Solving any quadratic equation with the quadratic formula. Includes stating and applying the formula, rewriting an equation into standard form before reading its coefficients, handling irrational solutions, using the formula to find the x-intercepts of a graph, and applying the vertical motion model for a thrown object.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Solving Quadratic Equations by the Quadratic Formula
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-539 — the lesson these objectives are drawn from
Warm-up
Lesson 9.2 solved equations with no x term. Lesson 9.5 estimated the rest from graphs. This lesson does all of them exactly.
Discussion prompt
Try to solve x squared minus nine x plus fourteen equals nought using square roots. What goes wrong?
Hint: Can the squared term be isolated?
Answer:
\[ x^2 - 9x + 14 = 0 \]
The x term blocks it. Isolating the squared term leaves nine x plus something on the other side, so taking a root of both sides gives an equation still containing x. Every method so far either needs the middle term to vanish or gives only an estimate, and the formula in this lesson is what removes both restrictions.
Concept
The quadratic formula gives the solutions of a x squared plus b x plus c equals nought in terms of the coefficients a, b and c. It works for every quadratic equation whose discriminant is not negative.
quadratic formula — The solutions of a x squared plus b x plus c equals nought are x equals the opposite of b, plus or minus the square root of b squared minus four a c, all divided by two a, provided a is not nought and b squared minus four a c is at least nought.
Lesson 12.5 shows where the formula comes from.
Figure (svg): The quadratic formula with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-533
Section
Section 1
Concept
Read a, b and c off the equation in standard form, substitute them into the formula, and simplify. The plus-or-minus sign produces the two solutions.
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Read aloud, it is the opposite of b, plus or minus the square root of b squared minus four a c, all over two a.
Figure (svg): The quadratic formula with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-533 — the Quadratic Formula box, its Reading Algebra note, and Example 1
Picture it
Opposite of b, the radical, the sign, the divisor.
Figure (svg): The quadratic formula with each part labelled
Saying the formula aloud in words is the surest way to memorise it, because the phrase all over two a keeps the division from being applied to the radical alone.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } x^2 - 9x + 14 = 0. \]
Identify the coefficients
Why: Each with its own sign.
\[ a = 1, \; b = -9, \; c = 14 \]
Substitute
Why: The opposite of negative nine is nine.
\[ x = \dfrac{9 \pm \sqrt{81 - 56}}{2} \]
Simplify the radical
Why: Eighty-one less fifty-six.
\[ x = \dfrac{9 \pm 5}{2} \]
Take both branches
Why: Fourteen over two and four over two.
\[ x = 7 \text{ or } x = 2 \]
Figure (svg): Substituting the coefficients into the formula
\[ x = 7 \text{ and } x = 2 \]
Verify: substitute both back
Why: Seven gives forty-nine minus sixty-three plus fourteen, which is nought, and two gives four minus eighteen plus fourteen, also nought. Both branches satisfy the original equation, which is what having two solutions means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-533
Faded example
b is negative here.
Fill in the blanks
a = 1, \; b = -9, \; c = 14: \quad x = \dfrac9 \pm \sqrt56}}}___
Why: The numerator's leading term is the opposite of b, so a negative b makes it positive. The four a c term keeps its own sign separately, and here everything is positive so it is subtracted as written.
Worked example
Guided Practice 1 to 3, each in standard form already.
\[ \text{Solve } x^2 - 4x + 3 = 0, \; 2x^2 + x - 10 = 0 \text{ and } x^2 + 3x - 4 = 0. \]
Do the first
Why: Sixteen less twelve is four.
\[ x = \tfrac{4 \pm 2}{2} = 3, \; 1 \]
Do the second
Why: One plus eighty is eighty-one.
\[ x = \tfrac{-1 \pm 9}{4} = 2, \; -\tfrac{5}{2} \]
Do the third
Why: Nine plus sixteen is twenty-five.
\[ x = \tfrac{-3 \pm 5}{2} = 1, \; -4 \]
Note the pattern
Why: All three discriminants were perfect squares.
