary quadratics.
Check your work — Ready 1–4 answer key
- a. 21; b. 24; c. 24; d. \(x=0,10\); e. Maximum 25 at \(x=5\), since \(A(x)=25-(x-5)^2\).
- a. 400; b. 600; c. 600; d. \(x=0,50\); e. Maximum 625 at \(x=25\), since \(A(x)=625-(x-25)^2\).
- a. 1100; b. 1400; c. 1400; d. \(x=0,75\); e. Maximum \(\frac{5625}{4}=1406.25\) at \(x=37.5\), since \(A(x)=1406.25-(x-37.5)^2\).
- a. 380; b. 560; c. 560; d. \(x=0,48\); e. Maximum 576 at \(x=24\), since \(A(x)=576-(x-24)^2\).
3. Set — solve efficiently, then explain the method
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 5–10 require choosing an effective method and using at least three different methods across the assignment. Problems 11–13 require a written description of completing the square, factoring, and the quadratic formula. The examples below model the level of explanation expected.
Because 5 and 8 multiply to 40 and add to 13, factoring is direct.
Here the coefficient of \(x^2\) is 1 and the middle coefficient is even, so a perfect square appears quickly.
Solve \(2x^2+5x-4=0\):
\[x=\frac{-5\pm\sqrt{5^2-4(2)(-4)}}{2(2)}=\boxed{\frac{-5\pm\sqrt{57}}4}.\]The discriminant 57 is not a perfect square. Exact radicals communicate the solutions without premature rounding.
Check your work — Set 5–13 answer key
- Factoring: \((x+5)(x+12)=0\), so \(x=-5,-12\).
- Factoring: \((x+3)(x+13)=0\), so \(x=-3,-13\).
- Quadratic formula: \(x=\frac{-7\pm\sqrt{69}}2\).
- Factoring: \((3x-1)(x+5)=0\), so \(x=\frac13,-5\).
- Completing the square: \((x-6)^2=28\), so \(x=6\pm2\sqrt7\).
- Completing the square: \((x+3)^2=16\), so \(x=1,-7\).
- Divide by the leading coefficient, isolate the constant, and add the square of half the x-coefficient to both sides. Factor the perfect square and take both square roots. Example: \(x^2+6x=7\) becomes \((x+3)^2=16\), giving \(x=1,-7\). Useful when a square is easy to form or vertex form is wanted.
- Move all terms to one side, factor, and set each factor equal to zero. Example: \(x^2+17x+60=(x+5)(x+12)=0\) gives \(x=-5,-12\). Best when factors are easy to recognize.
- Write \(ax^2+bx+c=0\), identify coefficients, and substitute into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Example: \(x^2+7x-5=0\) gives \(\frac{-7\pm\sqrt{69}}2\). Works for every quadratic; especially useful when factoring is difficult.
4. Go — graphs, intercepts, and systems
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.
To graph \(f(x)=ax^2+bx+c\), find its axis of symmetry \(x=-b/(2a)\), compute the vertex output, find the y-intercept \((0,c)\), and solve \(f(x)=0\) for x-intercepts. Sketch the parabola with symmetry and its opening direction.
Consider the new function \(f(x)=x^2-6x+5\):
\[f(x)=(x-3)^2-4=(x-1)(x-5).\]Axis: \(x=3\). Vertex: \((3,-4)\). x-intercepts: \((1,0)\), \((5,0)\). y-intercept: \((0,5)\). The parabola opens upward.
For the Go homework’s systems, distinguish linear–linear systems (usually eliminate or substitute) from linear–quadratic systems (substitute and then solve the quadratic). Always check ordered pairs.
Solve \(2x+y=11\) and \(x-y=1\).
Add equations to eliminate y: \(3x=12\), so \(x=4\). Substitute: \(4-y=1\), so \(y=3\).
\[\boxed{(4,3)}\]Check: \(2(4)+3=11\) and \(4-3=1\).
Solve \(y=x^2-6x+6\) and \(y=2x-9\).
\[x^2-6x+6=2x-9\] \[x^2-8x+15=0\] \[(x-3)(x-5)=0\quad\Longrightarrow\quad x=3,5.\]Substitute into the line: \(y(3)=-3\) and \(y(5)=1\). Thus:
\[\boxed{(3,-3)\ \text{and}\ (5,1)}.\]Check in the quadratic: \(9-18+6=-3\) and \(25-30+6=1\).
