Solving Quadratic Equations

🎯 Solving Quadratic Equations

Part 1 of 5 — What Is a Quadratic Equation?

Topics in This Part

🔑 Key Concept: A quadratic equation is any equation that can be written as $ax^2 + bx + c = 0$ with $a \ne 0$. "Solving" it means finding every value of $x$ that makes the equation true — these values are the roots (or solutions).

Standard Form

Every quadratic equation can be rearranged into standard form:

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

• $a$ is the leading coefficient (the number on $x^2$).
• $b$ is the linear coefficient (the number on $x$).
• $c$ is the constant term.

The requirement $a \ne 0$ is what makes it quadratic — if $a = 0$ there's no $x^2$ term and the equation is just linear.

Identifying $a$, $b$, and $c$

⚠️ Get to standard form first. $x^2 + 5x = -6$ is the same equation as $x^2 + 5x + 6 = 0$ — move every term to one side so the other side is $0$ before reading off $a$, $b$, $c$.

Concept Check 🎯

Why Two Solutions?

A linear equation like $2x = 6$ has exactly one solution ($x = 3$). A quadratic can have up to two.

Think about $x^2 = 9$. Both $x = 3$ and $x = -3$ work, because $3^2 = 9$ and $(-3)^2 = 9$. That $\pm$ behavior is the heart of every quadratic.

A quadratic equation has:

• two real solutions (the usual case),
• one repeated solution (a "double root"), or
• no real solutions.

💡 Graphically, the solutions are the $x$-values where the parabola $y = ax^2+bx+c$ crosses the $x$-axis. A parabola can cross twice, touch once, or miss entirely — matching the three cases above.

The Zero Product Property

This single idea powers the most common solving method.

🔑 Zero Product Property: If $A \cdot B = 0$, then $A = 0$ or $B = 0$ (or both).

The only way a product equals zero is if one of the factors is zero. So if we can write a quadratic as a product of factors that equals 0, we just set each factor to $0$ and solve.

Example

$(x - 2)(x + 5) = 0$

Set each factor to zero: $x - 2 = 0 \quad\Rightarrow\quad x = 2$ $x + 5 = 0 \quad\Rightarrow\quad x = -5$

The solutions are $x = 2$ and $x = -5$.

⚠️ The Zero Product Property only works when one side is exactly $0$. $(x-2)(x+5) = 8$ does not mean $x-2=8$ or $x+5=8$. Get a $0$ first.

Concept Check 🎯

Read Off the Roots 🧮

Each quadratic is already factored. Enter the two solutions, smaller value first.

1) $(x - 4)(x - 1) = 0 \Rightarrow x = \,?$ and $x = \,?$ 2) $x(x + 6) = 0 \Rightarrow x = \,?$ and $x = \,?$

🎯 Solving Quadratic Equations

Part 2 of 5 — Solving by Factoring

🔑 The Plan: (1) write the equation in standard form $= 0$, (2) factor the quadratic into two binomials, (3) apply the Zero Product Property. This is the fastest method when the quadratic factors nicely.

Factoring $x^2 + bx + c$

To factor $x^2 + bx + c$ (when $a = 1$), find two numbers that:

• multiply to $c$, and
• add to $b$.

Those two numbers go inside the binomials.

Example: $x^2 + 7x + 12$

We need two numbers that multiply to $12$ and add to $7$. Those are $3$ and $4$:

$x^2 + 7x + 12 = (x + 3)(x + 4)$

Sign guide

💡 Check by expanding (FOIL): $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ ✓

Find the Factor Pair 🔽

For each quadratic, choose the pair of numbers that multiply to $c$ and add to $b$.

Worked Example: Solve $x^2 + 7x + 12 = 0$

Step 1 — Standard form. Already done: $x^2 + 7x + 12 = 0$.

Step 2 — Factor. Two numbers multiplying to $12$, adding to $7$: that's $3$ and $4$. $(x + 3)(x + 4) = 0$

Step 3 — Zero Product Property. $x + 3 = 0 \Rightarrow x = -3 \qquad x + 4 = 0 \Rightarrow x = -4$

Solutions: $x = -3$ and $x = -4$.

✅ Check $x = -3$: $(-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0$ ✓

Worked Example: Solve $x^2 - 5x = 14$

Step 1 — Standard form. Subtract $14$ from both sides so one side is $0$: $x^2 - 5x - 14 = 0$

Step 2 — Factor. Product $-14$, sum $-5$: that's $-7$ and $+2$. $(x - 7)(x + 2) = 0$

Step 3 — Zero Product Property. $x - 7 = 0 \Rightarrow x = 7 \qquad x + 2 = 0 \Rightarrow x = -2$

⚠️ Don't skip Step 1. If you'd tried to factor "$x^2 - 5x = 14$" directly, the Zero Product Property would not apply — the right side must be $0$.

Concept Check 🎯

Solve by Factoring 🧮

Solve each by factoring. Enter the two solutions, smaller value first.

1) $x^2 + 6x + 8 = 0 \Rightarrow x = \,?$ and $x = \,?$ 2) $x^2 - 3x - 10 = 0 \Rightarrow x = \,?$ and $x = \,?$

🎯 Solving Quadratic Equations

Part 3 of 5 — The Square Root Method

🔑 When to use it: If the equation has no middle ($x$) term — it looks like $ax^2 + c = 0$ or $(x - h)^2 = k$ — you can isolate the square and take the square root of both sides. Fast and clean.