Figure (svg): The solution to Worked example three more with the formula shown as a ladder of expressions, one row per algebraic move
\[ 3, 1; \quad 2, -\tfrac{5}{2}; \quad 1, -4 \]
Verify: check one of the fractional answers
Why: For the second equation, negative five halves gives two times twenty-five quarters, which is twelve and a half, minus five halves, minus ten — and that is nought. Fractional answers are as legitimate as whole ones and are worth checking precisely because they look less familiar.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-533
Trap
\[ x^2 - 9x + 14 = 0 \;\Longrightarrow\; x = \dfrac{-9 \pm \sqrt{81 - 56}}{2} \]
Put the nine from the middle term into the numerator
Why: The visible number was nine, so nine went in.
The coefficient is negative nine, and the formula asks for its opposite, which is positive nine. This version gives negative two and negative seven, both of which fail the check.
\[ b = -9 \;\Longrightarrow\; -b = 9 \;\Longrightarrow\; x = \dfrac{9 \pm 5}{2} \]
Write b with its sign on its own line, then negate it as a separate step
Why: Two small steps beat one done in your head.
The b squared term is unaffected either way, which is exactly why the error survives to the end.
Matching
Read them with their signs.
Match the pairs
Why: In the second, the x term has no visible coefficient, which means b is one rather than nought. A missing number in front of a variable always means one, and a missing term entirely means nought.
Elimination
For x squared minus four x plus three equals nought.
Eliminate the wrong options
Which numerator is correct?
Survives elimination: A
Why: The opposite of negative four is four, b squared is sixteen whatever the sign of b, and four times one times three is twelve to be subtracted. Each wrong option changes exactly one of those three, and each gives an answer that looks plausible.
Socratic
Lesson 9.1 introduced the symbol.
Discussion prompt
Explain why the formula must produce two values rather than one. Then say what the two solutions have in common no matter what a, b and c are.
Hint: Think about the graph's symmetry.
Answer:
A parabola is symmetric about its axis, so if it crosses the axis once on one side it crosses again at the mirror point on the other. The formula reflects that by taking the axis position and stepping the same distance either way, which is precisely what the plus-or-minus does.
Whatever the coefficients, the two solutions average to negative b over two a, because the radical cancels when they are added — and that average is exactly the axis of symmetry from Lesson 9.4. So the formula is the vertex position plus and minus a distance, and the radical measures how far the roots sit from the turning point.
Section
Section 2
Concept
The formula reads coefficients from standard form only, so the equation must be rearranged before a, b and c are identified. When the discriminant is not a perfect square, the solutions are irrational and are rounded if the question asks.
A perfect-square discriminant gives rational solutions.
Figure (svg): An equation rearranged into standard form before the formula is used
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534 — Example 2, Write in Standard Form
Picture it
Zero on one side.
Figure (svg): An equation rearranged into standard form before the formula is used
Reading c off the unrearranged equation would give eight rather than negative eight, and the whole discriminant would change. The step is small and the consequence is not.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } 2x^2 - 3x = 8, \text{ rounding to the nearest hundredth.} \]
Write in standard form
Why: Subtract eight from each side.
\[ 2 x ^{2} - 3 x - 8 = 0 \]
Identify the coefficients
Why: c is negative.
\[ a = 2, \; b = -3, \; c = -8 \]
Substitute
Why: Nine minus negative sixty-four.
\[ x = \dfrac{3 \pm \sqrt{9 + 64}}{4} \]
Evaluate
Why: The root of seventy-three is about 8.54.
\[ x \approx 2.89, \; -1.39 \]
Figure (svg): An equation rearranged into standard form before the formula is used
\[ x = \dfrac{3 \pm \sqrt{73}}{4} \approx 2.89 \text{ and } -1.39 \]
Verify: substitute the positive root back
Why: Two times 2.89 squared is about 16.70, minus three times 2.89 which is 8.67, leaving about 8.03 — close to eight, with the small gap explained by the rounding. An exact check would use the radical form and come out exactly eight.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534
Faded example
The sign of c changes.
