Contents / Algebra / Quadratic Equations

Chapter 4

Quadratic Equations

Quadratic equations, the quadratic formula, the discriminant, and rational equations.

Introduction

Quadratic equations, the quadratic formula, the discriminant, and rational equations.

A quadratic equation contains a squared term as its highest power of the variable. In standard form it is ax² + bx + c = 0 with a ≠ 0.

Because the graph of the related function is a parabola, a quadratic equation can have zero, one, or two real solutions. This richer behavior compared with linear equations makes quadratics central to algebra.

4.1Standard Form and Terminology

Definition 4.1 (Quadratic equation in standard form). A quadratic equation in standard form is

a x^{2} + b x + c = 0, a \neq = 0,

where $a$ is the leading coefficient, $b$ the linear coefficient, and $c$ the constant term.

The solutions are often called roots, zeros, or x-intercepts.

Intuition. The x² term makes the sides of the "bowl" curve; the higher the power the faster it grows.

a tells whether the bowl opens upward (a > 0) or downward (a < 0).

Example 4.2. Put $3 x - 2 x^{2} = 7$ into standard form. First move every term to one side so the right side is $0$ : $3 x - 2 x^{2} - 7 = 0$ .

Solution.

Order the terms by descending power and, by convention, make the leading coefficient positive by multiplying through by $- 1$ : $2 x^{2} - 3 x + 7 = 0$ .
Now read off the coefficients: $a = 2$ , $b = - 3$ , $c = 7$ . Because $a = 2 \neq = 0$ this is a genuine quadratic.
Sanity check: expanding back, $2 x^{2} - 3 x + 7$ with the sign flip undone is $- 2 x^{2} + 3 x - 7 = 0$ , i.e. $3 x - 2 x^{2} = 7$ , the original. ✓

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Exercises for §4.1 →

4.2Solving by Factoring

Theorem 4.3 (Zero-product property). A product of two numbers is zero exactly when one of them is zero:

A B = 0 ⟺ A = 0 or B = 0.

So when a quadratic factors as $(p x + q) (r x + s) = 0$ , setting each factor to zero immediately gives the two roots.

Many quadratics are factorable over the integers after a little practice with looking for factor pairs of c that add to b.

Intuition. Factoring turns a single hard puzzle into two easy linear puzzles.

If two numbers multiply to zero, one of them has to be the zero.

Example 4.4 ((a = 1):). Solve $x^{2} - 7 x + 12 = 0$ . We need two numbers that multiply to $c = 12$ and add to $b = - 7$ .

Solution.

Factor pairs of $12$ : $(1, 12), (2, 6), (3, 4)$ . To get a sum of $- 7$ we take both negative: $- 3$ and $- 4$ , since $(- 3) (- 4) = 12$ and $(- 3) + (- 4) = - 7$ . ✓
So $x^{2} - 7 x + 12 = (x - 3) (x - 4) = 0$ .
Zero-product property: $x - 3 = 0 \Rightarrow x = 3$ , or $x - 4 = 0 \Rightarrow x = 4$ .
Check $x = 3$ : $9 - 21 + 12 = 0$ . ✓ Check $x = 4$ : $16 - 28 + 12 = 0$ . ✓

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Example 4.5 ((a ≠ 1):). Solve $2 x^{2} + x - 6 = 0$ . When $a \neq = 1$ , look for two numbers that multiply to $a \cdot c = 2 \cdot (- 6) = - 12$ and add to $b = 1$ . Those are $4$ and $- 3$ .

Solution.

Split the middle term using them: $2 x^{2} + 4 x - 3 x - 6 = 0$ .
Factor by grouping: $2 x (x + 2) - 3 (x + 2) = 0$ , so $(2 x - 3) (x + 2) = 0$ .
Set each factor to zero: $2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$ , or $x + 2 = 0 \Rightarrow x = - 2$ .
Check $x = \frac{3}{2}$ : $2 \cdot \frac{9}{4} + \frac{3}{2} - 6 = \frac{9}{2} + \frac{3}{2} - 6 = 6 - 6 = 0$ . ✓

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Exercises for §4.2 →

4.3Graph and Roots Relationship

Proposition 4.6 (Roots are x-intercepts). The real solutions of $a x^{2} + b x + c = 0$ are exactly the $x$ -values where the parabola

y = a x^{2} + b x + c

crosses or touches the x-axis.

If the parabola opens upward and the vertex is above the x-axis there are no real roots; if it touches there is one (repeated) root; if it crosses there are two.

Intuition. Picture throwing a ball; the times it is at ground level are the roots of the height equation.

