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3.8 — To Be Determined . . .
In the previous lessons you solved quadratic equations and found their x-intercepts. Today you will see why some quadratics factor neatly, some touch the x-axis only once, and others never reach it. This puzzle leads to a larger number system, which the next lesson explores in detail.
1. In class: A Develop Understanding Task
Begin with a quadratic table
The vertex is the turning point. Values at equal distances to either side of its x-coordinate agree; the line through the vertex is the axis of symmetry. For a quadratic table with consecutive x-values, constant second differences are a useful pattern. Here the packet tells us the tables represent quadratic functions; a finite table alone cannot establish that a function is quadratic everywhere.
| \(x\) | −4 | −3 | −2 | −1 | 0 |
|---|---|---|---|---|---|
| \(y\) | 4 | 1 | 0 | 1 | 4 |
The minimum is zero at \(x=-2\), so the vertex is \((-2,0)\). Begin with \(y=a(x+2)^2\). Substituting the point \((-1,1)\) gives \(1=a(1)^2\), so \(a=1\).
\[\boxed{y=(x+2)^2=x^2+4x+4.}\]This parabola just touches the x-axis at \((-2,0)\).
Write the same quadratic in three forms
Each form emphasizes a different feature. Vertex form displays the turning point. Standard form gives the y-intercept directly. Real factored form shows the x-intercepts—when real x-intercepts exist.
Vertex \((h,k)\); the sign of \(a\) tells whether the parabola opens up or down.
The y-intercept is \((0,c)\). Expand vertex form to obtain it.
Zeros \(r_1,r_2\); if the two roots coincide, the factor repeats.
The table has vertex \((-2,-1)\) and includes \((0,3)\). Start with vertex form:
\[y=a(x+2)^2-1,\qquad3=4a-1\Longrightarrow a=1.\]Now expand, then factor:
\[\boxed{y=(x+2)^2-1}\] \[\boxed{y=x^2+4x+3}\] \[\boxed{y=(x+3)(x+1).}\]Zeros are \(x=-3\) and \(x=-1\), symmetric about \(x=-2\). Check \(x=0\): \((3)(1)=3\), consistent with the table.
For the vertex \((-2,0)\) and the point \((0,4)\), \(a=1\).
\[\underbrace{(x+2)^2}_{\text{vertex and factored form}}=\underbrace{x^2+4x+4}_{\text{standard form}}.\]The equation \((x+2)^2=0\) has a repeated root, \(-2\). There is one distinct x-intercept but two equal algebraic roots counted with multiplicity.
Three ways a real quadratic can meet the x-axis
A real x-intercept corresponds to a real number \(x\) for which \(y=0\). An upward-opening parabola whose vertex lies above the x-axis never has \(y=0\); likewise a downward-opening parabola whose maximum is below the x-axis.
The table’s vertex is \((-2,1)\), and \((0,5)\) is on the graph, giving \(a=1\):
\[\boxed{y=(x+2)^2+1=x^2+4x+5.}\]Setting \(y=0\) would require \((x+2)^2=-1\). But the square of any real number is nonnegative, so there is no real solution and no factorization into real linear factors. Expanding a proposed real factored form and testing values will not rescue it.
For \(y=-(x+2)^2+2\), the maximum is \(2\), so the graph crosses the axis in two places. In contrast, \(y=-(x+2)^2-2\) has maximum \(-2\), so the entire graph lies below the x-axis.
Reflecting a quadratic across the x-axis changes the signs of its outputs but does not change its zeros. Moving its vertex vertically can change how many real zeros it has.
Predict real-root behavior without graphing
The quadratic formula from Section 3.5 is
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\quad a\ne0.\]The quantity inside the square root is called the discriminant, \(D=b^2-4ac\). Its sign tells how many distinct real roots a quadratic has.
The formula asks us to take a square root of a negative number, which cannot be done in the real number system.
For \(x^2+4x+5=0\), take \(a=1,b=4,c=5\):
\[D=4^2-4(1)(5)=16-20=-4.\]Because \(D<0\), there are no real roots. For \(x^2+4x+4=0\), \(D=16-16=0\), confirming the repeated root. And for \(x^2+4x+3=0\), \(D=16-12=4>0\), confirming two distinct real roots.
Compare \(x^2-6x+5=0\) and \(x^2-6x+2=0\). Both have \(D>0\), but:
\[x^2-6x+5=(x-1)(x-5)\Rightarrow x=1,5.\] \[x^2-6x+2=0\Rightarrow x=\frac{6\pm\sqrt{36-8}}2=3\pm\sqrt7.\]The first roots are integers. The second are irrational real numbers. When integer coefficients are used, a positive perfect-square discriminant leads to rational roots; a positive non-square discriminant leads to irrational roots.
