The Central Idea
Recall the quadratic formula:
\[ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]The expression under the radical,
\[ \boxed{b^2-4ac}, \]determines the nature of the roots.
We define
\[ \boxed{D=b^2-4ac}, \]where \(D\) is called the discriminant.
Why the Discriminant Matters
The quadratic formula may be written
\[ x= \frac{-b\pm\sqrt{D}}{2a}. \]Therefore, the kind of number represented by
\[ \sqrt{D} \]determines the kind of roots the quadratic has.
Three Main Cases
| Discriminant | Root Type | Graph Meaning |
|---|---|---|
| \(D>0\) | Two distinct real roots | Two real \(x\)-intercepts |
| \(D=0\) | One repeated real root | Parabola touches the \(x\)-axis once |
| \(D<0\) | Two nonreal complex roots | No real \(x\)-intercepts |
Case 1: Positive Discriminant
If
\[ D>0, \]then
\[ \sqrt{D} \]is a positive real number.
The plus and minus cases therefore produce two different real roots.
Thus,
\[ \boxed{D>0\Rightarrow\text{two distinct real roots}}. \]Example: Positive Discriminant
Consider
\[ x^2-5x+6=0. \]Here,
\[ a=1,\qquad b=-5,\qquad c=6. \]Compute:
\[ D=(-5)^2-4(1)(6). \] \[ D=25-24=1. \]Since
\[ D>0, \]there are two distinct real roots.
Case 2: Zero Discriminant
If
\[ D=0, \]then
\[ \sqrt{D}=0. \]The quadratic formula becomes
\[ x=\frac{-b}{2a}. \]The plus and minus cases collapse to the same value.
Therefore,
\[ \boxed{D=0\Rightarrow\text{one repeated real root}}. \]Example: Zero Discriminant
Consider
\[ x^2-6x+9=0. \]Here,
\[ a=1,\qquad b=-6,\qquad c=9. \]Compute:
\[ D=(-6)^2-4(1)(9). \] \[ D=36-36=0. \]Therefore, there is one repeated real root:
\[ \boxed{x=3}. \]Repeated Root and Perfect Square
The quadratic
\[ x^2-6x+9 \]factors as
\[ (x-3)^2. \]The repeated factor corresponds directly to the repeated root.
This is the algebraic reason the graph touches the axis instead of crossing at that point.
Case 3: Negative Discriminant
If
\[ D<0, \]then
\[ \sqrt{D} \]is not a real number.
To solve the equation, we extend the number system to include complex numbers.
Therefore,
\[ \boxed{D<0\Rightarrow\text{two nonreal complex roots}}. \]The Imaginary Unit
Define
\[ \boxed{i=\sqrt{-1}}. \]Therefore,
\[ i^2=-1. \]This allows square roots of negative numbers to be rewritten using real radicals and \(i\).
Square Roots of Negative Numbers
For \(k>0\),
\[ \sqrt{-k} = \sqrt{k}\sqrt{-1}. \]Therefore,
\[ \boxed{\sqrt{-k}=i\sqrt{k}}. \]For example,
\[ \sqrt{-9}=3i \]and
\[ \sqrt{-20}=2i\sqrt5. \]Complex Numbers
A complex number has the form
\[ \boxed{a+bi}, \]where \(a\) and \(b\) are real numbers.
The number \(a\) is the real part.
The number \(b\) is the coefficient of the imaginary part.
If \(b\neq0\), the number is nonreal complex.
Example: Negative Discriminant
Solve
\[ x^2+4x+8=0. \]Here,
\[ a=1,\qquad b=4,\qquad c=8. \]Compute:
\[ D=4^2-4(1)(8). \] \[ D=16-32=-16. \]Since \(D<0\), the roots are complex.
Finish the Complex-Root Example
Apply the quadratic formula:
\[ x= \frac{-4\pm\sqrt{-16}}{2}. \]Since
\[ \sqrt{-16}=4i, \]we obtain
\[ x= \frac{-4\pm4i}{2}. \]Therefore,
\[ \boxed{x=-2\pm2i}. \]Complex Conjugate Roots
The roots
\[ -2+2i \]and
\[ -2-2i \]are complex conjugates.
They have the same real part and opposite imaginary parts.
Quadratic polynomials with real coefficients produce nonreal complex roots in conjugate pairs.
