Last updated .
3.9 — My Irrational and Imaginary Friends
You already know that a quadratic may have two real x-intercepts, one repeated x-intercept, or no real x-intercepts. This lesson explains where irrational real roots fit and how we name the solutions that lie outside the real number system. We start with numbers you can place on a number line.
1. In class: A Solidify Understanding Task
Rational and irrational solutions are both real numbers
A rational number can be written as the quotient of two integers, with a nonzero denominator. An irrational number cannot. Both belong to the real numbers, meaning both can be located on the ordinary number line. A radical such as \(\sqrt2\) is an exact number, not merely a calculator approximation.
Expand \(4(x-\frac52)(x+\frac32)=0\):
\[\begin{aligned}4\left(x^2-x-\frac{15}{4}\right)&=0\\4x^2-4x-15&=0.\end{aligned}\]By the zero-product property, \(x=\frac52\) or \(x=-\frac32\). Both are rational. Multiplying the factored equation by a nonzero constant preserves its solutions.
Multiply the conjugate expressions, viewing \(x-2\) as one quantity:
\[\begin{aligned}(x-2+\sqrt2)(x-2-\sqrt2)&=(x-2)^2-(\sqrt2)^2\\&=x^2-4x+2.\end{aligned}\]Multiplying by 4 would give \(4x^2-16x+8\), so note that the worksheet’s quadratic is \(4x^2-4x+2\) only if the factorization is different. In the actual printed expression in the packet, the accompanying quadratic is \(4x^2-4x+2=0\), which does not have the same roots as the displayed factorization. For the correctly expanded factorization, the roots are:
\[\boxed{x=2-\sqrt2\quad\text{or}\quad x=2+\sqrt2.}\]Both are irrational because adding an integer to \(\sqrt2\) cannot make it rational. Approximately, these points lie at \(0.586\) and \(3.414\), but the exact answers are preferable.
The intended equivalent quadratic is \(x^2-4x+2=0\), since \((x-2+\sqrt2)(x-2-\sqrt2)=x^2-4x+2\). The source packet displays \(4x^2-4x+2=0\), which is not equivalent. Use the factored expression to identify the intended irrational roots \(2\pm\sqrt2\). This is a genuine mathematical inconsistency, not a change in method.
For a right triangle with legs of length 1 and 1, the Pythagorean theorem gives the hypotenuse:
\[c^2=1^2+1^2=2\quad\Rightarrow\quad c=\sqrt2.\]The perimeter is \(1+1+\sqrt2=2+\sqrt2\). A segment of length \(\sqrt2\), attached to a segment of length 2, locates \(2+\sqrt2\) exactly; transferring the \(\sqrt2\) length to the left of 2 locates \(2-\sqrt2\). Both are ordinary points on the real number line.
Real numbers consist of all rational numbers together with all irrational numbers. An irrational root is still a real x-intercept and can be located on a number line.
The imaginary unit extends the number system
No real number has square equal to \(-1\). To name solutions of equations requiring such a square, we introduce the imaginary unit \(i\).
Be careful: \(\sqrt{-4}\,\sqrt{-9}=(2i)(3i)=6i^2=-6\), not \(\sqrt{36}=6\). The familiar identity \(\sqrt a\sqrt b=\sqrt{ab}\) does not extend blindly to negative radicands.
The powers cycle in groups of four:
\[i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1.\]For example, \(i^7=i^4i^3=-i\). Use exponent laws to group powers, then replace every \(i^2\) by \(-1\).
A complex number has the form \(a+bi\), where \(a\) and \(b\) are real. Its real part is \(a\), and its imaginary part is \(b\). A pure imaginary number has \(a=0\) and \(b\ne0\). Every real number is also complex, since \(a=a+0i\).
\(7i\) is nonreal, pure imaginary, and complex. \(-3+2i\) is nonreal and complex, but not pure imaginary. The number \(-3\) is both real and complex, because it is \(-3+0i\).
Solve quadratic equations with complex roots
Now the quadratic formula can give an answer even when the discriminant is negative. Such a quadratic has two nonreal complex roots, which are not x-intercepts of its ordinary real graph.
The two solutions are complex conjugates: they have the same real part and opposite imaginary parts.
The product is a real quadratic even though the individual factors contain \(i\). This is a consequence of the opposite-sign imaginary terms canceling.
For a quadratic with real coefficients, if \(a+bi\) is a nonreal root, then \(a-bi\) is also a root.
Solve \(2x^2-4x+10=0\). Here \(a=2,b=-4,c=10\).
\[\begin{aligned}D&=(-4)^2-4(2)(10)=-64,\\x&=\frac{4\pm\sqrt{-64}}4=\frac{4\pm8i}{4}.\end{aligned}\] \[\boxed{x=1\pm2i.}\]Both solutions are exact. Writing “no solution” would be incorrect when working over the complex numbers.