Fill in the blanks
2x^2 - 3x = 8 \;\to\; 2x^2 - 3x - 8 = 0 \;\to\; c = -8, \quad b^2 - 4ac = 9 + 64
Why: A negative c always makes the discriminant larger, because subtracting four a c adds when c is negative. That is why equations with a negative constant term so often have two real solutions.
Worked example
Guided Practice 4 and 5.
\[ \text{Solve } x^2 + x = 1 \text{ and } x^2 = 2x + 3. \]
Rearrange the first
Why: Subtract one.
\[ x ^{2} + x - 1 = 0 \]
Apply the formula
Why: One plus four is five.
\[ x = \tfrac{-1 \pm \sqrt{5}}{2} \]
Rearrange the second
Why: Subtract two x and three.
\[ x ^{2} - 2 x - 3 = 0 \]
Apply the formula
Why: Four plus twelve is sixteen.
\[ x = 3, \; - 1 \]
Figure (svg): The solution to Worked example two more rearrangements shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{-1 \pm \sqrt{5}}{2} \approx 0.62, \; -1.62; \qquad x = 3, \; -1 \]
Verify: compare the two discriminants
Why: Five is not a perfect square, so the first pair is irrational and needs rounding; sixteen is, so the second pair is whole. Whether an answer is tidy is decided entirely by the discriminant, and it can be checked before the division is done.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534
Error analysis
The student was solving two x squared minus three x equals eight.
Annotate
On: \( \begin{aligned} a = 2, \; b = -3, \; c &= 8 \\ x &= \frac{3 \pm \sqrt{9 - 64}}{4} \\ &= \frac{3 \pm \sqrt{-55}}{4} \\ \text{so there is no } &\text{real solution} \end{aligned} \)
This error is particularly worth knowing because its output is a confident statement that no solution exists, which stops the work rather than producing a wrong number to check. A graph would have shown two crossings immediately.
Sorting
Look at the discriminant only.
Sort into buckets
Sort each equation by whether its solutions are rational.
Computing the discriminant first tells you what kind of answer to expect before any dividing is done. That preview is useful enough that Lesson 9.7 is devoted entirely to it.
Prediction
To the discriminant of any quadratic with positive a.
Predict first
If c is negative, what happens to b squared minus four a c?
Correct: It grows, because subtracting a negative adds.
\[ c = -8: \; b^2 - 4ac = 9 - 4(2)(-8) = 9 + 64 = 73 \]
Why: With a positive and c negative, four a c is negative, and subtracting a negative quantity increases the total. Since b squared is already at least nought, the discriminant is then strictly positive and two real solutions are guaranteed. Graphically this makes sense: a negative c puts the y-intercept below the axis, so an upward parabola must cross the axis on both sides of it.
Socratic
The formula often produces radicals.
Discussion prompt
Say when to give an exact radical and when to round. Then say what is lost by rounding partway through the formula rather than at the end.
Hint: Read what the question asks for.
Answer:
Give the exact radical whenever the question says solve, simplify or express exactly, and round only when it names a precision — to the nearest hundredth, say — or asks a practical question with units. Example 2 asks for rounding explicitly, which is why its answer is a decimal rather than three plus or minus the root of seventy-three over four.
Rounding partway means the division by two a is applied to an already-approximate numerator, so the error is carried through and often magnified. Worse, an approximate answer cannot be checked exactly: substituting 2.89 gives about 8.03 rather than eight, which leaves you unable to tell a rounding gap from a genuine mistake.
Section
Section 3
Concept
The x-intercepts of a quadratic function's graph occur where y is nought. Substituting nought for y gives a quadratic equation, and the formula solves it exactly.