Taking the Square Root

To solve $x^2 = k$ (with $k \ge 0$):

$x^2 = k \quad\Longrightarrow\quad x = \pm\sqrt{k}$

⚠️ The $\pm$ is mandatory. Every positive number has two square roots — one positive, one negative. Dropping the $\pm$ loses a solution.

Example: $x^2 = 49$

$x = \pm\sqrt{49} = \pm 7 \quad\Rightarrow\quad x = 7 \text{ or } x = -7$

Example: $3x^2 - 75 = 0$

Isolate $x^2$ first: $3x^2 = 75 \;\Rightarrow\; x^2 = 25 \;\Rightarrow\; x = \pm 5$

When the Square Is a Binomial

The same move works when a whole binomial is squared. Solve $(x - 3)^2 = 16$:

$x - 3 = \pm\sqrt{16} = \pm 4$ $x = 3 \pm 4 \quad\Rightarrow\quad x = 7 \text{ or } x = -1$

Example: $(x + 2)^2 = 9$

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

💡 This is exactly the last step of completing the square — so mastering it here pays off everywhere.

Take the Root 🔽

Walk through solving $(x + 4)^2 = 25$ one step at a time.

A Warning Sign: Negative Under the Root

What about $x^2 = -4$? There's no real number whose square is negative, since any real number squared is $\ge 0$.

$x^2 = -4 \quad\Rightarrow\quad \text{no real solutions}$

⚠️ If isolating the square leaves a negative number on the other side, the equation has no real solutions. (In later courses you'll meet imaginary numbers that handle this — but in Algebra 1 we report "no real solution.")

Concept Check 🎯

Square Root Method 🧮

Solve each. Enter the two solutions, smaller value first.

1) $x^2 = 36 \Rightarrow x = \,?$ and $x = \,?$ 2) $2x^2 = 50 \Rightarrow x = \,?$ and $x = \,?$ 3) $(x - 1)^2 = 25 \Rightarrow x = \,?$ and $x = \,?$

🎯 Solving Quadratic Equations

Part 4 of 5 — The Quadratic Formula & Discriminant

🔑 The universal tool: The Quadratic Formula solves every quadratic equation — even the ones that won't factor. Memorize it; it never fails.

The Quadratic Formula

For any equation in standard form $ax^2 + bx + c = 0$ (with $a \ne 0$):

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

To use it:

1. Write the equation in standard form and identify $a$, $b$, $c$.
2. Substitute carefully — watch signs, especially on $-b$ and on negative values of $c$.
3. Simplify the part under the root, then compute both the $+$ and $-$ versions.

⚠️ The whole numerator is $-b \pm \sqrt{\,\cdots\,}$, and the entire thing is divided by $2a$. Keep the fraction bar under everything.

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

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

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

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

Solutions: $x = -2$ and $x = -3$.

✅ Same answer as factoring $(x+2)(x+3)=0$ — the formula always agrees, it just always works too.

Worked Example: Solve $2x^2 - 4x - 3 = 0$

This one does not factor with whole numbers, so the formula is essential. Here $a = 2$, $b = -4$, $c = -3$.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-3)}}{2(2)} = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm \sqrt{40}}{4}$

Since $\sqrt{40} \approx 6.32$:

$x \approx \frac{4 + 6.32}{4} \approx 2.58 \qquad x \approx \frac{4 - 6.32}{4} \approx -0.58$

💡 Notice $-4(2)(-3) = +24$. A negative $c$ makes that term positive — the single most common sign slip in the formula.

Build the Formula 🔽

You're solving $x^2 - 6x + 5 = 0$ with the Quadratic Formula. Fill in each piece. ($a=1$, $b=-6$, $c=5$.)

The Discriminant: $b^2 - 4ac$

The expression under the root, $\,b^2 - 4ac\,$, is the discriminant. Its sign tells you how many real solutions exist — before you finish solving.

Example

For $x^2 + 5x + 6 = 0$: $\;b^2 - 4ac = 25 - 24 = 1 > 0$ → two real solutions. ✓

Concept Check 🎯

Compute the Discriminant 🧮

Find $b^2 - 4ac$ for each equation (it's already in standard form).

1) $x^2 + 3x - 10 = 0 \Rightarrow b^2-4ac = \,?$ 2) $x^2 - 6x + 9 = 0 \Rightarrow b^2-4ac = \,?$ 3) $2x^2 + x + 5 = 0 \Rightarrow b^2-4ac = \,?$

🎯 Solving Quadratic Equations

Part 5 of 5 — Choosing a Method, Mixed Practice & Mastery Check

You now have three ways to solve a quadratic: factoring, the square root method, and the Quadratic Formula. The last skill is knowing which to reach for.

Which Method Should I Use?

🔑 Golden rule: Always get the equation into standard form $= 0$ first. Then look for the shortcut; if none is obvious, the Quadratic Formula is your guaranteed fallback.

💡 The discriminant $b^2-4ac$ is a great quick check: compute it first to learn how many real solutions to expect.

Pick the Smartest Method 🔽

Choose the most efficient first method for each equation.

Mixed Practice 🎯

Solve It — Any Method 🧮

1) $x^2 - 16 = 0$. Enter both roots, smaller first: $x = \,?$ and $x = \,?$ 2) $x^2 + 2x - 15 = 0$. Enter both roots, smaller first: $x = \,?$ and $x = \,?$ 3) What is the discriminant of $x^2 - 4x + 4 = 0$? $\,?$

Quick Reference

⚠️ Top three pitfalls: dropping the $\pm$, forgetting to get a $0$ before using the Zero Product Property, and sign errors on $-b$ or a negative $c$ in the formula.

Exit Quiz ✅

Answer all three to finish the lesson.