The quadratic formula obtained by completing the square once in general, the three kinds of solution, the discriminant as a predictor of how many and what type, the link between the discriminant and the number of x-intercepts, and the vertical-motion model with an initial velocity.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Use the Quadratic Formula and the Discriminant
Objectives
Five outcomes. The first gives you a method that never fails; the third tells you what it will produce before you run it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 292-297 — the lesson these objectives are drawn from
Warm-up
Lesson 4.7 solved every quadratic by completing the square, one equation at a time.
Discussion prompt
Complete the square on the general equation a x squared plus bx plus c equals zero, keeping the letters. Divide by a first, then follow the usual steps. What do you get?
Hint: Half of b over a is b over 2a, and its square is b squared over 4a squared.
Answer:
\[ x^2 + \tfrac{b}{a}x = -\tfrac{c}{a} \;\Longrightarrow\; \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2} \]
\[ x + \tfrac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \;\Longrightarrow\; x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Doing the work once with letters means never doing it again with numbers. That result is the quadratic formula, and Exercise 67 asks you to reproduce this derivation.
Concept
Because completing the square works on every quadratic, carrying it out once on the general equation produces a formula that solves every quadratic. Substituting a, b and c gives the solutions directly, whether they are rational, irrational or imaginary.
quadratic formula — For real a, b and c with a not zero, the solutions of a x squared plus bx plus c equals zero are negative b, plus or minus the square root of b squared minus 4ac, all over 2a.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Everything the earlier methods could do, this can do. The earlier methods survive because when they apply they are faster, and because completing the square also produces vertex form.
Figure (svg): The quadratic formula with each letter labelled and the discriminant marked out
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 292-292
Section
Section 1
Concept
Write the equation in standard form so that a, b and c can be read off, then substitute them into the formula. The most common error is reading the coefficients from an equation that has not been rearranged yet.
\[ ax^2 + bx + c = 0 \;\Longrightarrow\; x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Take the signs with the coefficients: in x squared plus 3x minus 2, c is negative 2, not 2, and dropping that minus changes the discriminant completely.
Figure (svg): The quadratic formula with each letter labelled and the discriminant marked out
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 292-292 — The Quadratic Formula
Picture it
Example 1: x squared plus 3x equals 2.
Figure (svg): The quadratic formula substituted step by step for one equation, with the graph beside it
The algebra gives negative 3 plus or minus the root of 17, over 2. The graph's intercepts, about 0.56 and negative 3.56, agree — which is the textbook's own check.
Worked example
Example 1. The roots are irrational, so factoring was never going to work.
\[ \text{Solve } x^2 + 3x = 2. \]
Write in standard form
Why: Subtracting 2 puts every term on one side.
\[ x ^{2} + 3 x - 2 = 0 \]
Read a, b and c with their signs
Why: The leading coefficient is 1, the middle coefficient is 3, and the constant is negative 2.
\[ a = 1, b = 3, c = -2 \]
Substitute into the formula
Why: Negative 3 plus or minus the root of 3 squared minus 4 times 1 times negative 2, over 2.
\[ \frac{-3 + - \sqrt{9 + 8}}{2} \]
Simplify the radicand
Why: Nine minus negative 8 is 9 plus 8, which is 17, and 17 has no square factors.
\[ x = \frac{-3 + - \sqrt{17}}{2} \]
Figure (svg): The solution to Worked example two real solutions shown as a ladder of expressions, one row per algebraic move
\[ x = \frac{-3 \pm \sqrt{17}}{2} \]
Verify: compare with the graph
Why: The root of 17 is about 4.123, so the solutions are about 0.56 and about negative 3.56. Graphing y equals x squared plus 3x minus 2 shows x-intercepts at about 0.56 and negative 3.56. Note also that the two roots average to negative 1.5, which is negative b over 2a — the vertex, as symmetry requires.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 292-292
Matching
Rearrange first, then read.
Match the pairs
Why: Every one of these needed rearranging first, and in three of the four a coefficient changed sign in the process. The last one also required combining the two x terms, since negative 18x minus 12x is negative 30x.