The intercepts are also called the roots.
Figure (svg): The formula's answers appearing as x-intercepts on a graph
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534 — Example 3, Find the x-Intercepts of a Graph, and its Study Tip
Picture it
The formula names them.
Figure (svg): The formula's answers appearing as x-intercepts on a graph
Lesson 9.5 read these positions off a sketch and called them estimates. The formula makes them exact, and the sketch becomes the check rather than the method.
Worked example
This is Example 3 from the textbook.
\[ \text{Find the x-intercepts of } y = x^2 + 4x - 5. \]
Set y to nought
Why: Intercepts occur at height nought.
\[ 0 = x ^{2} + 4 x - 5 \]
Identify the coefficients
Why: c is negative.
\[ a = 1, \; b = 4, \; c = -5 \]
Substitute
Why: Sixteen plus twenty.
\[ x = \dfrac{-4 \pm \sqrt{36}}{2} \]
Take both branches
Why: Two over two and negative ten over two.
\[ x = 1, \; - 5 \]
Figure (svg): The formula's answers appearing as x-intercepts on a graph
\[ x = 1 \text{ and } x = -5 \]
Verify: check against the axis of symmetry
Why: The two roots average to negative two, and negative b over two a is negative four over two, which is also negative two. The roots are always symmetric about the axis, so their average must equal it — a check that costs one addition.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534
Faded example
Intercepts are where the height is nought.
Fill in the blanks
y = x^2 + 4x - 5 \;\to\; 0 = x^2 + 4x - 5 \;\to\; x = \dfrac-5}___ = 1, \; ___
Why: Substituting nought for y turns a function into an equation, which is what the formula solves. The two branches give the two crossings, one on each side of the axis of symmetry.
Worked example
Turning the answers back into a picture.
\[ \text{Sketch } y = x^2 + 4x - 5 \text{ using its roots and vertex.} \]
Plot the roots
Why: The two crossings.
\[ (-5, 0), \; (1, 0) \]
Find the axis
Why: The midpoint of the roots.
\[ x = -2 \]
Find the vertex
Why: Substitute negative two.
\[ (-2, -9) \]
Add the y-intercept
Why: The constant term.
\[ (0, -5) \]
Figure (svg): The formula's answers appearing as x-intercepts on a graph
\[ \text{roots } -5, 1; \; \text{vertex } (-2, -9) \]
Verify: confirm the vertex is below the axis
Why: The vertex height is negative nine, well below the x-axis, which is exactly the condition for two real roots on an upward parabola. The picture and the algebra agree, and either one predicts the other.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 534-534
Trap
The x-intercepts are (1, 0) and (-5, 0), so the solutions are (1, 0) and (-5, 0).
Report the intercepts as coordinate pairs throughout
Why: The graph gave points, so points were reported.
The solutions of an equation are numbers, not points. An x-intercept can be given either way, but a solution of x squared plus four x minus five equals nought is one and negative five.
\[ \text{intercepts at } (1, 0) \text{ and } (-5, 0); \quad \text{solutions } x = 1, \; x = -5 \]
Match the form of the answer to the form of the question
Why: Points for intercepts, numbers for solutions.
The distinction is small on paper and worth marks on a test.
Translation
Numbers or points?
Match the pairs
Why: All four questions have the same underlying answer and differ only in the form they expect. Roots and solutions are numbers, crossings are points, and an intercept is conventionally given either way.
Hypothesis
The roots are 1 and -5.
Predict first
What is the average of the two roots equal to?
Correct: The x-coordinate of the vertex, -2.
\[ \dfrac{1 + (-5)}{2} = -2 = \dfrac{-b}{2a} \]
Why: The two roots are the axis position plus and minus the same distance, so averaging them cancels that distance and leaves the axis itself. That gives a free check on any pair of roots: their average must equal negative b over two a, which here is negative four over two. It also means that knowing one root and the axis is enough to find the other without any further formula work.
