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Teacher Guide: The Quadratic Formula

Learn to solve any quadratic equation using the quadratic formula.

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Printable worksheet

All practice problems on paper, with a separate answer key.

Class quiz

10 questions on Quadratic Equations. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives
  • Apply the quadratic formula to solve equations of the form ax2+bx+c=0ax^2 + bx + c = 0
  • Correctly identify the coefficients aa, bb, and cc from any quadratic equation
  • Calculate and interpret the discriminant to predict the number of solutions
  • Simplify solutions including those with irrational numbers
Prerequisites
  • • Solving one-step and two-step equations
  • • Understanding square roots and simplifying radicals
  • • Order of operations (PEMDAS)
  • • Working with negative numbers
Discussion Starters
  • 1. Why do you think mathematicians developed the quadratic formula instead of just using trial and error?
  • 2. Can you think of situations where knowing exactly when something will happen (like a ball landing) is important?
  • 3. If an equation has no real solutions, what does that tell us about the real-world situation it models?
  • 4. How can you tell just by looking at a quadratic equation whether it will be easy or hard to solve?
Common Misconceptions

The quadratic formula only works for certain equations

Remediation: Emphasize that it works for ALL quadratic equations in standard form. Have students verify by trying both factoring and the formula on the same equation.

A negative discriminant means you made a calculation error

Remediation: Show examples where negative discriminants are correct (e.g., x2+1=0x^2 + 1 = 0). Relate to graphs that don't cross the x-axis.

Differentiation Ideas

For Struggling Students:

  • • Provide a formula template with blanks to fill in
  • • Start with equations where a=1a = 1 and bb, cc are small positive integers
  • • Use color-coding to match coefficients to formula positions

For On-Level Students:

  • • Practice with negative coefficients and non-unit leading coefficients
  • • Solve application problems requiring setting up the equation first
  • • Interpret discriminant values before solving

For Advanced Students:

  • • Derive the quadratic formula from completing the square
  • • Explore complex number solutions when discriminant is negative
  • • Analyze how changing coefficients affects the nature of solutions
Standards Alignment
  • A-REI.B.4 (CCSS.MATH.CONTENT.HSA.REI.B.4)

    Solve quadratic equations in one variable using the quadratic formula

  • A-REI.B.4b (CCSS.MATH.CONTENT.HSA.REI.B.4.B)

    Recognize when the quadratic formula gives complex solutions and write them as a±bia \pm bi

Lesson Resources
  • visualDiscriminant Explorer

    Interactive graph showing how discriminant affects parabola position

  • activityFormula Practice

    Step-by-step guided practice with immediate feedback

  • worksheetMixed Practice

    Problems ranging from simple to complex applications

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

The quadratic formula solves any equation of the form ax2+bx+c=0ax^2 + bx + c = 0:
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Where:
  • aa, bb, and cc are the coefficients from the equation
  • a≠0a \neq 0 (otherwise it's not quadratic)
  • The ±\pm symbol means there are usually two solutions
The expression under the square root, b2−4acb^2 - 4ac, is called the discriminant.

Worked Examples

Solve x2+6x+5=0x^2 + 6x + 5 = 0 using the quadratic formula.

1

Identify a, b, and c

Comparing to ax2+bx+c=0ax^2 + bx + c = 0: a=1a = 1, b=6b = 6, c=5c = 5

2

Calculate the discriminant

b2−4ac=62−4(1)(5)=36−20=16b^2 - 4ac = 6^2 - 4(1)(5) = 36 - 20 = 16 → D=16D = 16

3

Apply the formula

x=−6±162(1)=−6±42x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2} → x=−6±42x = \frac{-6 \pm 4}{2}

4

Find both solutions

x1=−6+42=−22=−1x_1 = \frac{-6 + 4}{2} = \frac{-2}{2} = -1 x2=−6−42=−102=−5x_2 = \frac{-6 - 4}{2} = \frac{-10}{2} = -5 → x=−1x = -1 or x=−5x = -5

Common Mistakes

Forgetting to make b negative when it's already negative

Why it's wrong: If b=−7b = -7, then −b=−(−7)=7-b = -(-7) = 7, not −7-7. The formula says −b-b, so you negate whatever b is.