Why “two roots” can mean more than two x-intercepts
The packet introduces the Fundamental Theorem of Algebra: a polynomial of degree \(n\) has \(n\) roots when roots are counted with multiplicity and complex numbers are allowed. A degree-two polynomial therefore has two complex roots, though the graph may have two, one, or zero real x-intercepts.
The equation \((x+2)^2=-1\) has no real solution. The equation \(u^2=-1\) motivates a new number, called the imaginary unit, written \(i\), with \(i^2=-1\). Thus we may write the two roots of \((x+2)^2+1=0\) as \(x=-2+i\) and \(x=-2-i\).
These are nonreal complex roots, not x-intercepts on an ordinary real coordinate plane. Today we only identify why such numbers are needed; operations with them are developed in the following lessons.
The packet’s abbreviated wording says an nth-degree polynomial has n roots. This counts complex roots with multiplicity. It does not assert n different real x-intercepts. For example, \((x+2)^2\) has one distinct real zero counted twice, while \((x+2)^2+1\) has none on the real x-axis.
2. Ready — simplify square roots
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.
Look for the largest perfect-square factor under the radical. For nonnegative \(a,b\), \(\sqrt{ab}=\sqrt a\sqrt b\). A number like \(\sqrt8\) is irrational, yet it can be expressed in simpler radical form.
The six expressions in Ready (problems 1–6) all simplify to an integer multiple of \(\sqrt2\) or \(\sqrt5\), illustrating how apparently different radicals can share a common irrational factor.
They are not equal, but both are multiples of \(\sqrt2\). Do not combine unlike radicals by adding the numbers inside them: \(\sqrt2+\sqrt3\ne\sqrt5\).
Check your work — Ready 1–7 answer key
- \(2\sqrt2\).
- \(3\sqrt2\).
- \(4\sqrt2\).
- \(2\sqrt5\).
- \(3\sqrt5\).
- \(4\sqrt5\).
- The first three are 2, 3, and 4 times \(\sqrt2\); the last three are 2, 3, and 4 times \(\sqrt5\). Extracting the square factors 4, 9, and 16 reveals these common radical parts.
3. Set — identify the type of x-intercept
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.
The Set questions use graphs, formulas, and polynomial expressions. Read graphs to count real intersections; use factors or the discriminant to classify exact values. Number categories are nested: whole numbers are integers, integers are rational, and rational and irrational numbers are both real.
| Given information | What you can conclude |
|---|---|
| Graph crosses the x-axis twice | Two distinct real x-intercepts; without exact labels, a sketch may not establish whether they are rational. |
| Graph touches the x-axis once | One distinct real x-intercept (double root). |
| Graph stays above or below x-axis | No real x-intercepts. |
| Exact factored form \((2x-3)(x+4)\) | Rational zeros \(3/2\) and \(-4\), with \(-4\) an integer. |
The first root is rational but not an integer; the second is a whole number, an integer, and rational. When choosing the most specific category, classify each root separately.
For \(g(x)=2x^2-4x+7\), \(D=(-4)^2-4(2)(7)=16-56=-40\). There are no real x-intercepts.
Set problems 16–18 ask for the number of roots of higher-degree polynomials. The degree is the largest exponent with a nonzero coefficient. A degree-five polynomial has five complex roots counted with multiplicity, even if its real graph shows fewer x-intercepts.
The largest exponent is 6, so this polynomial has six complex roots counted with multiplicity. You cannot infer how many of these are real merely from the degree.
Check your work — Set 8–18 answer key
- Graph shows x-intercepts −2 and 3: both integers and rational; 3 is also whole.
- No real x-intercepts: the upward parabola stays above the x-axis. Its two roots are nonreal complex numbers.
- Irrational roots \(-2\pm2\sqrt7\) because the discriminant 112 is positive but not a perfect square.
- Rational, noninteger roots \(\frac12,-\frac25\), from the two linear factors.
- Rational roots \(1,-\frac52\), since \((2x+5)(x-1)=0\). The root 1 is also whole and an integer.
- No real x-intercepts: the downward parabola stays below the x-axis. Its roots are nonreal complex numbers.
- One distinct whole-number x-intercept, \(t=4\); \((t-4)^2\) has root 4 with multiplicity 2 (also integer and rational).
- No real x-intercepts: \(b^2-4ac=25-108=-83<0\). The roots are nonreal complex numbers.
- 5 complex roots, counting multiplicity (degree 5).
- 3 complex roots, counting multiplicity (degree 3).
- 7 complex roots, counting multiplicity (degree 7).
4. Go — find exact x-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.
Set the quadratic equal to zero. First look for an easy common factor or integer binomial factors. Otherwise apply the quadratic formula, keeping radicals exact. Report real x-intercepts as ordered pairs \((r,0)\). If the discriminant is negative, write “no real x-intercepts,” rather than plotting imaginary numbers.