Socratic
Lesson 9.5 found intercepts too.
Discussion prompt
Say what the formula gives that a sketch cannot. Then say what the sketch still contributes once the formula is available.
Hint: Think about irrational roots.
Answer:
The formula gives exact values, including irrational ones that no sketch can resolve — three plus or minus the root of seventy-three over four is a precise answer that a graph could only place near 2.9 and near negative 1.4. It also works when the roots are close together or far outside a convenient window, where a sketch is unreadable.
The sketch still catches gross errors instantly: if the formula returns two positive roots but the curve plainly crosses on both sides of the y-axis, something is wrong. It also shows the behaviour between and beyond the roots, which the formula says nothing about, and that is what most word problems actually ask about.
Section
Section 4
Concept
A dropped object's height is negative sixteen t squared plus s. An object thrown up or down has an extra term v t, where v is the initial velocity in feet per second, positive upwards and negative downwards.
\[ h = -16t^2 + vt + s \]
Velocity carries a direction; speed is its size.
Figure (svg): The two vertical motion models
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 535-535 — the Vertical Motion Models box and its Study Tip on velocity
Picture it
One term is the whole difference.
Figure (svg): The two vertical motion models
Lesson 9.2's model is this one with v equal to nought, which is what dropping means. Nothing new is being introduced, only a case that was previously excluded.
Worked example
This is Example 4 from the textbook.
\[ \text{Thrown downwards at } 30 \text{ ft/s from } 200 \text{ feet, when does the marker land?} \]
Set up the model
Why: Downwards makes v negative.
\[ h = -16 t ^{2} - 30 t + 200 \]
Set the height to nought
Why: Landing means height nought.
\[ 0 = -16 t ^{2} - 30 t + 200 \]
Apply the formula
Why: Nine hundred plus twelve thousand eight hundred.
\[ t = \dfrac{30 \pm \sqrt{13\,700}}{-32} \]
Evaluate both branches
Why: Keep the positive time.
\[ t \approx 2.72 \]
Figure (svg): A marker thrown downwards from a balloon
\[ t \approx 2.72 \text{ seconds} \]
Verify: substitute the time back
Why: Sixteen times 2.72 squared is about 118.4, and thirty times 2.72 is 81.6, so the height is two hundred less 118.4 less 81.6, which is about nought. The marker is at ground level at that instant, confirming the root.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 535-535
Faded example
Thrown downwards from a balloon.
Fill in the blanks
\text-30: \quad h = -16t^2 + (200)t + ___
Why: The initial height is always positive because it is measured up from the ground, while the velocity carries the direction of the throw. Mixing those two conventions is what makes the sign of v easy to get wrong.
Worked example
Guided Practice 7, with a prediction to test.
\[ \text{Thrown downwards at } 60 \text{ ft/s from } 200 \text{ feet, when does it land?} \]
Set up the model
Why: Double the downward velocity.
\[ 0 = -16 t ^{2} - 60 t + 200 \]
Simplify
Why: Divide through by negative four.
\[ 4 t ^{2} + 15 t - 50 = 0 \]
Apply the formula
Why: Two hundred and twenty-five plus eight hundred.
\[ t = \dfrac{-15 \pm \sqrt{1025}}{8} \]
Evaluate
Why: Keep the positive branch.
\[ t \approx 2.13 \]
Figure (svg): A marker thrown downwards from a balloon
\[ t \approx 2.13 \text{ seconds} \]
Verify: compare with the prediction
Why: Doubling the throwing speed cut the time from about 2.72 seconds to about 2.13, a reduction of only about a fifth rather than a half. Gravity is still accelerating the marker throughout, so the initial velocity is only part of what determines the fall time.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 535-535
Trap
\[ h = -16t^2 + 30t + 200 \]
Substitute the thirty feet per second as given
Why: The problem said thirty, so thirty went in.