The picture of the parabola immediately tells you how many times it hits the ground.

Example 4.7. Where does $y = x^{2} - 4$ meet the x-axis? Points on the x-axis have $y = 0$ , so set $x^{2} - 4 = 0$ .

Solution.

Solve: $x^{2} = 4 \Rightarrow x = \pm 2$ . The parabola crosses at $x = - 2$ and $x = 2$ — two real roots.
Sanity check: at $x = 2$ , $y = 4 - 4 = 0$ ; at $x = - 2$ , $y = 4 - 4 = 0$ . Both lie on the x-axis. ✓

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Example 4.8 ((one vs. no crossings):). Compare $y = x^{2}$ and $y = x^{2} + 1$ . For $y = x^{2}$ , setting $x^{2} = 0$ gives the single root $x = 0$ : the parabola's vertex sits exactly on the axis, so it touches once.

Solution.

For $y = x^{2} + 1$ , setting $x^{2} + 1 = 0$ gives $x^{2} = - 1$ , which has no real solution. The vertex $(0, 1)$ is above the axis and the bowl opens upward, so the curve never reaches $y = 0$ — zero real roots.
This illustrates the full range: a parabola can cross twice, touch once, or miss entirely.

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Exercises for §4.3 →

4.4Real-World Applications

Quadratic equations appear in physics (projectile motion), geometry (areas, Pythagorean problems), business (profit maximization), and whenever a quantity has a squared relationship.

Method 4.9 (Modelling with a quadratic).

Translate the situation into symbols.
Form the equation.
Solve it — by factoring, completing the square, or the quadratic formula.
Discard any physically meaningless roots.
State the answer in context.

Intuition. Many "how far?" or "when?" questions involving acceleration or area produce a squared term.

The parabola shape often matches the real curve of the situation.

Example 4.10. A rectangular garden has area $60 m^{2}$ ; one side is $4$ m longer than the other. Let the width be $x$ ; then the length is $x + 4$ . Area $=$ length $\times$ width, so $x (x + 4) = 60$ , which expands to $x^{2} + 4 x - 60 = 0$ . Factor: two numbers multiplying to $- 60$ and adding to $4$ are $10$ and $- 6$ , giving $(x + 10) (x - 6) = 0$ . Roots: $x = - 10$ or $x = 6$ . A width cannot be negative, so we discard $x = - 10$ as physically meaningless and keep $x = 6$ m; the length is $x + 4 = 10$ m. Sanity check: $6 \times 10 = 60 m^{2}$ , and $10 - 6 = 4$ m. ✓

Solution.

Discarding a mathematically valid but physically impossible root is the step beginners most often skip in applied problems.

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Exercises for §4.4 →

Common Mistakes

Forgetting that a must be nonzero: If a = 0 the equation collapses to linear.
Stopping after finding one root: Quadratics frequently have two roots; check both.
Sign errors when factoring: The pair of numbers that multiply to c must add to b; watch negatives.
Not checking by substitution: Especially important when using the quadratic formula later.
Assuming every quadratic has two real roots: The discriminant can be negative (no real roots) or zero (one real root).

The quadratic formula gives the solutions to any quadratic equation ax² + bx + c = 0 (a ≠ 0):

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

It is one of the most famous and useful formulas in algebra. Derived by completing the square on the general quadratic, it works whether the roots are real or complex and whether or not the quadratic factors nicely over the integers.

4.5Derivation and Statement

Theorem 4.11 (Quadratic formula). If $a \neq = 0$ , the solutions of $a x^{2} + b x + c = 0$ are

x = \frac{- b \pm b ^{2} - 4 a c}{2 a} .

Proof. Start with $a x^{2} + b x + c = 0$ and divide by $a$ , which is legal precisely because $a \neq = 0$ :

x^{2} + \frac{b}{a} x + \frac{c}{a} = 0, so x^{2} + \frac{b}{a} x = - \frac{c}{a} .

Add $(\frac{b}{2 a})^{2}$ to both sides to complete the square:

x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + \frac{b ^{2}}{4 a ^{2}},

(x + \frac{b}{2 a})^{2} = \frac{- 4 a c + b ^{2}}{4 a ^{2}} = \frac{b ^{2} - 4 a c}{4 a ^{2}} .

Now take square roots, writing $D = b^{2} - 4 a c$ :

x + \frac{b}{2 a} = \pm \frac{D}{2 a}, so x = - \frac{b}{2 a} \pm \frac{D}{2 a} = \frac{- b \pm D}{2 a} .

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Intuition. Completing the square is like rearranging an L-shaped pile of tiles into a full square, then adding the one small corner tile that's missing so it becomes a perfect square you can "un-square" with a square root.