Worked example
Guided Practice 1. Standard form is the first move again.
\[ \text{Solve } x^2 = 6x - 4. \]
Rearrange into standard form
Why: Subtracting 6x and adding 4 gives every term on the left.
\[ x ^{2} - 6 x + 4 = 0 \]
Read the coefficients
Why: One, negative 6 and 4.
\[ a = 1, b = -6, c = 4 \]
Substitute
Why: Negative negative 6 is positive 6; the radicand is 36 minus 16.
\[ \frac{6 + - \sqrt{20}}{2} \]
Simplify the radical and reduce
Why: The root of 20 is 2 root 5, and dividing both terms by 2 leaves 3 plus or minus root 5.
\[ x = 3 + - \sqrt{5} \]
Figure (svg): The solution to Worked example another with two real roots shown as a ladder of expressions, one row per algebraic move
\[ x = 3 \pm \sqrt{5} \]
Verify: check the sum of the roots
Why: The two roots sum to 6, and for x squared plus bx plus c the roots always sum to negative b, which here is 6. Their product is 9 minus 5, or 4, matching c. Both relations hold, so the answer is right. The reducing step at the end matters: leaving it as 6 plus or minus 2 root 5, over 2, is correct but unfinished.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 293-293
Trap
\[ x^2 + 3x = 2 \]
Read a, b and c straight off the equation
Why: The 2 on the right is taken as c.
\[ x = \frac{-3 \pm \sqrt{9 - 8}}{2} = \frac{-3 \pm 1}{2} \;\Longrightarrow\; x = -1 \text{ or } -2 \quad \text{(wrong)} \]
Substituting negative 1 gives 1 minus 3, which is negative 2, not 2. Both answers fail.
\[ x^2 + 3x - 2 = 0 \;\Longrightarrow\; a = 1, \; b = 3, \; c = -2 \]
Move every term to one side before reading anything
Why: The formula is stated for an equation equal to zero, and c is whatever constant remains on that side.
\[ x = \frac{-3 \pm \sqrt{17}}{2} \]
The book flags this in an Avoid Errors note. The failure is quiet — the wrong version even produces tidy-looking answers — so the check by substitution is what catches it.
Fill the middle
Example 1, at the radicand.
Fill in the blanks
x = \frac17}___ = \frac___}}}___
Why: Three squared is 9, and negative 4 times 1 times negative 2 is positive 8, so the radicand is 17. The double negative is where sign errors live: subtracting a negative c adds to the discriminant, which is why a negative constant always guarantees two real roots.
Ranking
Solving with the quadratic formula.
Put in order
Why: Steps one and two are where nearly all the errors are, because the formula itself is mechanical once the letters are right. Keeping brackets around negative coefficients when substituting prevents the second most common error, which is losing a sign inside the radicand.
Prediction
Commit before reasoning.
Predict first
The quadratic formula is derived by which method?
Correct: Completing the square on the general equation.
\[ \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2-4ac}{4a^2} \;\Longrightarrow\; x = \tfrac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Why: Dividing by a, moving the constant, adding b over 2a squared to both sides and taking roots produces exactly the formula, with b squared minus 4ac appearing as the numerator of the right side over 4a squared. That is why the formula inherits completing the square's universality: it works on every quadratic because the method it came from does.
Section
Section 2
Concept
If the expression under the radical comes out zero, the plus-or-minus contributes nothing and there is a single solution. If it comes out negative, Lesson 4.6 turns it into an imaginary number and the two solutions form a conjugate pair.
\[ \sqrt{0} = 0 \;\Longrightarrow\; \text{one root}; \qquad \sqrt{-4} = 2i \;\Longrightarrow\; \text{two imaginary roots} \]
Nothing about the procedure changes in either case. The formula reports what is there, and the three outcomes are not three methods but three results of the same one.
Figure (svg): Three parabolas showing two x-intercepts, one, and none, each labelled with the sign of its discriminant
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 293-293 — Examples 2 and 3
Picture it
The same a and b with constants 15, 16 and 17.
Figure (svg): Three parabolas showing two x-intercepts, one, and none, each labelled with the sign of its discriminant
As the constant rises the parabola lifts, crossing the axis twice, then touching once, then missing entirely. The solutions change type at exactly the moment the graph stops crossing.
Worked example
Example 2. The discriminant comes out zero.
\[ \text{Solve } 25x^2 - 18x = 12x - 9. \]
Write in standard form
Why: Subtracting 12x and adding 9 gives negative 30x in the middle.
\[ 25 x ^{2} - 30 x + 9 = 0 \]
Read the coefficients
Why: Twenty-five, negative 30 and 9.
\[ a = 25, b = -30, c = 9 \]
Substitute and evaluate the radicand
Why: Negative 30 squared is 900, and 4 times 25 times 9 is also 900, so the radicand is zero.
\[ \frac{30 + - \sqrt{0}}{50} \]
Simplify
Why: The plus-or-minus adds nothing, leaving 30 over 50.
\[ x = \frac{3}{5} \]
Figure (svg): The solution to Worked example one real solution shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{3}{5} \]
Verify: solve it a second way
Why: The left side factors as the quantity 5x minus 3, squared, whose repeated root is three fifths — the book's Another Way note. Graphing y equals 25x squared minus 30x plus 9 shows a single x-intercept at 0.6. A zero discriminant always means a perfect square trinomial and a parabola that touches the axis without crossing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 293-293
Sorting
Evaluate the radicand and look at its sign.