Why Conjugate Pairs Appear
In the quadratic formula,
\[ \frac{-b\pm\sqrt{D}}{2a}, \]if \(D<0\), then
\[ \sqrt{D}=i\sqrt{|D|}. \]Therefore, the two solutions have the form
\[ \frac{-b}{2a} \pm \frac{\sqrt{|D|}}{2a}i. \]The real parts match, while the imaginary parts differ only by sign.
Positive Discriminant: Rational or Irrational?
A positive discriminant guarantees two real roots, but those roots may be rational or irrational.
The key question is whether
\[ D \]is a perfect square.
Positive Perfect-Square Discriminant
If
\[ D>0 \]and \(D\) is a perfect square, then
\[ \sqrt{D} \]is rational.
If \(a\), \(b\), and \(c\) are integers, the resulting roots are rational.
Therefore,
\[ \boxed{ D>0\text{ and perfect square} \Rightarrow \text{two rational real roots} }. \]Example: Rational Real Roots
Consider
\[ 2x^2-7x+3=0. \]Compute:
\[ D=(-7)^2-4(2)(3). \] \[ D=49-24=25. \]Since \(25\) is a positive perfect square, the roots are rational.
Indeed,
\[ x= \frac{7\pm5}{4}, \]giving
\[ \boxed{x=3,\ \frac12}. \]Positive Non-Square Discriminant
If
\[ D>0 \]but \(D\) is not a perfect square, then
\[ \sqrt{D} \]is irrational.
Therefore,
\[ \boxed{ D>0\text{ and not a perfect square} \Rightarrow \text{two irrational real roots} }. \]Example: Irrational Real Roots
Consider
\[ x^2-2x-1=0. \]Compute:
\[ D=(-2)^2-4(1)(-1). \] \[ D=4+4=8. \]Since \(8>0\) but is not a perfect square, the roots are real and irrational.
The solutions are
\[ \boxed{x=1\pm\sqrt2}. \]A More Detailed Classification
| Discriminant | Number of Roots | Type |
|---|---|---|
| \(D>0\), perfect square | Two | Distinct rational real roots |
| \(D>0\), not a perfect square | Two | Distinct irrational real roots |
| \(D=0\) | One repeated | Rational real root when coefficients are rational |
| \(D<0\) | Two | Nonreal complex conjugate roots |
Graph Interpretation
The discriminant also describes how the parabola relates to the \(x\)-axis.
- \(D>0\): the parabola crosses the axis twice.
- \(D=0\): the parabola touches the axis once.
- \(D<0\): the parabola does not meet the axis.
The discriminant therefore connects symbolic algebra to graph geometry.
Why \(D=0\) Means the Vertex Is on the Axis
When
\[ D=0, \]the two quadratic-formula roots collapse to
\[ x=-\frac{b}{2a}. \]This is exactly the \(x\)-coordinate of the vertex.
Therefore, the repeated root occurs at the axis of symmetry, and the vertex lies on the \(x\)-axis.
Why \(D<0\) Means No Real \(x\)-Intercepts
A real \(x\)-intercept would require some real \(x\) satisfying
\[ f(x)=0. \]But if \(D<0\), the quadratic formula produces no real solutions.
Therefore, the graph cannot intersect the real \(x\)-axis.
Example: Classify Without Solving
Consider
\[ 3x^2+4x+7=0. \]Compute:
\[ D=4^2-4(3)(7). \] \[ D=16-84=-68. \]Since
\[ D<0, \]we can immediately conclude:
\[ \boxed{\text{two nonreal complex roots}} \]without solving the equation completely.
Example: Determine Root Type Only
Consider
\[ 4x^2-12x+9=0. \]Compute:
\[ D=(-12)^2-4(4)(9). \] \[ D=144-144=0. \]Therefore, the equation has
\[ \boxed{\text{one repeated real root}}. \]No further solving is required if classification is the only goal.
Example: Predict Rationality
Consider
\[ 5x^2+x-1=0. \]Compute:
\[ D=1^2-4(5)(-1)=21. \]Since \(21>0\) but is not a perfect square, the equation has
\[ \boxed{\text{two irrational real roots}}. \]Complex Roots from the Quadratic Formula
Solve
\[ x^2-6x+13=0. \]Compute:
\[ D=(-6)^2-4(1)(13). \] \[ D=36-52=-16. \]Then
\[ x= \frac{6\pm\sqrt{-16}}{2}. \] \[ x= \frac{6\pm4i}{2}. \]Therefore,
\[ \boxed{x=3\pm2i}. \]Verify a Complex Root by Substitution
Let
\[ P(x)=x^2-6x+13. \]Test
\[ x=3+2i. \]Compute
\[ (3+2i)^2=9+12i+4i^2. \]Since
\[ i^2=-1, \]we get
\[ (3+2i)^2=5+12i. \]Then
\[ P(3+2i) = (5+12i)-6(3+2i)+13. \] \[ =5+12i-18-12i+13=0. \]The root is verified.