Solve \(y=x^2+2x-4\) and \(y=x+2\).
\[x^2+2x-4=x+2\quad\Longrightarrow\quad x^2+x-6=0\] \[(x+3)(x-2)=0\quad\Longrightarrow\quad x=-3,2.\]The corresponding y-values from \(y=x+2\) are −1 and 4. Hence \(\boxed{(-3,-1),(2,4)}\).
Sketch \(g(x)=x^2-4x-2\) and name its features.
\[g(x)=(x-2)^2-6.\]Its axis is \(x=2\), vertex \((2,-6)\), y-intercept \((0,-2)\). Solve \((x-2)^2=6\): the x-intercepts are \((2-\sqrt6,0)\) and \((2+\sqrt6,0)\). Their approximate x-values are −0.449 and 4.449.
Check your work — Go 14–19 answer key
- a. \(x=-4\); b. \((-4-\sqrt3,0),(-4+\sqrt3,0)\); c. \((0,13)\); d. \((-4,-3)\). Opens upward. Plot not shown.
- a. \(x=2\); b. \((2-\sqrt5,0),(2+\sqrt5,0)\); c. \((0,-1)\); d. \((2,-5)\). Opens upward. Plot not shown.
- Elimination gives \(7y=9\), then \(x=\frac{20}{7}\). Solution: \((\frac{20}{7},\frac97)\). Substitution gives 15 and 6 in the two equations.
- Set \(-7x+12=5x-36\): \(x=4,y=-16\). Solution \((4,-16)\) satisfies both equations.
- Set \(2x+12=10x-x^2\): \((x-2)(x-6)=0\). Solutions \((2,16),(6,24)\) satisfy both equations.
- Set \(24x-x^2=8x+48\): \((x-4)(x-12)=0\). Solutions \((4,80),(12,144)\) satisfy both equations.
5. Additional learning resources
The following are supplementary references, not prerequisites for today’s packet.
| Direct resource | What it helps with | Scope / what to skip |
|---|---|---|
| Paul’s Online Math Notes — quadratic equations, Part II practice | Worked solutions for completing the square and quadratic formula; practice for Set. | Some extra material: focus on real-root examples that resemble the packet; skip exercises needing complex-number arithmetic, introduced later. |
| Khan Academy — completing the square review | Brief written explanations for Tia’s strategy, including sample equations. | Close match: use basic square-completion examples. |
| Professor Leonard — Intermediate Algebra 11.1, Completing the Square | Optional full-classroom-style explanation of the completion method. | Some extra material: select sections demonstrating quadratic equations; a precise timestamp was not verified. Skip any discussion involving complex solutions. |
| Professor Leonard — Intermediate Algebra 11.2, Quadratic Formula | Optional extended lecture explaining formula substitutions and solving. | Some extra material: focus on real solutions. Timestamps were not verified; avoid complex-number applications until the upcoming lessons. |
Teacher’s review (teachers and parents)
Main task, use of tables and graphs: The packet asks students to “demonstrate” all five methods on equations whose roots are sometimes irrational. Without a stated precision standard, numerical tables and hand-sketched graphs cannot supply the exact irrational roots that the algebraic methods do. A clearer direction is: “Use tables and graphs to estimate the solutions, and algebraic methods to find exact solutions when possible.”
No other serious mathematical issue was identified in this section.
6. Curriculum coverage
| Packet section | Content addressed here |
|---|---|
| Main task, pp. 43–45 | Five solving strategies and how to choose among them (In class); quadratic–linear system challenge (In class). |
| Ready, 1–4 | Evaluate \(A(x)=x(S-x)\), find zeros, symmetry, maxima, interpret rectangle constraints (Section 2). |
| Set, 5–13 | Choose efficient factoring, square-completion, and formula methods and explain each strategy (Section 3, with In class). |
| Go, 14–15 | Axis, intercepts, vertex, quadratic graphing (Section 4). |
| Go, 16–19 | Algebraic linear–linear and linear–quadratic systems, substitution/elimination, checking (In class and Section 4 and additional Go problems). |
This is a daily supplement for Section 3.7 only. The next section’s exploration of nonreal roots and complex numbers is intentionally not taught here.
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