Sort into buckets
Sort each equation.
Every one of these has two solutions counted properly; only the sorting into real and imaginary changes, and one of them has its two solutions coincide.
Worked example
Example 3. The radicand is negative, so Lesson 4.6 finishes it.
\[ \text{Solve } -x^2 + 4x = 5. \]
Write in standard form
Why: Subtracting 5 leaves a negative leading coefficient, which is allowed.
\[ -x ^{2} + 4 x - 5 = 0 \]
Read the coefficients
Why: Negative 1, 4 and negative 5.
\[ a = -1, b = 4, c = -5 \]
Substitute
Why: The radicand is 16 minus 4 times negative 1 times negative 5, which is 16 minus 20.
\[ \frac{-4 + - \sqrt{-4}}{-2} \]
Rewrite using i
Why: The square root of negative 4 is 2i.
\[ \frac{-4 + - 2 i}{-2} \]
Divide both terms by negative 2
Why: Negative 4 over negative 2 is 2, and plus or minus 2i over negative 2 is minus or plus i, which is still plus or minus i.
\[ x = 2 + - i \]
Figure (svg): The solution to Worked example two imaginary solutions shown as a ladder of expressions, one row per algebraic move
\[ x = 2 \pm i \]
Verify: substitute the complex solution
Why: For 2 plus i: the square is 4 plus 4i plus i squared, which is 3 plus 4i, so negative that is negative 3 minus 4i; adding 4 times 2 plus i gives 8 plus 4i; and negative 3 minus 4i plus 8 plus 4i is 5, matching the right side exactly. The imaginary parts cancel, which is what always happens with a conjugate pair.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 293-293
Error analysis
A student solves an equation whose leading coefficient is negative.
Annotate
On: \( -x^2 + 4x - 5 = 0 \;\Longrightarrow\; a = 1, \; b = 4, \; c = -5 \)
Either read a as negative 1, or multiply the whole equation by negative 1 first so that a is positive. Both are correct; mixing them is not.
Fill the middle
Example 2, at the simplification.
Fill in the blanks
x = \frac3/5}___ = \frac______ = ___
Why: Thirty over 50 reduces to three fifths, or 0.6. When the radicand is zero the two branches of the formula collapse into one, and the single root is always negative b over 2a — the vertex, sitting exactly on the axis.
Two truths and a lie
All three are about the three cases.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. It has no REAL solutions, which is a different claim: in the complex numbers of Lesson 4.6 it has exactly two, and they are perfectly good numbers. The distinction matters because it is the difference between a parabola that misses the axis and an equation that is unsolvable, and only the first of those is true.
Prediction
Commit before reasoning.
Predict first
Can a quadratic with real coefficients have exactly one imaginary solution and one real one?
Correct: No — the plus-or-minus produces a conjugate pair or nothing.
\[ x = \tfrac{-b}{2a} \pm \tfrac{\sqrt{b^2-4ac}}{2a}: \; \text{one expression, two signs} \]
Why: The two solutions differ only by the sign in front of one radical, so either both are real or both are imaginary, and in the imaginary case they are conjugates. This is a general fact about polynomials with real coefficients: their non-real roots always pair up, which is why a cubic with real coefficients must have at least one real root.
Section
Section 3
Concept
The expression b squared minus 4ac inside the radical is called the discriminant of the equation. Computing it takes one line and settles how many solutions there are and what kind, before any of the rest of the formula is evaluated.
discriminant — The expression b squared minus 4ac from the quadratic formula. It is positive when there are two real solutions, zero when there is one, and negative when the two solutions are imaginary.
\[ b^2 - 4ac > 0, \; = 0, \; < 0 \]
The word discriminant means the thing that tells them apart, which is exactly its job: it distinguishes the three cases without doing the work of finding the solutions.
Figure (svg): Three equations differing only in the constant, each with its discriminant and its solutions
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294 — Using the Discriminant
Picture it
Example 4: constants of 17, 16 and 15 with everything else fixed.
Figure (svg): Three equations differing only in the constant, each with its discriminant and its solutions
Moving the constant by one unit at a time walks the discriminant from negative through zero to positive, and the solutions from a conjugate pair through a repeated root to a real pair.