Why Real-Coefficient Quadratics Produce Conjugates
Suppose
\[ a,\ b,\ c \]are real and
\[ D<0. \]Then the roots have the form
\[ p+qi \]and
\[ p-qi. \]Their imaginary parts cancel when the corresponding linear factors are multiplied, leaving real polynomial coefficients.
Build a Quadratic from Complex Conjugate Roots
Suppose the roots are
\[ 2+3i \]and
\[ 2-3i. \]The corresponding factors are
\[ x-(2+3i) \]and
\[ x-(2-3i). \]Rewrite:
\[ (x-2-3i)(x-2+3i). \]Multiply the Conjugate Factors
Let
\[ u=x-2. \]Then
\[ (u-3i)(u+3i)=u^2-(3i)^2. \]Since
\[ (3i)^2=-9, \]we get
\[ u^2+9. \]Therefore,
\[ (x-2)^2+9. \]Expand:
\[ \boxed{x^2-4x+13}. \]Complex Roots and Factorization
Over the real numbers,
\[ x^2-4x+13 \]does not factor into real linear factors.
Over the complex numbers, it factors as
\[ \boxed{ [x-(2+3i)][x-(2-3i)] }. \]The number system determines what kinds of factors are available.
Real Versus Complex Factorization
Consider
\[ x^2+4. \]Over the real numbers, this has no real linear factors.
But since
\[ x^2=-4 \]gives
\[ x=\pm2i, \]over the complex numbers,
\[ \boxed{x^2+4=(x-2i)(x+2i)}. \]The Discriminant as a Decision Tool
Before solving a quadratic, computing \(D\) can answer several questions:
- Will the roots be real or complex?
- Will there be one real root or two?
- If the coefficients are integers, are the real roots likely rational or irrational?
- Will the graph cross, touch, or miss the \(x\)-axis?
This makes the discriminant useful even when the exact roots are not required.
Common Error: Forgetting the Sign of \(b\)
For
\[ 2x^2-7x+4=0, \]we have
\[ b=-7, \]so
\[ b^2=(-7)^2=49. \]The coefficient should be substituted with its sign before squaring.
Common Error: Misreading \(-4ac\)
Suppose
\[ a=2,\qquad c=-3. \]Then
\[ -4ac=-4(2)(-3)=24. \]The product is positive because two negatives are involved.
Parentheses help preserve the signs.
Common Error: Saying \(D<0\) Means No Solutions
A negative discriminant means no real solutions.
It does not mean the equation has no solutions at all.
Over the complex numbers, a quadratic with \(D<0\) has two nonreal complex roots.
Common Error: Treating \(\sqrt{-k}\) as \(-\sqrt{k}\)
For example,
\[ \sqrt{-25} \]is not
\[ -5. \]Since
\[ i=\sqrt{-1}, \]we have
\[ \boxed{\sqrt{-25}=5i}. \]Common Error: Forgetting the Conjugate Pair
If one root of a real-coefficient quadratic is
\[ 4+3i, \]the other must be
\[ \boxed{4-3i}. \]The two roots come from the \(\pm\) in the quadratic formula.
A Reliable Procedure for Classifying Roots
- Write the quadratic in standard form \(ax^2+bx+c=0\).
- Identify \(a\), \(b\), and \(c\) with signs.
- Compute \(D=b^2-4ac\).
- If \(D>0\), expect two distinct real roots.
- If \(D=0\), expect one repeated real root.
- If \(D<0\), expect two nonreal complex conjugate roots.
- If \(D>0\), inspect whether \(D\) is a perfect square to distinguish rational from irrational roots.
A Reliable Procedure for Solving Complex Quadratic Roots
- Compute the discriminant.
- If it is negative, write \(D=-k\) with \(k>0\).
- Rewrite \(\sqrt{-k}=i\sqrt{k}\).
- Simplify the radical.
