The Quadratic Formula

🧮 The Quadratic Formula

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

Topics in This Part

🔑 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:

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

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.

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.

⚠️ 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.

Concept Check 🎯

Identify the Coefficients 🧮

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

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

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

Recap

To get ready for the formula:

1. Rearrange so one side is $0$: $ax^2 + bx + c = 0$.
2. Read off $a$, $b$, $c$ — keeping every sign.
3. 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.

🧮 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: "negative $b$, plus or minus the square root of $b$ squared minus $4ac$, all over $2a$." Memorizing the rhythm makes it stick.

The Substitution Routine

A clean process prevents sign errors:

1. Identify $a$, $b$, $c$ from standard form.
2. Substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, wrapping each value in parentheses.
3. Simplify under the radical first (that's the discriminant, $b^2 - 4ac$).
4. 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 grows to $49$. 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, $x = \,?$

🧮 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 $\sqrt{\;\;}$. What you can take the square root of decides everything:

💡 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.

$x^2 + 6x + 9 = 0$: $\Delta = 6^2 - 4(1)(9) = 36 - 36 = 0$ → one solution ($x = -3$).

$x^2 - 4x + 5 = 0$: $\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0$ → no real 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.

🧮 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 not a perfect square, simplify the radical and leave the answer exact.

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 $a = -16$, $b = 32$, $c = 5$.

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

$\sqrt{1344} \approx 36.66$, 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) $\sqrt{\Delta} = \,?$ 3) The valid (positive) value of $x = \,?$

🧮 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

⚠️ 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.