The marker was thrown downwards, and down is negative, so v is negative thirty. This version models a marker thrown upwards, which stays in the air longer and lands at about 4.60 seconds rather than 2.72.
\[ h = -16t^2 - 30t + 200 \]
Read the direction of the throw and sign v accordingly
Why: Up positive, down negative.
The two wrong and right answers are the same pair of numbers with their signs swapped, which is worth noticing.
Sorting
Up is positive, down is negative.
Sort into buckets
Sort each description by the sign of its initial velocity.
The two zero cases are exactly Lesson 9.2's dropped-object model, which is this model with the middle term gone. Recognising them as the same formula saves memorising two.
Elimination
The formula gives 2.72 and -4.60 for a landing time.
Eliminate the wrong options
Which should be reported?
Survives elimination: A
Why: The equation is satisfied by both values, and the situation admits only the positive one. Saying explicitly why the negative root was discarded is part of a complete answer rather than an optional remark.
Socratic
Sixty feet per second instead of thirty.
Discussion prompt
Explain why throwing the marker twice as fast does not halve its fall time. Then say what would happen if the balloon were much higher.
Hint: Ask what else is speeding the marker up.
Answer:
Gravity accelerates the marker throughout the fall, adding speed that has nothing to do with the throw. The initial velocity only contributes a head start, and by the time the marker is near the ground it is moving far faster than either throwing speed — so doubling the head start changes the total time by much less than half.
From a much greater height the effect would be smaller still, because the fall would be dominated by the accelerating part and the initial push would matter proportionally less. In the limit, two markers thrown at different speeds from a very great height would land at almost the same moment, which is why the squared term rather than the linear term controls a long fall.
Section
Section 5
Concept
Nearly every wrong answer from the formula comes from a sign or a grouping, not from the arithmetic. Four checks catch almost all of them.
The last one is invisible on paper and fatal on a calculator.
Figure (svg): The four sign checks to run before evaluating the formula
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-535 — the sign handling throughout Examples 1 to 4
Picture it
Four checks, all quick.
Figure (svg): The four sign checks to run before evaluating the formula
Each of these produces a plausible answer rather than an obviously silly one, which is exactly what makes them dangerous and why substitution at the end is worth the time.
Worked example
An equation with every sign working against you.
\[ \text{Solve } -x^2 - 5x + 6 = 0 \text{ with the formula.} \]
Read the coefficients
Why: All three signs matter.
\[ a = -1, \; b = -5, \; c = 6 \]
Compute the numerator's first term
Why: The opposite of negative five.
\[ 5 \]
Compute the discriminant
Why: Twenty-five minus four times negative one times six.
\[ 25 + 24 = 49 \]
Divide everything by two a
Why: Two times negative one.
\[ x = \dfrac{5 \pm 7}{-2} \]
Figure (svg): The four sign checks to run before evaluating the formula
\[ x = -6 \text{ and } x = 1 \]
Verify: substitute both back
Why: At one, negative one minus five plus six is nought, and at negative six, negative thirty-six plus thirty plus six is also nought. Both work, and note that the negative denominator swapped which branch gave which root — the plus branch produced the more negative answer.
Elimination
Grouping matters.
Eliminate the wrong options
Which is the quadratic formula?
Survives elimination: A
Why: The fraction bar covers the whole numerator and the radical covers the whole discriminant. Options B and C are grouping errors and D is a sign error, and all three produce numbers rather than nonsense, which is why they survive to the end of a problem.
Worked example
The grouping error, shown in full.
\[ \text{Why does } 9 + \sqrt{25} \div 2 \text{ not give a solution of } x^2 - 9x + 14 = 0? \]
Read what the calculator does
Why: Division before addition.
\[ 9 + (5 \div 2) \]
Evaluate
Why: Nine plus two and a half.
\[ 11.5 \]
Compare with the true root
Why: The correct branch is seven.