- Substitute into the quadratic formula.
- Simplify both conjugate solutions.
Practice 1: Classify by Discriminant
Determine the number and type of roots:
\[ x^2-8x+12=0. \]Practice 2: Repeated Root
Determine the number and type of roots:
\[ 4x^2+12x+9=0. \]Practice 3: Complex Roots
Determine the number and type of roots:
\[ x^2+2x+5=0. \]Practice 4: Rational or Irrational?
Classify the real roots:
\[ 3x^2-8x+4=0. \]Practice 5: Another Irrational Case
Classify the roots:
\[ 2x^2+x-2=0. \]Practice 6: Solve Complex Roots
Solve:
\[ x^2-4x+8=0. \]Practice 7: Solve Another Complex Quadratic
Solve:
\[ 2x^2+4x+5=0. \]Practice 8: Graph Meaning
A quadratic has discriminant
\[ D=-11. \]What does this tell you about its real \(x\)-intercepts?
Practice 9: Build a Quadratic
Find a monic quadratic with roots
\[ 1+2i \]and
\[ 1-2i. \]Practice 10: Error Analysis
A student says:
“The discriminant is negative, so the quadratic has no solutions.”
Correct the statement.
Practice Answers
Practice 1
\[ D=(-8)^2-4(1)(12)=64-48=16. \]Since \(16>0\) and is a perfect square,
\[ \boxed{\text{two distinct rational real roots}}. \]Practice 2
\[ D=12^2-4(4)(9)=144-144=0. \]Therefore,
\[ \boxed{\text{one repeated real root}}. \]Practice 3
\[ D=2^2-4(1)(5)=4-20=-16. \]Therefore,
\[ \boxed{\text{two nonreal complex conjugate roots}}. \]Practice 4
\[ D=(-8)^2-4(3)(4)=64-48=16. \]Since \(16\) is a positive perfect square,
\[ \boxed{\text{two rational real roots}}. \]Practice 5
\[ D=1^2-4(2)(-2)=1+16=17. \]Since \(17>0\) but is not a perfect square,
\[ \boxed{\text{two irrational real roots}}. \]Practice 6
\[ D=(-4)^2-4(1)(8)=16-32=-16. \] \[ x= \frac{4\pm\sqrt{-16}}{2} = \frac{4\pm4i}{2}. \]Therefore,
\[ \boxed{x=2\pm2i}. \]Practice 7
\[ D=4^2-4(2)(5)=16-40=-24. \]Then
\[ x= \frac{-4\pm\sqrt{-24}}{4}. \]Since
\[ \sqrt{-24}=2i\sqrt6, \]we obtain
\[ \boxed{x=-1\pm\frac{\sqrt6}{2}i}. \]Practice 8
Since
\[ D<0, \]the quadratic has no real roots and therefore
\[ \boxed{\text{no real }x\text{-intercepts}}. \]Practice 9
Use the factors
\[ [x-(1+2i)][x-(1-2i)]. \]Rewrite:
\[ (x-1-2i)(x-1+2i). \]Let \(u=x-1\):
\[ (u-2i)(u+2i)=u^2+4. \]Therefore,
\[ (x-1)^2+4 = x^2-2x+5. \]A suitable quadratic is
\[ \boxed{x^2-2x+5}. \]Practice 10
A negative discriminant means the quadratic has
\[ \boxed{\text{no real solutions but two nonreal complex conjugate solutions}}. \]What to Remember
The discriminant is
\[ \boxed{D=b^2-4ac}. \]It classifies quadratic roots:
- \(D>0\): two distinct real roots;
- \(D=0\): one repeated real root;
- \(D<0\): two nonreal complex conjugate roots.
When \(D>0\), a perfect-square discriminant gives rational roots, while a non-square discriminant gives irrational roots when the coefficients are rational.
The imaginary unit
\[ \boxed{i=\sqrt{-1}} \]allows negative discriminants to be handled naturally.
Real-coefficient quadratics with nonreal roots produce those roots in conjugate pairs.
Where We Go Next
Quadratic equations provide a complete model of degree-\(2\) behavior. Higher-degree polynomial equations require a broader collection of tools: factoring, synthetic division, known roots, substitution patterns, and degree reduction.
The next lesson, Higher-Degree Polynomial Equations, extends the solving process beyond quadratics and develops strategies for reducing more complicated polynomial equations to simpler ones.