Quadratic Equations

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Quick Reference

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Try This First

Suppose a rectangle is $4$ units longer than it is wide, and its area is $60$ square units. Call the width $x$. Then the length is $x + 4$, and the area condition is

\[ x(x + 4) = 60. \]

Before any formula, try to find a width that works by testing whole numbers.

Key idea

The quadratic formula is not a memorization exercise. It is completing the square, done once in general.

Every quadratic equation $ax^2 + bx + c = 0$ can be solved by completing the square: rewrite the left side as a perfect square plus a constant, then isolate $x$. The process always works. It is just arithmetic and algebra.

The quadratic formula IS that process, carried out symbolically for arbitrary $a$, $b$, and $c$. Once someone has done that work (the derivation you will see in Section 4), the result is a single compact formula you can use forever without repeating the derivation. The formula is not separate knowledge; it is summarized understanding.

This means: if you understand completing the square, you understand the formula. You are not memorizing a mystery; you are using a shortcut whose origin you know.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• Factor a trinomial: $x^2 - 7x + 10 = (x-2)(x-5)$
• Apply the Zero Product Property: if $(x-2)(x-5) = 0$, then $x = 2$ or $x = 5$
• Compute a square root: $\sqrt{25} = 5$, $\sqrt{12} = 2\sqrt{3}$
• Identify perfect square trinomials: $x^2 + 6x + 9 = (x+3)^2$
• Solve a one-step equation involving a square: if $x^2 = 16$, then $x = \pm 4$

If any of these is shaky, review Factoring Techniques and Linear Equations first.

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Quick Reference

Standard form. A quadratic equation in one variable has the form \[ ax^2 + bx + c = 0 \quad \text{where } a \neq 0. \]

Three solution methods.

Quadratic formula. \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Discriminant. The expression $\Delta = b^2 - 4ac$ controls the number of real solutions:

• $\Delta > 0$: two distinct real solutions
• $\Delta = 0$: exactly one real solution (a repeated root)
• $\Delta < 0$: no real solutions (two complex solutions, which are outside the scope of this lesson)

Key Concepts

1. What Makes an Equation Quadratic?

Official definition (OpenStax College Algebra 2e, Section 2.5):

A quadratic equation is an equation containing a second-degree polynomial, written in the standard form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers and $a$ is not equal to $0$.

The condition $a \neq 0$ is what makes the equation quadratic. If $a$ were $0$, the $x^2$ term would vanish and the equation would be linear. After collecting all terms on one side, a quadratic can always be written in this standard form, and the coefficients $a$, $b$, $c$ are read directly off it.

The degree of the equation is the highest power of $x$ that appears after simplification. Degree 1 is linear (one solution). Degree 2 is quadratic (zero, one, or two real solutions).

2. Solving by Factoring (Zero Product Property)

If you can factor the quadratic, factoring is the fastest method.

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

This works because zero is the only number with the property that its product with anything is zero. If neither $A$ nor $B$ is zero, their product cannot be zero.

The method:

1. Move all terms to one side so the equation equals zero.
2. Factor the left side completely.
3. Set each factor equal to zero.
4. Solve each resulting linear equation.

Example 1. Solve $x^2 - 7x + 10 = 0$.

Predict first. The constant $10$ is positive and the middle coefficient $-7$ is negative, so predict two solutions that are both positive.

Factor: need $p + q = -7$, $pq = 10$. Try $-2$ and $-5$: correct. \[ (x - 2)(x - 5) = 0 \] Set each factor to zero: \[ x - 2 = 0 \implies x = 2 \qquad x - 5 = 0 \implies x = 5 \]

Check: $2^2 - 7(2) + 10 = 4 - 14 + 10 = 0$. $5^2 - 7(5) + 10 = 25 - 35 + 10 = 0$. Both correct.

Example 2. Solve $6x^2 + x - 2 = 0$.

This has leading coefficient $6 \neq 1$. Use the AC method: $ac = 6(-2) = -12$. Need $mn = -12$, $m + n = 1$. Try $4$ and $-3$: $4(-3) = -12$, $4 + (-3) = 1$. Rewrite: \[ 6x^2 + 4x - 3x - 2 = (2x)(3x + 2) - 1(3x + 2) = (3x + 2)(2x - 1) \] Set each factor to zero: \[ x = -\tfrac{2}{3} \qquad x = \tfrac{1}{2} \]

Check: $6\left(\frac{1}{4}\right) + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 0$. Correct.

Important: Move everything to one side first. Students sometimes factor $x^2 - 7x = -10$ as $x(x - 7) = -10$ and then write $x = -10$ or $x - 7 = -10$. This is wrong. The Zero Product Property requires the product to equal ZERO. Until both terms are on the same side, the property does not apply.