Fix the grouping
Why: Bracket the whole numerator.
\[ (9 + 5) \div 2 = 7 \]
Figure (svg): The four sign checks to run before evaluating the formula
\[ \dfrac{9 + 5}{2} = 7 \]
Verify: check the fixed version
Why: Fourteen divided by two is seven, which does satisfy the equation. The fraction bar in the written formula groups the entire numerator, and brackets have to do that job when the expression is typed on one line.
Trap
\[ x = -b \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \]
Write the formula with the division under the radical only
Why: The fraction bar was remembered as belonging to the root.
For x squared minus nine x plus fourteen this gives nine plus or minus two and a half, which is 11.5 and 6.5 — neither of which solves the equation. The opposite of b has to be divided as well.
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Say the formula aloud, ending with all over two a
Why: The words carry the grouping that the layout shows.
This is why the Reading Algebra note spells the formula out in words in the first place.
Faded example
Two a is negative here.
Fill in the blanks
a = -1, \; b = -5, \; c = 6: \quad x = \dfrac-21} \;\Longrightarrow\; x = -6 \text___ ___
Why: A negative denominator swaps which branch gives the larger root, so the plus branch produced negative six. That is harmless as long as both branches are computed, and confusing only if you expect the plus branch to be the bigger answer.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Part of the formula | Does the sign of b matter? | Does the sign of c matter? |
|---|---|---|
| the opposite of b | yes, it is negated | no, c does not appear |
| b squared | no, squaring removes it | no, c does not appear |
| four a c | no, b does not appear | yes, a negative c makes the discriminant larger |
The sign of b matters in exactly one place and the sign of c in exactly one other. Knowing which is which turns four vague warnings into two specific checks.
Socratic
The working looked clean.
Discussion prompt
Explain why re-reading your own formula work is a weak check. Then describe a check that costs almost nothing and catches most errors.
Hint: Ask what a second reading is likely to reproduce.
Answer:
Re-reading tends to reproduce the same interpretation that produced the error, because the mistake usually lies in how the expression was read rather than in the arithmetic. A dropped sign on b looks correct on a second pass for exactly the reason it looked correct on the first.
Substituting each answer into the original equation is independent of how the formula was applied, so it catches sign errors, grouping errors and arithmetic slips alike. Cheaper still is averaging the two roots and comparing with negative b over two a: one addition and one division, and it detects any error that moved the roots asymmetrically.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Method | Works when | Answers are |
|---|---|---|
| Square roots (9.2) | there is no x term | exact |
| Graphing (9.5) | always | estimates |
| The formula (9.6) | always | exact |
The formula is the only row that is both general and exact, which is why it becomes the default. The other two remain quicker in the cases they cover.
Pattern
To solve any quadratic equation with the formula, these five moves cover it.
Doing step three separately means the radical is a single number before it enters the fraction, which removes most of the grouping errors at a stroke.
OpenStax Elementary Algebra 2e, §10.3 Solve Quadratic Equations Using the Quadratic Formula §10.3
Check
Mind the sign of b.
Check your understanding
Solve x squared minus 9x plus 14 = 0.
Answer: A
Why: With b equal to negative nine, the opposite of b is nine, and the discriminant is eighty-one minus fifty-six, which is twenty-five. That gives nine plus or minus five, over two.
Check
Standard form first.
Check your understanding
For 2x squared minus 3x = 8, what is c?
Answer: A
Why: Subtracting eight from both sides gives two x squared minus three x minus eight equals nought, so c is negative eight and the discriminant is nine plus sixty-four.
Check
Down is negative.
Check your understanding
An object is thrown downwards at 30 ft/s from 200 feet. What is the model?
Answer: A
Why: Downwards makes the initial velocity negative thirty, while the initial height of two hundred feet stays positive because it is measured up from the ground.
Real world
This is the baseball question from the lesson opener. A batter hits a ball upwards at 80 feet per second from a height of 3 feet.
Discussion prompt
Write the vertical motion model, find when the ball hits the ground to the nearest hundredth of a second, and say which root you discarded and why.