Worked example
Example 4, all three parts.
\[ \text{Give the number and type of solutions of } x^2 - 8x + 17 = 0, \; x^2 - 8x + 16 = 0, \; x^2 - 8x + 15 = 0. \]
First: compute b squared minus 4ac
Why: Negative 8 squared is 64, and 4 times 1 times 17 is 68.
\[ 64 - 68 = -4 \]
Conclude for the first
Why: The discriminant is negative, so there are two imaginary solutions, namely 4 plus i and 4 minus i.
Second and third
Why: Four times 16 is 64, giving a discriminant of zero; 4 times 15 is 60, giving 4.
\[ 0\text{ and } 4 \]
Conclude for the other two
Why: Zero means one real solution, 4; positive means two real solutions, 3 and 5.
Figure (svg): The solution to Worked example use the discriminant shown as a ladder of expressions, one row per algebraic move
\[ -4, \quad 0, \quad 4 \]
Verify: check the middle case by factoring
Why: The second trinomial is x squared minus 8x plus 16, which is the quantity x minus 4, squared — a perfect square with the repeated root 4, exactly as a zero discriminant predicts. The third factors as the quantity x minus 3, times x minus 5, giving the roots 3 and 5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294
Matching
Rearrange first, then compute b squared minus 4ac.
Match the pairs
Why: The last two both come out positive, so both have two real solutions, but only one of them has a discriminant that is a perfect square. That extra detail predicts something the sign alone does not: 144 is 12 squared, so those roots are rational and the equation would have factored.
Worked example
Guided Practice 4 to 9. Standard form first each time.
\[ \text{Classify } 2x^2 + 4x - 4 = 0, \; 3x^2 + 12x + 12 = 0, \; 8x^2 = 9x - 11, \; 7x^2 - 2x = 5, \; 4x^2 + 3x + 12 = 3 - 3x, \; 3x - 5x^2 + 1 = 6 - 7x. \]
The first two
Why: Sixteen minus 4 times 2 times negative 4 is 16 plus 32, or 48; and 144 minus 144 is 0.
\[ 48 > 0; 0 \]
The next two, after rearranging
Why: Eight x squared minus 9x plus 11 gives 81 minus 352, or negative 271; 7x squared minus 2x minus 5 gives 4 plus 140, or 144.
\[ -271 < 0; 144 > 0 \]
The fifth, after collecting
Why: Four x squared plus 6x plus 9 equals zero gives 36 minus 144, or negative 108.
\[ -108 < 0 \]
The sixth, after collecting
Why: Negative 5x squared plus 10x minus 5 equals zero gives 100 minus 100, or 0.
\[ 0 \]
Report each verdict
Why: Two real, one real, two imaginary, two real, two imaginary, one real.
Figure (svg): The solution to Worked example six more discriminants shown as a ladder of expressions, one row per algebraic move
\[ 48, \; 0, \; -271, \; 144, \; -108, \; 0 \]
Verify: look at the two zeros
Why: Both zero-discriminant equations are constant multiples of perfect squares: 3x squared plus 12x plus 12 is 3 times the quantity x plus 2, squared, and the sixth is negative 5 times the quantity x minus 1, squared. That is a reliable pattern — a zero discriminant and a perfect square are the same fact stated two ways.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294
Error analysis
A student classifies an equation without collecting terms first.
Annotate
On: \( 4x^2 + 3x + 12 = 3 - 3x: \quad b^2 - 4ac = 3^2 - 4(4)(12) = -183 \)
Getting the right answer from the wrong coefficients is luck, not method. Standard form is the first step for the discriminant exactly as it is for the formula.
Fill the middle
Example 4a.
Fill in the blanks
x^2 - 8x + 17 = 0: \quad (-8)^2 - 4(1)(17) = 64 - 68 = -4
Why: Sixty-four minus 68 is negative 4, so the solutions are imaginary. Note that negative 8 must be squared as a whole, brackets included: writing minus 8 squared without brackets would give negative 64 and reverse the verdict entirely.
Prediction
Commit before reasoning.
Predict first
The discriminant of 7x squared minus 2x minus 5 equals zero is 144. What extra fact does that give?
Correct: The roots are rational, so the trinomial factors over the integers.
\[ x = \tfrac{2 \pm 12}{14} = 1 \text{ or } -\tfrac{5}{7} \;\Longrightarrow\; 7x^2 - 2x - 5 = (7x+5)(x-1) \]
Why: The root of 144 is 12, a whole number, so the formula produces fractions rather than radicals: the roots are 1 and negative five sevenths, and the trinomial factors as the quantity 7x plus 5, times the quantity x minus 1. This gives a fast test worth remembering: compute the discriminant, and if it is a perfect square, go back and factor instead, because factoring will be quicker.