We do the rearranging once on the general letters $a$ , $b$ , $c$ , so we never have to redo it — the formula is the finished recipe.

The ± is there because when you take a square root you get two possibilities (positive and negative), which is exactly why a quadratic can have two roots.

Example 4.12 ((applying the just-derived formula):). Solve $x^{2} - 5 x + 6 = 0$ . Identify $a = 1$ , $b = - 5$ , $c = 6$ .

Solution.

Compute the discriminant $D = b^{2} - 4 a c = (- 5)^{2} - 4 (1) (6) = 25 - 24 = 1$ .
Substitute into $x = \frac{- b \pm D}{2 a} = \frac{- ( - 5 ) \pm 1}{2 ( 1 )} = \frac{5 \pm 1}{2}$ .
Split the $\pm$ : $x = \frac{5 + 1}{2} = 3$ or $x = \frac{5 - 1}{2} = 2$ .
Check against factoring: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ , whose roots are indeed $2$ and $3$ . ✓

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Exercises for §4.5 →

4.6Using the Formula

Method 4.13 (Using the quadratic formula).

Identify $a$ , $b$ , $c$ carefully, signs included.
Compute the discriminant $D = b^{2} - 4 a c$ first; it tells you the nature of the roots before you finish.
Substitute into the formula, simplifying the square root where possible.
Verify by plugging the solutions back into the original equation — arithmetic errors are common.

If $D < 0$ the roots involve $i$ (see the complex number lessons).

Intuition. a is the coefficient of x² (the "curvy" part), b of x, c the constant.

D is your "preview": positive two real, zero one real (touch), negative two complex.

Example 4.14. $2 x^{2} - 3 x - 2 = 0$ . $a = 2$ , $b = - 3$ , $c = - 2$ .

Solution.

$D = 9 + 16 = 25$ . $x = [3 \pm 5] / (4) \to (8) /4 = 2$ or $(- 2) /4 = - 0.5$ .
Check: $2 (4) - 3 (2) - 2 = 8 - 6 - 2 = 0$ ; $2 (0.25) - 3 (- 0.5) - 2 = 0.5 + 1.5 - 2 = 0$ . Good.

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Exercises for §4.6 →

4.7Number and Nature of Roots (via Discriminant)

Theorem 4.15 (The discriminant decides the roots). Write $D = b^{2} - 4 a c$ . Then for $a x^{2} + b x + c = 0$ with real coefficients and $a \neq = 0$ :

$D > 0$ : two distinct real roots.
$D = 0$ : one real root, repeated with multiplicity two.
$D < 0$ : two complex conjugate roots, and no real roots.

When the coefficients are real, complex roots always appear in conjugate pairs.

Proposition 4.16 (Vieta's formulas). The two roots of $a x^{2} + b x + c = 0$ satisfy

sum of roots = - \frac{b}{a}, product of roots = \frac{c}{a} .

Intuition. The discriminant measures whether the parabola crosses, touches, or misses the x-axis.

Positive D: crosses twice; zero: touches at vertex; negative: entirely above or below (never crosses).

Example 4.17. $x^{2} + 1 = 0$ . $D = 0 - 4 = - 4 < 0$ . Roots $\pm i$ . No real roots, graph never crosses x-axis.

Example 4.18. $x^{2} - 2 x + 1 = 0$ . $D = 4 - 4 = 0$ . Root $x = 1$ (double). $(x - 1)^{2} = 0$ .

Exercises for §4.7 →

Common Mistakes

Sign errors on b (the formula has -b): Forgetting the minus in front of b is the most common slip.
Computing 4ac as 4a + c or 4ac wrong order: Must be b squared minus 4ac.
Dividing by 2a only at the end incorrectly: The entire numerator is divided by 2a.
Assuming two real roots always: Check D first; many problems are designed with D<0 or 0.
Not simplifying √D when it is a perfect square: Leaves ugly answers.
Plugging a,b,c into the formula without writing them down first: Easy to grab the wrong number.
Forgetting that if a is negative the parabola opens down, but the formula still works the same.
Using the formula when the equation is not quadratic (a=0): It is not valid; solve as linear instead.

The discriminant of the quadratic equation ax² + bx + c = 0 is the expression D = b² − 4 a c that appears under the square root in the quadratic formula.

Although it looks like a minor piece of the formula, the sign and value of D completely determine the number and nature of the solutions without having to compute the roots themselves. It is a powerful diagnostic tool.

