🎯⭐ INTERACTIVE LESSON

The Quadratic Formula

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The Quadratic Formula - Complete Interactive Lesson

Part 1: Standard Form & Identifying a, b, c

🧮 The Quadratic Formula

Part 1 of 5 — Standard Form & Identifying $a$ , $b$ , $c$

Topics in This Part

Section
What the Quadratic Formula Does
Standard Form: $ax^2 + bx + c = 0$
Reading Off $a$ , $b$ , and

🔑 Key Concept: The quadratic formula solves every quadratic equation. No matter how messy the numbers are — whole, fraction, or irrational — the formula always finds the solutions. But it only works once the equation is in standard form and you've correctly identified $a$ , $b$ , and $c$ . That setup is Part 1.

What the Quadratic Formula Does

A quadratic equation is any equation you can write as:

$ax^2 + bx + c = 0 \qquad (a \ne 0)$

Standard Form First

To use the formula you must have zero on one side:

$ax^2 + bx + c = 0$

If an equation isn't in this form, rearrange it first.

Given equation	Standard form	$a$

Concept Check 🎯

Identify the Coefficients 🧮

First rewrite each equation in standard form $ax^2+bx+c=0$ , then enter $a$ , $b$ , and .

Recap

To get ready for the formula:

Rearrange so one side is $0$ : $ax^2 + bx + c = 0$ .
Read off $a$ , , — keeping every sign.

Part 2: Plugging Into the Formula

🧮 The Quadratic Formula

Part 2 of 5 — Plugging Into the Formula

🔑 The Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Say it out loud: "" Memorizing the rhythm makes it stick.

Part 3: The Discriminant: How Many Solutions?

🧮 The Quadratic Formula

Part 3 of 5 — The Discriminant: How Many Solutions?

🔑 The discriminant is the part under the radical: $\Delta = b^2 - 4ac$ Its sign tells you how many real solutions a quadratic has — before you finish solving.

Reading the Discriminant

The discriminant $\Delta = b^2 - 4ac$ lives inside . What you can take the square root of decides everything:

Part 4: Irrational Roots & Word Problems

🧮 The Quadratic Formula

Part 4 of 5 — Irrational Roots & Word Problems

🔑 Big Picture: Not every quadratic factors with whole numbers. When the discriminant isn't a perfect square, the quadratic formula gives exact irrational answers using $\sqrt{\;}$ — and it's the tool that turns real-world setups into solutions.

When the Roots Are Irrational

If $\Delta$ is positive but a perfect square, simplify the radical and leave the answer exact.

Part 5: Mixed Practice & Mastery Check

🧮 The Quadratic Formula

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) set up standard form and read $a$ , $b$ , $c$ , (2) substitute into the formula, (3) use the discriminant to count solutions, and (4) handle irrational roots and word problems. Let's pull it all together.

Quick Reference

Goal	Key move
Set up	Write $ax^2+bx+c=0$ ; read

The solutions — the values of $x$ that make it true — are called the roots or zeros. The quadratic formula gives them directly:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Other methods (factoring, completing the square) only work nicely on some quadratics. The quadratic formula works on all of them — that's why it's worth memorizing cold.

💡 Where it comes from: The formula is just completing the square done once, in general, on $ax^2+bx+c=0$ . You don't have to re-derive it — but knowing it's not magic helps you trust it.

⚠️ Watch the signs. $a$ , $b$ , and $c$ carry their signs. In $x^2 - 4x - 12 = 0$ , $b = -4$ (not $4$ ) and $c = -12$ (not $12$ ). Getting a sign wrong here breaks everything downstream.

with their signs

For $x^2 = 6x - 8$ :

1) $a = \,?$ 2) $b = \,?$ 3) $c = \,?$

A missing term means that coefficient is

0

(e.g.

3x^2 - 7 = 0

has

b = 0

🔑 You now have $a$ , $b$ , $c$ ready to drop into the formula. In Part 2 we'll learn the formula itself and plug them in carefully.

negative $b$ , plus or minus the square root of $b$ squared minus $4ac$ , all over $2a$ .

The Substitution Routine

A clean process prevents sign errors:

Identify $a$ , $b$ , $c$ from standard form.
Substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , wrapping each value in parentheses.
Simplify under the radical first (that's the discriminant, $b^2 - 4ac$ ).
Finish the arithmetic, splitting into the $+$ case and the $-$ case.

Worked Example: $x^2 + 5x + 6 = 0$

Here $a = 1$ , $b = 5$ , $c = 6$ .

$x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(6)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2}$

$x = \frac{-5 \pm 1}{2} \;\Rightarrow\; x = \frac{-4}{2} = -2 \;\text{ or }\; x = \frac{-6}{2} = -3$

✅ Check: $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$ ✓

Worked Example: $2x^2 + 5x - 3 = 0$

Here $a = 2$ , $b = 5$ , $c = -3$ . Notice $c$ is negative, so $-4ac$ becomes positive:

$x = \frac{-(5) \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4}$

$x = \frac{-5 \pm 7}{4} \;\Rightarrow\; x = \frac{2}{4} = \frac{1}{2} \;\text{ or }\; x = \frac{-12}{4} = -3$

⚠️ The $-4ac$ trap: When $c$ is negative, $-4ac$ flips to a $+$ . Here $-4(2)(-3) = +24$ , so the discriminant to . Mishandling this double-negative is the #1 quadratic-formula mistake.

Build the Substitution 🔽

You're solving $x^2 - 3x - 10 = 0$ , so $a=1$ , $b=-3$ , $c=-10$ . Choose each piece as you substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Plug and Simplify 🧮

Solve $x^2 - 6x + 8 = 0$ with the quadratic formula ( $a=1$ , $b=-6$ , $c=8$ ).