Comparison
Fill the blanks. One is much cheaper than the other.
Comparison matrix
| Question | Discriminant alone | The whole formula |
|---|---|---|
| Effort | one line | four or five lines |
| Tells you how many solutions | yes | yes |
| Tells you what the solutions are | no | yes |
| Tells you the number of x-intercepts | yes | yes, but the long way |
When a question asks only how many or what type, computing the discriminant and stopping is the complete answer, not a shortcut.
Section
Section 4
Concept
The solutions of a x squared plus bx plus c equals zero are the x-intercepts of the graph of y equals that quadratic. So a positive discriminant means two x-intercepts, a zero discriminant means exactly one, and a negative discriminant means none.
\[ b^2 - 4ac > 0, = 0, < 0 \;\Longleftrightarrow\; 2, 1, 0 \text{ x-intercepts} \]
This joins together everything since Lesson 4.1: the zeros of Lesson 4.3, the intercept form of Lesson 4.2 and the vertex of Lesson 4.7 are all visible in the same picture, and the discriminant says which of the three arrangements you are looking at.
Figure (svg): Three parabolas showing two x-intercepts, one, and none, each labelled with the sign of its discriminant
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294 — Graph of y = ax^2 + bx + c
Picture it
The same three equations as Example 4, drawn.
Figure (svg): Three parabolas showing two x-intercepts, one, and none, each labelled with the sign of its discriminant
The zero-discriminant case is the boundary between the other two, which is why it is exactly the case where the vertex sits on the axis.
Worked example
Example 4, reinterpreted geometrically.
\[ \text{How many times does each of } y = x^2-8x+17, \; y = x^2-8x+16, \; y = x^2-8x+15 \text{ meet the x-axis?} \]
Note what all three share
Why: All have a equal to 1 and b equal to negative 8, so all have their vertex at x equal to 4.
Find each vertex height
Why: At x equal to 4 the values are 17 minus 16, or 1; 16 minus 16, or 0; and 15 minus 16, or negative 1.
\[ \text{vertex heights } 1, 0, -1 \]
Match against the discriminants
Why: The vertex above the axis goes with a negative discriminant; on the axis with zero; below with positive.
\[ -4, 0, 4 \]
State the intercept counts
Why: None, one and two respectively.
\[ 0, 1, 2\text{ intercepts} \]
Figure (svg): The solution to Worked example read the graph from the discriminant shown as a ladder of expressions, one row per algebraic move
\[ 0, \quad 1, \quad 2 \text{ x-intercepts} \]
Verify: check the sign relationship
Why: For an upward parabola, the vertex being below the axis is exactly what makes two crossings possible, and the vertex height turned out to be the negative of the discriminant divided by 4a in each case: negative 4 over 4 is negative 1, matching the third vertex. So the discriminant and the vertex height are the same information scaled.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294
Sorting
Compute the discriminant and read its sign.
Sort into buckets
Sort each parabola by its number of x-intercepts.
Three of these are the same parabola shifted vertically by one unit at a time, which is the cleanest possible demonstration of the boundary case.
Worked example
Reading the argument in reverse.
\[ \text{A parabola opens upward with vertex } (3, -5). \text{ What is the sign of its discriminant?} \]
Locate the vertex relative to the axis
Why: The vertex height is negative 5, so the vertex is below the horizontal axis.
Use the direction of opening
Why: The parabola opens upward, so it rises from the vertex in both directions and must cross the axis on each side.
Translate crossings into solutions
Why: Two x-intercepts means the equation has two real solutions.
State the sign
Why: Two real solutions correspond to a positive discriminant.
\[ b ^{2} - 4 a c > 0 \]
Figure (svg): The solution to Worked example from a graph back to the discriminant shown as a ladder of expressions, one row per algebraic move
\[ b^2 - 4ac > 0 \]
Verify: build an example and check
Why: Take y equals the quantity x minus 3, squared, minus 5, which expands to x squared minus 6x plus 4. Its discriminant is 36 minus 16, or 20, which is positive as predicted. Changing the vertex height to positive 5 gives x squared minus 6x plus 14 and a discriminant of negative 20 — the sign flips exactly when the vertex crosses the axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 294-294
Error analysis
A student describes a parabola that misses the horizontal axis.
Annotate
On: \( y = -x^2 + 4x - 5 \text{ has no x-intercepts, so } -x^2 + 4x - 5 = 0 \text{ has no solutions} \)
Say which number system you mean. Every quadratic has two solutions in the complex numbers, counted with repetition, and the graph only shows the real ones.