Example 3. Solve $x(x - 4) = 12$.

Wrong approach: write $x = 12$ or $x - 4 = 12$. The product is $12$, not $0$.

Correct approach: Expand, then move everything to one side. \[ x^2 - 4x = 12 \] \[ x^2 - 4x - 12 = 0 \] \[ (x - 6)(x + 2) = 0 \] \[ x = 6 \quad \text{or} \quad x = -2 \]

Check: $6(6-4) = 6 \cdot 2 = 12$. $(-2)(-2-4) = (-2)(-6) = 12$. Both correct.

3. Solving by Completing the Square

Completing the square works for any quadratic, even those that do not factor over the integers. It also reveals the geometric meaning of the quadratic: the vertex of the corresponding parabola.

The core idea. A perfect square trinomial $x^2 + 2kx + k^2$ equals $(x + k)^2$. If you add $k^2$ to the first two terms, you complete the square.

The method (when $a = 1$):

1. Write $x^2 + bx = -c$ (move the constant to the right).
2. Take half of the $x$-coefficient, square it: $\left(\dfrac{b}{2}\right)^2$.
3. Add that number to both sides.
4. The left side is now a perfect square. Factor it.
5. Solve by taking square roots.

Example 4. Solve $x^2 + 6x - 7 = 0$ by completing the square.

Step 1. Move constant: $x^2 + 6x = 7$.

Step 2. Half of 6 is 3; $3^2 = 9$.

Step 3. Add 9 to both sides: $x^2 + 6x + 9 = 16$.

Step 4. Factor the left side: $(x + 3)^2 = 16$.

Step 5. Take square roots: $x + 3 = \pm 4$.

\[ x = -3 + 4 = 1 \qquad \text{or} \qquad x = -3 - 4 = -7 \]

Check: $1^2 + 6(1) - 7 = 0$. $(-7)^2 + 6(-7) - 7 = 49 - 42 - 7 = 0$. Both correct.

Example 5. Solve $x^2 - 4x + 7 = 0$ by completing the square.

Step 1. $x^2 - 4x = -7$.

Step 2. Half of $-4$ is $-2$; $(-2)^2 = 4$.

Step 3. $x^2 - 4x + 4 = -3$.

Step 4. $(x - 2)^2 = -3$.

Step 5. $x - 2 = \pm\sqrt{-3}$.

Since $\sqrt{-3}$ is not a real number, this equation has no real solutions.

The discriminant appears here. Notice that we ended up with a negative number under the square root. Before you complete the square, you can predict this using $\Delta = b^2 - 4ac = 16 - 28 = -12 < 0$. Negative discriminant means no real solutions.

When $a \neq 1$: Divide every term by $a$ first, then complete the square.

Example 6. Solve $2x^2 + 12x + 10 = 0$.

Divide by 2: $x^2 + 6x + 5 = 0$.

Now factor: $(x+1)(x+5) = 0$, so $x = -1$ or $x = -5$.

(Completing the square would also work: move 5, halve 6 to get 3, add 9 to both sides, etc. Here factoring is faster once the leading coefficient is removed.)

4. The Quadratic Formula (Derived from Completing the Square)

Apply completing the square to the general equation $ax^2 + bx + c = 0$ with $a \neq 0$:

\[ ax^2 + bx + c = 0 \]

Divide by $a$:

\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Move the constant to the right:

\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]

Half of $\dfrac{b}{a}$ is $\dfrac{b}{2a}$; its square is $\dfrac{b^2}{4a^2}$. Add to both sides:

\[ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a} \]

Left side is a perfect square. Right side: combine using common denominator $4a^2$:

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \]

Take square roots:

\[ x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \]

Subtract $\dfrac{b}{2a}$:

\[ \boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \]

Key theorem: the Quadratic Formula (OpenStax College Algebra 2e, Section 2.5). The solutions to a quadratic equation $ax^2 + bx + c = 0$ (with $a \neq 0$) are given by $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

This formula is exact, complete, and works for every quadratic. The derivation above is worth reading once in full so the formula never feels arbitrary.

Using the formula.

Example 7. Solve $3x^2 - 2x - 8 = 0$.

Predict first. Compute the discriminant before solving: $\Delta = (-2)^2 - 4(3)(-8) = 4 + 96 = 100$, a perfect square. Predict two rational solutions (which means this equation could also have been factored).

Identify: $a = 3$, $b = -2$, $c = -8$.