Find the x-intercepts of \(f(x)=x^2-x-20\).
\[x^2-x-20=(x-5)(x+4)=0\]\[x=5\quad\text{or}\quad x=-4.\]Therefore the x-intercepts are \(\boxed{(5,0)\text{ and }(-4,0)}\). Check: both substitutions make the output zero.
Find the x-intercepts of \(g(x)=3x^2-2x-6\).
\[x=\frac{2\pm\sqrt{(-2)^2-4(3)(-6)}}{2(3)}\] \[=\frac{2\pm\sqrt{76}}6=\frac{1\pm\sqrt{19}}3.\]Hence the intercepts are \(\boxed{\left(\frac{1+\sqrt{19}}3,0\right)\text{ and }\left(\frac{1-\sqrt{19}}3,0\right)}\). They are irrational. No rounding is needed.
For \(h(x)=4x^2+12x+9\), factor:
\[h(x)=(2x+3)^2.\]The only distinct x-intercept is \(\boxed{(-3/2,0)}\); its root has multiplicity two. The graph touches the axis there.
For \(k(x)=2x^2+4x+8\):
\[D=4^2-4(2)(8)=16-64=-48.\]There are \(\boxed{\text{no real x-intercepts}}\). The quadratic formula produces nonreal roots, but the ordinary graph does not cross the real x-axis.
Check your work — Go 19–27 answer key
- \((x+7)(x-3)\); intercepts \((-7,0),(3,0)\).
- \((5x+1)(x+3)\); intercepts \((-\frac15,0),(-3,0)\).
- Quadratic formula: intercepts \((2-\sqrt3,0),(2+\sqrt3,0)\).
- Quadratic formula: intercepts \((8-2\sqrt{15},0),(8+2\sqrt{15},0)\).
- \((x+8)(x-5)\); intercepts \((-8,0),(5,0)\).
- \((2x+3)(x-3)\); intercepts \((-\frac32,0),(3,0)\).
- \(x(x-3)\); intercepts \((0,0),(3,0)\).
- \((x+2)(x+4)\); intercepts \((-2,0),(-4,0)\).
- Quadratic formula: intercepts \((-\frac{\sqrt{33}}3,0),(\frac{\sqrt{33}}3,0)\).
5. Additional learning resources
| Resource and direct link | What it covers / why use it | Scope / what to skip |
|---|---|---|
| Paul’s Online Math Notes — Quadratic Equations: A Summary practice | Practice recognizing equation-solving methods and classifying roots using the discriminant. | Some extra material. Use discriminant and real-root exercises; postpone calculations with complex roots. |
| OpenStax Intermediate Algebra 2e — 9.3 Quadratic Formula | Formula, discriminant, and examples of two, one, and no real solutions. | Some extra material. Read the discriminant discussion; skip detailed complex-number calculations until 3.9. |
| Professor Leonard — Intermediate Algebra Lecture 11.2: Quadratic Formula | Optional extended traditional lecture reviewing the formula and different root types. | Significantly broader. Focus on discriminant and real roots; skip complex arithmetic. Exact timestamps could not be verified. |
Teacher’s review (teachers and parents)
Section 3.8, main-task directions, especially problems 4 and 8: Students are asked to write every quadratic in all three forms, including real factored form. But \((x+2)^2+1\) and \((x+2)^2+2\) have no real linear factorization. Clarify: “Write vertex and standard forms for every function; give real factored form when possible, and otherwise explain why no real linear factorization exists.” Nonreal factors belong to the next lesson.
Section 3.8, problem 10: The theorem statement omits “over the complex numbers, counting multiplicity.” Without this qualifier, graphs with zero or one real intercept appear to contradict it. State the complete theorem explicitly.
Section 3.8, Set problems 8, 9, 12, 13: Unlabeled or hand-plotted graph crossings may establish the number of real intercepts but do not reliably establish the rationality of exact x-coordinates. If number types are intended, provide equations or exact labeled intercepts; otherwise ask only for the number of real intercepts.
6. Curriculum coverage
| Packet portion | Instruction in this lesson |
|---|---|
| Main task 1–8 | Infer quadratics from tables; convert between standard, vertex, and feasible real factored forms; recognize two/one/no real zeros (In class). |
| Main task 9–10 | Effect of reflections and vertex position; Fundamental Theorem of Algebra and multiplicity (In class). |
| Ready 1–7 | Extract perfect squares and identify common simplified radical factors (Section 2). |
| Set 8–15 | Count x-intercepts and classify whole, integer, rational, irrational, or nonreal behavior (Section 3). |
| Set 16–18 | Determine polynomial degree and number of complex roots with multiplicity (In class and Section 3). |
| Go 19–27 | Factor where possible; otherwise use the quadratic formula to determine exact real x-intercepts (Section 4). |
This daily supplement covers Section 3.8 only. Detailed arithmetic of imaginary and complex numbers belongs to Section 3.9 and later.
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