Fill the middle
A downward parabola with vertex above the axis.
Fill in the blanks
\text2 (2, 7) \;\Longrightarrow\; ___ \text___
Why: A downward parabola falls in both directions from its vertex, and its vertex is above the axis, so it must cross on each side. Two intercepts means a positive discriminant. Compare with an upward parabola whose vertex is above the axis, which has none: direction and vertex position have to be read together.
Prediction
Commit before reasoning.
Predict first
Is there a quadratic whose graph meets the horizontal axis at three points?
Correct: No — a quadratic equation has at most two solutions.
\[ x = \tfrac{-b + \sqrt{D}}{2a} \text{ and } x = \tfrac{-b - \sqrt{D}}{2a}: \; \text{exactly two expressions} \]
Why: The formula produces exactly two values, which may coincide, so there is no way to get three. Geometrically a parabola is a single smooth curve turning once, and a horizontal line can meet it at most twice. This is the degree-two case of a general fact: a polynomial of degree n has at most n real roots, which Chapter 5 takes up for higher degrees.
Comparison
Fill the blanks. Direction and vertex together decide everything.
Comparison matrix
| Opens | Vertex | Solutions |
|---|---|---|
| Upward | below the axis | two real |
| Upward | above the axis | two imaginary |
| Downward | above the axis | two real |
| Either | on the axis | one real, repeated |
The rule is that the vertex must be on the opposite side of the axis from the direction the parabola opens, or the curve never reaches it.
Section
Section 5
Concept
Lesson 4.5's dropped-object model gains a term when the object is thrown: the height is negative 16t squared plus the initial vertical velocity times t, plus the initial height. That middle term is why these equations need the quadratic formula rather than a square root.
\[ h = -16t^2 + v_0 t + h_0 \]
The initial velocity is positive for an upward throw, negative for a downward one, and zero for a drop — in which case the model collapses back to the one from Lesson 4.5.
Figure (svg): The height of a tossed ball over time, with the model's three forms of initial velocity shown
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 295-295 — Modeling launched objects
Picture it
Example 5: released at 4 feet with an upward velocity of 40 feet per second, caught at 3 feet.
Figure (svg): The height of a tossed ball over time, with the model's three forms of initial velocity shown
The ball is caught slightly below where it left the hand, so the catch happens a little after the symmetric point. Solving gives about 2.5 seconds in the air.
Worked example
Example 5. Substitute, rearrange, apply the formula, reject.
\[ \text{A ball leaves a hand at } 4 \text{ ft with } v_0 = 40 \text{ ft/s and is caught at } 3 \text{ ft. Find the time.} \]
Write the model and substitute
Why: The height is 3, the initial velocity 40 and the initial height 4.
\[ 3 = -16 t ^{2} + 40 t + 4 \]
Write in standard form
Why: Subtracting 3 leaves a constant of 1.
\[ 0 = -16 t ^{2} + 40 t + 1 \]
Apply the formula
Why: Negative 40 plus or minus the root of 1600 minus 4 times negative 16 times 1, over negative 32.
\[ \frac{-40 + - \sqrt{1664}}{-32} \]
Evaluate both roots
Why: The root of 1664 is about 40.79, giving about negative 0.025 and about 2.525.
\[ -0.025\text{ or } 2.5 \]
Reject the negative time
Why: A time before the throw has no meaning here.
\[ \text{about } 2.5\text{ seconds} \]
Figure (svg): The solution to Worked example how long is the ball in the air shown as a ladder of expressions, one row per algebraic move
\[ t \approx 2.5 \text{ seconds} \]
Verify: substitute the time back
Why: At t equal to 2.5, negative 16 times 6.25 is negative 100, plus 40 times 2.5, or 100, plus 4 gives 4 — close to the target 3, and the small gap is the rounding from 2.525 to 2.5. The exact root gives exactly 3. The tiny negative root is also meaningful: it is when the ball would have been at 3 feet if it had been thrown from lower down.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 295-295
Matching
Ask what the initial velocity is.
Match the pairs
Why: Two of these have no initial vertical velocity, so they reduce to the Lesson 4.5 model — including the ball rolled off a table, which leaves horizontally and therefore has zero VERTICAL velocity. The downward throw has a negative velocity term, which is the only case where that middle coefficient is negative.