1) Value under the radical, $b^2 - 4ac = \,?$ 2) The smaller solution, $x = \,?$ 3) The larger solution, $x = \,?$

Discriminant	Square root gives	Number of real solutions	Graph (parabola)
$\Delta > 0$	a real, nonzero number	two distinct real solutions	crosses $x$ -axis twice
$\Delta = 0$	$\sqrt{0} = 0$	one repeated real solution	touches $x$ -axis once
$\Delta < 0$	$\sqrt{\text{negative}}$ — not real	no real solutions	never touches -axis

💡 When $\Delta = 0$ , the $\pm$ adds and subtracts $0$ , so both branches collapse to the same value $x = \dfrac{-b}{2a}$ . That single answer is called a double root.

Three Quick Examples

$x^2 - 6x + 8 = 0$ : $\Delta = (-6)^2 - 4(1)(8) = 36 - 32 = 4 > 0$ → two solutions.

$x^2 + 6x + 9 = 0$ : $\Delta = 6^2 - 4(1)(9) = 36 - 36 = 0$ → solution ().

$x^2 - 4x + 5 = 0$ : $\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0$ → solutions.

⚠️ A negative discriminant does not mean "no answer" — it means no real answer. In Algebra 1 we report "no real solutions"; later courses use imaginary numbers for these.

Concept Check 🎯

Compute the Discriminant 🧮

Find $\Delta = b^2 - 4ac$ for each (enter the number, with sign).

1) $x^2 - 5x + 6 = 0$ , $\;\Delta = \,?$ 2) $4x^2 - 12x + 9 = 0$ , $\;\Delta = \,?$ 3) $x^2 + x + 4 = 0$ , $\;\Delta = \,?$

Match the Count 🔽

Use the sign of the discriminant to state how many real solutions each equation has.

Worked Example: $x^2 + 4x + 1 = 0$

Here $a=1$ , $b=4$ , $c=1$ :

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2}$

Simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$ :

$x = \frac{-4 \pm 2\sqrt{3}}{2} = \frac{2(-2 \pm \sqrt{3})}{2} = -2 \pm \sqrt{3}$

💡 Factor out, then cancel. Both terms on top share a factor of $2$ , which cancels the $2$ in the denominator. Always check whether the numerator and denominator share a common factor.

Concept Check 🎯

A Word Problem

A ball is thrown upward. Its height in feet after $t$ seconds is $h = -16t^2 + 32t + 5$ . When does it hit the ground ( $h = 0$ )?

Set $h = 0$ : $-16t^2 + 32t + 5 = 0$ , so $a = - 16$ , , .

$t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(5)}}{2(-16)} = \frac{-32 \pm \sqrt{1024 + 320}}{-32} = \frac{-32 \pm \sqrt{1344}}{-32}$

$\sqrt{1344} \approx 36.66$ , so:

$t \approx \frac{-32 + 36.66}{-32} \approx -0.15 \quad\text{or}\quad t \approx \frac{-32 - 36.66}{-32} \approx 2.15$

⚠️ Reject the negative time. A negative $t$ is meaningless here, so the ball lands at $t \approx 2.15$ seconds. Always sanity-check word-problem answers against what's physically possible.

Apply It 🧮

A rectangle's area is modeled by $x^2 + 3x - 28 = 0$ , where $x$ is a length (so $x > 0$ ).

1) Discriminant $b^2 - 4ac = \,?$ 2) $\sqrt{\Delta} = \,?$ The valid (positive) value of

⚠️ Top three mistakes to avoid: (1) forgetting the $\pm$ , (2) mishandling $-4ac$ when $c$ is negative, (3) dropping a sign on $b$ when computing $-b$ .

Solve Step by Step 🔽

Solve $3x^2 - 2x - 1 = 0$ ( $a=3$ , $b=-2$ , $c=-1$ ). Choose each piece.

Mixed Drill 🧮

1) Solve $x^2 - 2x - 15 = 0$ ; enter the larger root: $x = \,?$ 2) For $5x^2 + 2x + 1 = 0$ , compute $\Delta = b^2 - 4ac = \,?$ (its sign tells you the solution count) 3) Solve $x^2 - 9 = 0$ ; enter the positive root: $x = \,?$

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

−32−32±1024+320=

−32−32−36.66≈

$2x^2 + 5x - 3 = 0$	already standard	$2$	$5$	$-3$
$x^2 = 4x + 12$	$x^2 - 4x - 12 = 0$
$3x^2 - 7 = 0$	$3x^2 + 0x - 7 = 0$
$-x^2 + 6x = 9$	$-x^2 + 6x - 9 = 0$

The Quadratic Formula

The Quadratic Formula - Complete Interactive Lesson

Part 1: Standard Form & Identifying a, b, c

🧮 The Quadratic Formula

Topics in This Part

What the Quadratic Formula Does

Standard Form First

Recap

Part 2: Plugging Into the Formula

🧮 The Quadratic Formula

Part 3: The Discriminant: How Many Solutions?

🧮 The Quadratic Formula

Reading the Discriminant

Part 4: Irrational Roots & Word Problems

🧮 The Quadratic Formula

When the Roots Are Irrational

Part 5: Mixed Practice & Mastery Check

🧮 The Quadratic Formula

Quick Reference

The Substitution Routine

Worked Example: x2+5x+6=0x^2 + 5x + 6 = 0x2+5x+6=0

Worked Example: 2x2+5x−3=02x^2 + 5x - 3 = 02x2+5x−3=0

Three Quick Examples

Worked Example: x2+4x+1=0x^2 + 4x + 1 = 0x2+4x+1=0

A Word Problem

Worked Example: $x^2 + 5x + 6 = 0$

Worked Example: $2x^2 + 5x - 3 = 0$

Worked Example: $x^2 + 4x + 1 = 0$