Correct: Always write −b-b first, then substitute: if b=−7b = -7, write −(−7)=7-(-7) = 7

Calculating b2b^2 incorrectly when b is negative

Why it's wrong: Students sometimes write −72=−49-7^2 = -49 instead of (−7)2=49(-7)^2 = 49

Correct: The entire coefficient is squared: (−7)2=(−7)×(−7)=49(-7)^2 = (-7) \times (-7) = 49

Dividing only part of the numerator by 2a

Why it's wrong: Writing −b2a±b2−4ac\frac{-b}{2a} \pm \sqrt{b^2-4ac} instead of −b±b2−4ac2a\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Correct: The entire numerator (−b±...)(-b \pm \sqrt{...}) is divided by 2a2a

Stopping at one solution

Why it's wrong: The ±\pm means plus OR minus, giving two potential solutions

Correct: Always calculate both: x1=−b+D2ax_1 = \frac{-b + \sqrt{D}}{2a} and x2=−b−D2ax_2 = \frac{-b - \sqrt{D}}{2a}

Why It Matters

The quadratic formula is one of the most powerful tools in algebra because:
  • Universal: It works for ANY quadratic equation, even when factoring doesn't
  • Predictable: Follow the same steps every time
  • Real-world applications: Physics (projectile motion), engineering (parabolic structures), finance (profit optimization)
While factoring is faster when it works, many real-world problems have "messy" numbers that don't factor nicely. The quadratic formula handles them all!

Real World Applications

Projectile Motion

When you throw a ball, its height follows a quadratic equation. The formula helps find when it lands.

Example:

A ball thrown upward has height h=−5t2+20t+1h = -5t^2 + 20t + 1 meters after tt seconds. To find when it hits the ground (h=0h = 0), solve −5t2+20t+1=0-5t^2 + 20t + 1 = 0 using the formula.

1Try It Yourself

A rocket's height is modeled by h=−4t2+32th = -4t^2 + 32t meters.

When does the rocket return to the ground?

Step 1: Write the mathematical expression

Set h=0h = 0 and identify the coefficients:

Business Profit Optimization

Companies use quadratic equations to model profit and find break-even points.

Example:

A company's profit is P=−2x2+100x−800P = -2x^2 + 100x - 800 dollars, where xx is units sold. To find break-even points (where P=0P = 0), use the quadratic formula.

2Try It Yourself

A shop's weekly profit is P=−x2+50x−400P = -x^2 + 50x - 400 dollars for selling xx items.

At what sales levels does the shop break even?

Step 1: Write the mathematical expression

Solve −x2+50x−400=0-x^2 + 50x - 400 = 0:

Key Takeaways

  • 1The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • 2It works for any equation in the form ax2+bx+c=0ax^2 + bx + c = 0
  • 3The discriminant D=b2−4acD = b^2 - 4ac tells you how many solutions exist
  • 4If D>0D > 0: two distinct real solutions; if D=0D = 0: one repeated solution; if D<0D < 0: no real solutions
  • 5Always check your answers by substituting back into the original equation

Frequently Asked Questions

When should I use the quadratic formula instead of factoring?

Use the quadratic formula when: (1) the equation doesn't factor easily, (2) the coefficients are large or decimals, or (3) you want a reliable method that always works. Factoring is faster when it works, but the formula is more universal.

What does it mean when the discriminant is zero?

When b2−4ac=0b^2 - 4ac = 0, the equation has exactly one solution (called a repeated or double root). Graphically, the parabola just touches the x-axis at one point.

Why is there a plus-minus sign in the formula?

The ±\pm comes from taking the square root. Since both 52=255^2 = 25 and (−5)2=25(-5)^2 = 25, the square root of 25 could be either +5 or -5. This gives us two possible values for x.

Glossary

Quadratic equation
An equation of the form ax2+bx+c=0ax^2 + bx + c = 0 where a≠0a \neq 0
Discriminant
The expression b2−4acb^2 - 4ac that determines the number and type of solutions
Coefficient
The numerical factor of a term (e.g., in 3x23x^2, the coefficient is 3)
Root/Solution
A value of xx that makes the equation true (where the parabola crosses the x-axis)

Formula Card

Quadratic Formula

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Solves any equation $ax^2 + bx + c = 0$

Discriminant

D=b2−4acD = b^2 - 4ac

Determines number and type of solutions

Two solutions

D>0D > 0

Discriminant positive: two distinct real roots

One solution

D=0D = 0

Discriminant zero: one repeated root

No real solutions

D<0D < 0

Discriminant negative: no real roots

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