Since $\Delta > 0$, there are two real solutions.

\[ x = \frac{-(-2) \pm \sqrt{100}}{2(3)} = \frac{2 \pm 10}{6} \]

\[ x = \frac{12}{6} = 2 \qquad \text{or} \qquad x = \frac{-8}{6} = -\frac{4}{3} \]

Check: $3(4) - 2(2) - 8 = 12 - 4 - 8 = 0$. $3\left(\frac{16}{9}\right) - 2\left(-\frac{4}{3}\right) - 8 = \frac{16}{3} + \frac{8}{3} - 8 = 8 - 8 = 0$. Both correct.

Example 8. Solve $x^2 - 6x + 9 = 0$.

$a = 1$, $b = -6$, $c = 9$.

$\Delta = 36 - 36 = 0$.

\[ x = \frac{6 \pm 0}{2} = 3 \]

There is exactly one solution: $x = 3$ (a repeated root). Indeed $x^2 - 6x + 9 = (x-3)^2$, which confirms this: the square is zero only when $x = 3$.

5. The Discriminant: Counting Solutions Before You Solve

The discriminant $\Delta = b^2 - 4ac$ is a number you can compute instantly from the coefficients. Its sign tells you everything about the real solutions.

Example 9. Without solving, determine how many real solutions $5x^2 - 3x + 4 = 0$ has.

$\Delta = (-3)^2 - 4(5)(4) = 9 - 80 = -71 < 0$.

No real solutions.

Example 10. Without solving, determine how many real solutions $4x^2 - 12x + 9 = 0$ has.

$\Delta = (-12)^2 - 4(4)(9) = 144 - 144 = 0$.

Exactly one (repeated) real solution.

Why is the discriminant useful? Computing $\Delta$ takes three arithmetic operations. If $\Delta < 0$, you can stop: the equation has no real solutions. This saves time on exams and prevents students from hunting for real solutions that do not exist.

6. Choosing the Right Method

Use this decision flow:

1. Factor out any common constant. Divide through to simplify.
2. Check for special forms:
   • $x^2 = k$ (no middle term): take square roots directly, $x = \pm\sqrt{k}$.
   • Perfect square: $(x + h)^2 = k$: take square roots.
3. Check whether the trinomial factors over the integers: try a small number of factor pairs. If yes, factor and apply the Zero Product Property.
4. If factoring is not obvious within 30 seconds: use the quadratic formula.

Example 11. Solve $4x^2 - 36 = 0$.

No middle term. Factor out 4: $x^2 - 9 = 0$. Difference of squares: $(x-3)(x+3) = 0$, so $x = \pm 3$.

Alternatively: $4x^2 = 36 \implies x^2 = 9 \implies x = \pm 3$. Same answer, less work.

Example 12. Solve $x^2 + 2x - 1 = 0$.

Try to factor: need $pq = -1$, $p+q = 2$. Integer pairs of $-1$: $(1,-1)$ with sum $0$. No integer factorization.

Use the formula: $a=1$, $b=2$, $c=-1$. $\Delta = 4 + 4 = 8$. \[ x = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

7. Setting Up Quadratic Word Problems

Many geometric and physical situations lead to quadratic equations.

A four-step word-problem process.

1. Identify the unknown. Name it with a variable.
2. Write the equation using the geometric or physical relationship given.
3. Rearrange into standard form $ax^2 + bx + c = 0$.
4. Solve, then answer the original question (include units; discard any solution that is not physically meaningful).

Example 13. A rectangular garden has perimeter 36 meters and area 80 square meters. Find its dimensions.

Let $w$ = width in meters. Since the perimeter is 36, the length is $\dfrac{36 - 2w}{2} = 18 - w$.

The area condition gives: \[ w(18 - w) = 80 \] \[ 18w - w^2 = 80 \] \[ w^2 - 18w + 80 = 0 \] \[ (w - 8)(w - 10) = 0 \] \[ w = 8 \quad \text{or} \quad w = 10 \]

Both are positive, so both are geometrically valid. If $w = 8$, then the length is $18 - 8 = 10$. If $w = 10$, then the length is $8$. These describe the same rectangle (width and length are just labels). The garden is 8 meters by 10 meters.

Common Misconceptions

Common Errors Summary

Leveled Practice

Level 1: Direct Application

Problem 1. Solve by factoring: $x^2 - 3x - 18 = 0$.

Problem 2. Solve: $4x^2 = 64$.

Problem 3. Use the discriminant to determine the number of real solutions of $2x^2 - 3x + 5 = 0$.

Level 2: Multiple Methods

Problem 4. Solve by completing the square: $x^2 + 10x + 3 = 0$.

Problem 5. Solve using the quadratic formula: $2x^2 + 5x - 12 = 0$.

Problem 6. Solve: $x^2 + 4x + 7 = 0$.