Worked example
Guided Practice 10. Only the velocity changes.
\[ \text{Repeat with } v_0 = 50 \text{ feet per second.} \]
Substitute the new velocity
Why: The model becomes 3 equals negative 16t squared plus 50t plus 4.
\[ 0 = -16 t ^{2} + 50 t + 1 \]
Apply the formula
Why: The radicand is 2500 minus 4 times negative 16 times 1, which is 2500 plus 64.
\[ \sqrt{2564} \]
Evaluate
Why: The root of 2564 is about 50.64, so the roots are about negative 0.02 and about 3.15.
\[ -0.02\text{ or } 3.1 \]
Reject and report
Why: The ball is in the air for about 3.1 seconds.
\[ \text{about } 3.1\text{ seconds} \]
Figure (svg): The solution to Worked example a faster throw shown as a ladder of expressions, one row per algebraic move
\[ t \approx 3.1 \text{ seconds} \]
Verify: compare the two throws
Why: Raising the velocity from 40 to 50, a 25 percent increase, raised the time from about 2.5 to about 3.1 seconds, also about 25 percent. That near-proportionality is expected: the time in the air is close to twice the initial velocity divided by 32, and the small constant term barely disturbs it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 295-295
Trap
\[ h = -16t^2 + h_0 \;\Longrightarrow\; 3 = -16t^2 + 4 \]
Use Lesson 4.5's model because the object ends up falling
Why: The initial velocity is ignored because the ball comes back down anyway.
\[ 16t^2 = 1 \;\Longrightarrow\; t = 0.25 \quad \text{(wrong)} \]
A quarter of a second is far too short: the ball was thrown upward at 40 feet per second and has to rise first.
\[ h = -16t^2 + v_0 t + h_0 \;\Longrightarrow\; 3 = -16t^2 + 40t + 4 \]
Include the initial velocity term whenever the object is thrown
Why: The dropped model is the special case where that velocity happens to be zero.
\[ t \approx 2.5 \text{ seconds} \]
The missing term is also what makes the equation genuinely quadratic in the sense of needing the formula: without it there is no first-power term and Lesson 4.5's square-root method would suffice.
Fill the middle
Example 5, at the substitution.
Fill in the blanks
3 = -16t^2 + 40t + 4 \;\Longrightarrow\; 0 = -16t^2 + 40t + 1
Why: Subtracting 3 from both sides leaves 4 minus 3, which is 1. That small constant is what makes the equation not quite symmetric: had the ball been caught at exactly the height it left, the constant would be zero and one root would be exactly t equal to zero.
Ranking
Answering a thrown-object question.
Put in order
Why: Choosing the model comes first because the two models differ by a whole term, and using the wrong one produces an answer that is not merely inaccurate but about a different situation. The rejection at the end is the same physical judgement made in every model since Lesson 4.3.
Prediction
Commit before reasoning.
Predict first
The juggling equation gives t about negative 0.025 as well as about 2.5. What is that first value?
Correct: A time before the throw, when the model does not apply.
\[ t \approx -0.025 \text{ or } 2.525; \; \text{only } t \ge 0 \text{ is real time} \]
Why: The parabola described by the model extends backwards in time as well as forwards, and it passes through a height of 3 feet a fraction of a second before t equals zero. That moment is fictional: the ball was not yet in the air. Rejecting it is a statement about the situation, not about the algebra, which is perfectly happy with both roots.
Comparison
Fill the blanks. Four methods, in the order you should consider them.
Comparison matrix
| Method | Use when | Lesson |
|---|---|---|
| Take square roots | there is no first-power term | 4.5 |
| Factor | the discriminant is a perfect square | 4.3 and 4.4 |
| Complete the square | you also want vertex form | 4.7 |
| Quadratic formula | always, and when the others are awkward | 4.8 |
The second row is the useful new fact: the discriminant tells you in one line whether factoring is worth trying at all.
Pattern
One routine for any quadratic equation.
Step two is worth doing even when you intend to use the formula anyway, because it tells you what the answer should look like before you compute it.
OpenStax Algebra and Trigonometry 2e, §2.5 Quadratic Equations §2.5
Check
The formula. Standard form first.
Check your understanding
Solve x^2 + 3x = 2.
Answer: A
Why: In standard form c is -2, so the radicand is 9 + 8 = 17.
Check
The discriminant. Note the sign carefully.
Check your understanding
Find the discriminant of x^2 - 8x + 17 = 0 and say what it means.
Answer: A
Why: The discriminant is 64 - 68 = -4, so the solutions are the conjugate pair 4 + i and 4 - i.
Check
Vertical motion. Include the velocity term.
Check your understanding
A ball leaves a hand at 4 ft with an upward velocity of 40 ft/s and is caught at 3 ft. How long is it in the air?