Hint: Upwards makes the velocity positive.
Answer:
\[ 0 = -16t^2 + 80t + 3 \;\Longrightarrow\; t = \dfrac{-80 \pm \sqrt{6400 + 192}}{-32} \]
The discriminant is six thousand five hundred and ninety-two, whose root is about 81.19, so the roots are about negative 0.04 and about 5.04 seconds. The ball lands about 5.04 seconds after being hit.
The negative root refers to an instant before the ball was struck, when the model does not apply, so it is discarded. It is very close to nought here because the ball started only three feet up — a hit from ground level would give a root of exactly nought, and the small negative value is the model saying the ball would have left the ground a fraction of a second earlier had it been travelling this path all along.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
For 2x squared minus 3x = 8, what is the discriminant?
Correct: 9 + 64, which is 73.
\[ b^2 - 4ac = (-3)^2 - 4(2)(-8) = 9 + 64 = 73 \]
Why: The equation must first be rearranged to two x squared minus three x minus eight equals nought, which makes c negative eight rather than eight. Then four a c is four times two times negative eight, or negative sixty-four, and subtracting a negative adds — giving nine plus sixty-four. Reading c as positive eight before rearranging gives nine minus sixty-four and the false conclusion that there are no real solutions, which is the single most common way this problem is failed. The b squared term is nine either way, since squaring removes the sign of b.
Explain it
They typed the formula into a calculator as negative b plus the root, divided by two a, and got the wrong answer.
Discussion prompt
In no more than four sentences, explain what the calculator did and how to fix the keystrokes. Then give them the words that carry the grouping.
Hint: Where does the fraction bar reach?
Answer:
A usable answer: the calculator divided only the radical by two a, because typing it in a line loses the fraction bar that covers the whole numerator. Bracket everything above the bar before dividing, so the opposite of b gets divided as well.
Say the formula as the opposite of b, plus or minus the square root of b squared minus four a c, all divided by two a. The words all divided by are what tell you where the brackets go.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Signs are fixed by writing the three coefficients on their own line before touching the formula. Rearranging is fixed by making the nought the first thing you write. Grouping is fixed by computing the discriminant as a single number first, then bracketing the numerator. Motion models are fixed by asking whether the object was thrown up, thrown down or released, which sets the sign of v. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the quadratic formula large, and underneath it write the same formula out in words, ending with the phrase all divided by two a. Label the opposite of b, the discriminant, the plus-or-minus and the divisor with arrows. Underneath, solve x squared minus nine x plus fourteen equals nought in full, writing a, b and c on their own line with signs before substituting anything. In the middle, take two x squared minus three x equals eight through the whole method, showing the rearrangement as its own line and marking clearly where c changes sign, then round both answers to two decimal places. Beneath that, write both vertical motion models side by side, label every letter with its meaning and units, and work the balloon problem from a downward throw of thirty feet per second, marking which root you discarded. In the lower corner, write the four sign checks as a numbered list. Finally, in the margin, write the check that the two roots average to negative b over two a.
Your two roots should always average to the axis of symmetry. If they do not, the error moved one root without moving the other, which points at the numerator rather than at the discriminant.
Recap
Five things, and the first one solves every quadratic equation there is.
| If the question says | Your first move is |
|---|---|
| Solve a quadratic equation | Rearrange to standard form, then use the formula |
| Find the x-intercepts | Substitute 0 for y, then use the formula |
| Round to the nearest hundredth | Keep exact values until the last line |
| An object is thrown downwards | Make v negative in the motion model |
| When does it hit the ground? | Set h to nought and keep the positive root |
Lesson 9.7 looks at just one part of the formula. The expression under the radical decides how many solutions there are before any of them is computed, which turns the counting you have been doing graphically into a single calculation.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.6 Solving Quadratic Equations by the Quadratic Formula §9.6, pp. 533-539 — everything on these slides traces back here
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