Level 3: Applications and Deeper Problems

Problem 7. A rectangular deck has area $60$ square feet. One side is 7 feet longer than the other. Find the dimensions.

Problem 8. For what values of $k$ does $x^2 + kx + 9 = 0$ have exactly one real solution?

Problem 9. Solve $\sqrt{x + 3} = x - 3$.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• Write any quadratic equation in standard form $ax^2 + bx + c = 0$
• Factor a quadratic and apply the Zero Product Property to find solutions
• Complete the square on a quadratic with leading coefficient 1
• Complete the square on a quadratic with leading coefficient $a \neq 1$ (by dividing first)
• Derive the quadratic formula by completing the square on $ax^2 + bx + c = 0$
• Apply the quadratic formula correctly, including the $\pm$ and the $-b$
• Compute the discriminant and state how many real solutions the equation has
• Check every solution by substituting back into the original equation
• Discard extraneous solutions when they arise

Mental Model

Think of a quadratic equation as a question about the height of a parabola.

The equation $ax^2 + bx + c = 0$ asks: “At what $x$-values does the parabola $y = ax^2 + bx + c$ touch the $x$-axis (where height is zero)?”

• Two real solutions: the parabola cuts through the $x$-axis at two points.
• One solution: the parabola is tangent to the $x$-axis at its vertex.
• No real solutions: the parabola floats entirely above (or below) the $x$-axis.

The discriminant $\Delta = b^2 - 4ac$ is the number inside the square root in the formula. If that number is negative, you would need to take the square root of a negative number, which has no real value. The parabola does not touch the axis. If it is zero, the $\pm$ adds or subtracts nothing; there is one crossing point. If it is positive, $\pm\sqrt{\Delta}$ gives two different values; two crossing points.

The three methods correspond to three different ways of locating those crossing points:

• Factoring finds the crossings by recognizing the parabola as a product of two lines.
• Completing the square relocates the parabola’s vertex to the origin, then reads off the crossings.
• The formula does completing the square once in general and hands you the crossings for any parabola.

Resources

Connections

Within Precalculus

• Quadratic functions and graphing: The solutions of $ax^2 + bx + c = 0$ are the $x$-intercepts of the parabola $y = ax^2 + bx + c$. Completing the square gives the vertex form $y = a(x-h)^2 + k$, which locates the vertex at $(h, k)$.
• Solving inequalities: Once you know where a quadratic is zero, you know where it is positive and where it is negative. The solution of $x^2 - 7x + 10 > 0$ uses the roots $x = 2$ and $x = 5$ as boundary points.
• Rational equations: Clearing denominators often produces a quadratic. After solving, check for extraneous solutions (values that make a denominator zero in the original).

Toward Calculus (MATH161)

• Finding zeros and critical points: Calculus involves setting derivatives equal to zero to find maxima, minima, and inflection points. Derivatives of polynomials are polynomials, often quadratic. Solving those equations uses exactly these methods.
• Limits involving $\frac{0}{0}$: After factoring a quadratic numerator, you can cancel the factor that causes the indeterminate form. For example:

\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x+3)(x-3)}{x-3} = \lim_{x \to 3} (x + 3) = 6. \]

• Quadratic approximation (MATH162, Taylor series): Any smooth function looks like a quadratic near a point where its second derivative is nonzero. The solutions of that local quadratic approximate the zeros of the original function.

Audience Notes

For students who find math intimidating: You only need to memorize two things in this lesson: the quadratic formula, and the fact that you must set the equation equal to zero before factoring. Everything else is a procedure you can re-derive with a few minutes of algebra. Write the formula on a card and use it; the derivation shows you why it works, so you can trust it.

For students interested in proof: The discriminant classifies quadratics over an arbitrary field. Over the real numbers, $\Delta < 0$ means no solution. Over the complex numbers, $\sqrt{\Delta}$ is always defined, so every quadratic has exactly two complex solutions (counted with multiplicity). This is a special case of the Fundamental Theorem of Algebra.

For students interested in careers: Quadratic equations appear in projectile motion (physics), break-even analysis (business), and signal processing (the characteristic equation of a second-order system is quadratic). The quadratic formula is the last closed-form solution formula that exists; degree 5 and above have no general formula, a result proved in the nineteenth century.

For gifted and curious students: Completing the square is not just a solution technique; it is a change of coordinates. Writing $u = x + \frac{b}{2a}$ shifts the parabola so its vertex sits at the origin, where it has the form $au^2 + \text{constant}$. This same idea (completing the square to “center” a conic) appears in analytic geometry, multivariable calculus, and the analysis of quadratic forms in linear algebra.

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