Answer: A
Why: Solving 0 = -16t^2 + 40t + 1 gives t about -0.025 or 2.525; the negative time is rejected.
Real world
A firework is launched from a platform 6 feet above the ground with an initial vertical velocity of 80 feet per second. Its fuse is set to burst at a height of 100 feet on the way up.
Discussion prompt
Will it burst? If so, when? Use the discriminant before solving, and say what it tells you in advance.
Hint: Set the height to 100 and look at the discriminant of the resulting equation.
Answer:
\[ 100 = -16t^2 + 80t + 6 \;\Longrightarrow\; 16t^2 - 80t + 94 = 0 \;\Longrightarrow\; 8t^2 - 40t + 47 = 0 \]
\[ b^2 - 4ac = 1600 - 1504 = 96 > 0 \;\Longrightarrow\; \text{two real times} \]
\[ t = \frac{40 \pm \sqrt{96}}{16} = \frac{10 \pm \sqrt{6}}{4} \approx 1.89 \text{ or } 3.11 \]
It does reach 100 feet, at about 1.89 seconds on the way up and again at about 3.11 seconds on the way down, so the fuse should be set for about 1.9 seconds.
The discriminant did real work here. Had it been negative, the firework would never have reached 100 feet at all and no fuse setting would have worked — and one line of arithmetic would have told you that before any solving. That is the practical value of computing it first.
Commit first
Answer, then rate your confidence honestly.
Predict first
If the discriminant of a quadratic with integer coefficients is 49, what do you know?
Correct: Two real rational solutions, so it factors over the integers.
\[ \sqrt{49} = 7 \;\Longrightarrow\; x = \tfrac{-b \pm 7}{2a}, \text{ both rational} \]
Why: Forty-nine is a perfect square, so the radical evaluates to the whole number 7 and the formula produces two fractions rather than two radicals. Rational roots of a quadratic with integer coefficients correspond exactly to a factorisation over the integers, so a perfect-square discriminant is a reliable signal to go back and factor. A positive non-square discriminant, by contrast, guarantees irrational roots and no integer factorisation.
Explain it
They have memorised the formula and are unsure what the discriminant is for.
Discussion prompt
In four sentences or fewer, explain what the discriminant is, why it predicts the number of solutions, and why anyone would compute it separately.
Hint: Point at where it sits in the formula.
Answer:
The discriminant is the expression under the radical, b squared minus 4ac. If it is positive its square root is a real number, so the plus-or-minus gives two different answers; if it is zero the plus-or-minus adds nothing and there is one; if it is negative the root is imaginary and the two answers are a conjugate pair.
It is worth computing on its own because it takes one line, whereas the whole formula takes four or five. If a question only asks how many solutions there are, or whether a graph crosses the axis, the discriminant is the complete answer.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For reading coefficients, always write the standard form on its own line before touching the formula. For signs, put brackets around every negative when you substitute. For reducing, look for a common factor in all three parts of the fraction, not just two. For interpreting, memorise the three cases as pictures of parabolas rather than as words. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the three equations x squared minus 8x plus 15 equals zero, x squared minus 8x plus 16 equals zero and x squared minus 8x plus 17 equals zero and put them side by side across the page. For each, in a column: compute the discriminant, state the number and type of solutions, solve with the formula in full, and sketch the parabola marking the vertex and any x-intercepts. Below the three columns, write one sentence saying what changed and one saying what stayed the same as you moved from left to right. At the bottom of the page, derive the quadratic formula by completing the square on a x squared plus bx plus c equals zero, keeping the letters throughout. In a margin, write which of the four solving methods in this chapter you would try first on each of the three equations, and why.
If your three sketches have different vertices, check them: all three have b equal to negative 8 and a equal to 1, so all three vertices sit on the line x equals 4 and differ only in height.
Recap
Five things, and the first one means no quadratic equation can defeat you.
| If you see | Then |
|---|---|
| Any quadratic equation | The formula will solve it |
| Terms on both sides | Rearrange before reading a, b and c |
| A positive discriminant | Two real roots, two x-intercepts |
| A zero discriminant | One repeated root, vertex on the axis |
| A negative discriminant | Two imaginary roots, no x-intercepts |
| A perfect-square discriminant | It factors; go back and factor |
| A thrown object | Include the v0 t term |
Lesson 4.9 asks a different question about the same parabolas: not where the quadratic equals zero, but where it is positive or negative — which turns these solutions into the endpoints of intervals.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.8 Use the Quadratic Formula and the Discriminant §4.8, pp. 292-297 — everything on these slides traces back here
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