What graphs tell us about real and complex roots
For a quadratic \(y=ax^2+bx+c\), the vertex’s horizontal coordinate is \(-b/(2a)\). The two root expressions are centered on that x-coordinate:
\[x=\underbrace{\frac{-b}{2a}}_{\text{center}}\;\pm\;\underbrace{\frac{\sqrt{b^2-4ac}}{2a}}_{\text{signed offset}}.\]For two real roots, the magnitude of the offset is their horizontal distance from the symmetry line. If the discriminant is negative, this offset involves \(i\), and no real intercepts appear.
For \(y=x^2-6x+7\), the vertex is \((3,-2)\). Set the output to zero:
\[(x-3)^2-2=0\quad\Rightarrow\quad x=3\pm\sqrt2.\]The symmetry line is \(x=3=-b/(2a)\). The roots are \(\sqrt2\) units to the left and right. The graph confirms two real intercepts, but their exact irrational values come from algebra.
In problem 12 the shown parabola has vertex \((-2,3)\) and includes the y-intercept \((0,7)\). Thus its equation is \(y=(x+2)^2+3\).
\[(x+2)^2+3=0\quad\Rightarrow\quad\boxed{x=-2\pm i\sqrt3}.\]Reflect across the horizontal line \(y=3\). Reflection sends an output \(y\) to \(6-y\), so the reflected parabola is \(y=3-(x+2)^2\). Its roots satisfy \((x+2)^2=3\), giving:
\[\boxed{x=-2\pm\sqrt3}.\]The reflected graph has two irrational real intercepts; the original has no real intercepts. Both sets of roots arise by isolating a squared expression and taking a square root, but one asks for \(\sqrt{-3}\) while the other asks for \(\sqrt3\).
Every quadratic \(ax^2+bx+c\) with \(a\ne0\) has two complex roots counted with multiplicity. They may be two distinct real numbers, one repeated real number, or two nonreal conjugates. A graph shows only the real roots.
A graph that never intersects the x-axis shows there are no real roots, but the picture alone is not enough to calculate the exact complex roots. For problem 12, use the visible vertex and another clear point to reconstruct the equation first, then solve it algebraically. The graph is not displaying imaginary coordinates as x-intercepts.
2. Ready — Classify numbers by all applicable sets
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 Ready problems (1–9) ask for every set a number belongs to. The sets overlap: whole numbers \(\subset\) integers \(\subset\) rational numbers \(\subset\) real numbers \(\subset\) complex numbers. Irrational numbers belong to the real and complex numbers, but not to the rational numbers.
| Number type | Meaning | Example and applicable sets |
|---|---|---|
| Whole | \(0,1,2,\dots\) | \(0\): whole, integer, rational, real, complex |
| Integer | Positive and negative whole numbers and zero | \(-13\): integer, rational, real, complex |
| Rational | Ratio of integers | \(\frac49\): rational, real, complex |
| Irrational | Real but not rational | \(\sqrt{75}=5\sqrt3\): irrational, real, complex |
| Nonreal complex | Nonzero imaginary part | \(\sqrt{-16}=4i\): complex only among these listed sets |
Classify \(\sqrt{\frac49}\) and \(5+\sqrt2\).
\[\sqrt{\frac49}=\frac23\quad\Rightarrow\quad\text{rational, real, complex}.\]The number \(5+\sqrt2\) is irrational, real, and complex. A rational number added to an irrational number stays irrational.
Check your work — Ready 1–9 answer key
- Irrational, real, complex.
- Integer, rational, real, complex.
- \(4i\): complex (nonreal).
- Whole, integer, rational, real, complex.
- \(5\sqrt3\): irrational, real, complex.
- \(3\): whole, integer, rational, real, complex.
- \(\frac23\): rational, real, complex.
- Irrational, real, complex.
- \(2i\sqrt{10}\): complex (nonreal).
3. Set — Combine like radicals and simplify imaginary expressions
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.
Problems 10–15 combine square-root terms; first simplify each radical, then combine coefficients only when the radicals match. Problems 16–21 use \(i^2=-1\) and the powers of \(i\).
Do not add numbers inside the radicals: \(\sqrt{27}+\sqrt{12}\ne\sqrt{39}\).
Always simplify \(i^2\) first. The sign changes because \(i^2=-1\), not \(+1\).
Check your work — Set 10–25 answer key
- \(-4+4\sqrt2\).
- \(2+8\sqrt5\).
- \(6\sqrt3\).
- \(3\sqrt2\).
- \(6\sqrt7\).
- \(4\sqrt7+6\sqrt3\).
- \(-2\).
- \(7+5i\).
- \(-16\).
- \(i\).
- \(-8i\).
- \(-50i\).
- \(x=\pm10i\).
- \(t=\pm2i\sqrt6\).
- \(x=3\pm2i\).
- \(r=1\pm2i\).
4. Go — Use the discriminant to describe the 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.
For Go problems 26–31, calculate \(D=b^2-4ac\), then describe the number and kind of roots. You do not always need to find their exact values to classify them.
Determine the nature of the roots of \(x^2-8x+15=0\).
\[D=(-8)^2-4(1)(15)=64-60=4.\]Since \(D\) is a positive perfect square, the roots are two distinct rational real numbers. Indeed, \((x-3)(x-5)=0\), so \(x=3,5\).
Determine the nature of the roots of \(3x^2+2x+4=0\), and then find the exact roots to verify.
\[D=2^2-4(3)(4)=4-48=-44.\]The negative discriminant gives two nonreal conjugate roots. Continue:
\[\begin{aligned}x&=\frac{-2\pm\sqrt{-44}}6\\&=\frac{-2\pm2i\sqrt{11}}6\\&=\boxed{-\frac13\pm\frac{\sqrt{11}}3i}.\end{aligned}\]Classify the roots of \(2x^2+2x-3=0\).
\[D=2^2-4(2)(-3)=28.\]Positive but not a perfect square: two distinct irrational real roots. The exact roots are \(\frac{-2\pm\sqrt{28}}4=\frac{-1\pm\sqrt7}{2}\).
Classify and solve \(4t^2-12t+9=0\).
\[D=(-12)^2-4(4)(9)=144-144=0.\]There is one repeated rational real root: \(t=\frac{12}{8}=\frac32\). It counts twice in the Fundamental Theorem of Algebra.
Check your work — Go 26–38 answer key
- \(D=-3\): two nonreal complex roots.
- \(D=1\): two distinct rational (integer) roots.
- \(D=-15\): two nonreal complex roots.
- \(D=41\): two distinct irrational real roots.
- \(D=1\): two distinct rational real roots.
- \(D=-7\): two nonreal complex roots.
- \(D=1\): two distinct rational real roots.
- \(D=97\): two distinct irrational real roots.
- \(D=0\): one distinct rational (integer) real root, with multiplicity 2.
- Factoring \((m+7)(m+8)=0\) gives \(m=-7,-8\).
- Quadratic formula gives \(x=\frac{3\pm i\sqrt{131}}{10}\).
- Factoring \((x-3)(x-7)=0\) gives \(x=3,7\).
- Factoring \((3x+5)(2x-1)=0\) gives \(x=-\frac53,\frac12\).
5. Additional learning resources
| Resource and direct link | What it covers | Scope / What to skip |
|---|---|---|
| Paul’s Online Math Notes — Section 1.7, Complex Numbers | Negative square roots, \(i^2=-1\), complex form, and algebraic examples. | Some extra material: use the definition and multiplication examples; skip division by complex numbers and rationalizing complex denominators, which belong later. |
| Khan Academy — Intro to the imaginary numbers | Short written introduction to \(i\) and square roots of negative numbers. | Close match: strong introduction for main task problems 6–8 and Set 16–21. |
| Paul’s Online Math Notes — Complex Numbers practice with solutions | Worked practice with imaginary terms and products. | Some extra material: select multiplication questions 1, 4–6; skip division questions 7–9 for now. |
| Professor Leonard — Intermediate Algebra Lecture 10.7: An Introduction to Operations with Complex Numbers | Optional extended classroom-style explanation of imaginary and complex numbers. | Significantly broader: a long lecture (about 1 h 42 min); use the introduction and multiplication portion only. Skip complex division and other operations not needed for today. Precise timestamps could not be verified. |
6. Teacher’s review — Serious packet issues
Main Task, Problem 2: The displayed factors expand to \(x^2-4x+2\), not to the printed \(4x^2-4x+2\). This materially changes the roots. Correct the second equation to \(x^2-4x+2=0\) or change the factors and subsequent number-line discussion accordingly.
Main Task, Problem 12: The instruction “find the complex roots ... represented by its graph” omits the algebraic reconstruction step. The graph can establish that there are no real x-intercepts and (when scale is legible) suggest the vertex/another point; exact nonreal roots must be calculated from the reconstructed formula. A clearer instruction: “Use the graph to write the quadratic equation, then solve it over the complex numbers.”
Fundamental Theorem wording (Main Task, Problem 15): A degree-\(n\) polynomial has exactly \(n\) complex roots counting multiplicity. State that qualification explicitly to avoid equating the number of roots with distinct real x-intercepts.
7. Curriculum coverage
| Packet section | Skills taught here | Where |
|---|---|---|
| Main Task 1–5 | Equivalent factorizations, rational/irrational roots, exact numbers on the real line, geometric construction of \(\sqrt2\) | In class |
| Main Task 6–8 | Imaginary unit, powers of \(i\), complex factorization and conjugates | In class |
| Main Task 9–15 | Symmetry of roots, graph interpretation, complex roots, horizontal reflection, fundamental theorem | In class |
| Ready 1–9 | Number-set classification and simplified radical values | Section 2 |
| Set 10–25 | Like radicals, imaginary products and powers, quadratic equations solved over complex numbers | Section 3 |
| Go 26–31 | Discriminant classification; rational, irrational, repeated and complex roots | Section 4 and four new worked problems |
This companion teaches Section 3.9 only. Additional complex-number operations are intentionally not assigned ahead of Section 3.10.
Last updated .