4.8Definition and Role in the Formula

Definition 4.19 (Discriminant). The discriminant of $a x^{2} + b x + c = 0$ is the expression under the square root in the quadratic formula,

D = b^{2} - 4 a c .

In the quadratic formula the term $D$ tells us two things: whether we stay in the reals or must enter the complex numbers, and whether the two potential roots are the same or different.

D is also equal to (root1 - root2)² * a² (up to sign), so |D| measures how far apart the roots are.

Intuition. D is the "decider" inside the formula: positive → two different real answers; zero → exactly one; negative → complex pair.

You can look at D first and know what kind of answer the whole formula will give.

Example 4.20. For $x^{2} - 3 x + 2 = 0$ , $D = 9 - 8 = 1 > 0 \to$ two real roots (indeed 1 and 2).

Solution.

For $x^{2} - 3 x + 3 = 0$ , $D = 9 - 12 = - 3 < 0 \to$ two complex roots.

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Exercises for §4.8 →

4.9Interpreting the Sign of D

Assuming a, b, c real and a ≠ 0:

D > 0 : two distinct real roots. The parabola crosses the x-axis twice.
D = 0 : exactly one real root (a double root). The parabola touches the x-axis at its vertex.
D < 0 : no real roots; two complex conjugate roots. The parabola lies entirely above or entirely below the x-axis (depending on the sign of a).

The value of a tells the direction the parabola opens; D tells whether and how it meets the axis.

Intuition. Positive D: the "mystery" under the root is a real number you can take ± of → two hits on the x-axis.

Zero: exactly touches once.

Negative: the square root is imaginary → the curve never reaches the x-axis in the real world.

Example 4.21. $2 x^{2} + 4 x + 2 = 0 \to D = 16 - 16 = 0$ . Root $x = - 1$ (double). Graph touches at one point.

Example 4.22. $x^{2} + 1 = 0$ , $D = - 4 < 0$ . No real roots. Graph of $y = x^{2} + 1$ never crosses x-axis.

Exercises for §4.9 →

4.10Using the Discriminant Without Fully Solving

Often we only need to know how many real solutions exist or whether the quadratic expression is always positive/negative.

Compute D only; no need for the full ± formula.

For inequalities (e.g. ax² + bx + c > 0) the sign of D plus the sign of a tells the solution set without finding the roots.

Vieta's formulas (sum and product of roots) can be used together with D for other relations.

Intuition. D is like a quick test: "Will this quadratic equation have real answers?" before doing all the arithmetic.

For "is this quadratic always above the x-axis?" D and a give the answer instantly.

Example 4.23. Does $3 x^{2} - 5 x + 1 = 0$ have real solutions? $D = 25 - 12 = 13 > 0 \to$ yes, two.

Example 4.24. For which $k$ does $x^{2} + k x + 9 = 0$ have exactly one real root? Set $D = 0$ : $k^{2} - 36 = 0 \to k = \pm 6$ .

Example 4.25 ((inequality without finding roots):). Show that $x^{2} + x + 1 > 0$ for every real $x$ , without solving.

Solution.

Here $a = 1 > 0$ , so the parabola opens upward. Compute $D = 1^{2} - 4 (1) (1) = 1 - 4 = - 3 < 0$ .
Since $D < 0$ the parabola never touches the x-axis, and since it opens upward it lies entirely above it. Therefore $x^{2} + x + 1 > 0$ for all real $x$ — the solution set is all real numbers.
Spot check at the vertex $x = - \frac{b}{2 a} = - \frac{1}{2}$ : $\frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} > 0$ . ✓ (The vertex is the lowest point, so if even it is positive, everything is.)

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Exercises for §4.10 →

Common Mistakes

Computing 4ac incorrectly (order of operations or sign): Must be b² minus (4 times a times c).
Forgetting that D=0 still counts as "real roots" (one of them): It does have a real solution (repeated).
Assuming D tells the roots themselves: D only classifies; you still need the formula or factoring to find the actual numbers.
Ignoring a when describing the graph: D= positive with a>0 still crosses, but the shape (width) depends on a.
Using D for non-quadratic equations: Only valid when the equation really is quadratic (a≠0).
Sign error when D appears in inequalities: The test intervals depend on both D and sign(a).
Thinking complex roots mean "no solutions at all": They are perfectly valid solutions in the complex numbers; just not real.

A rational equation sets two rational expressions equal (or a rational expression equal to a number). The presence of variables in denominators means we must exclude values that make any denominator zero; those values can never be solutions.

Clearing denominators by multiplication produces a polynomial equation that is easier to solve, but the resulting candidates must be checked in the original equation.

4.11Definition and Domain Restrictions

Definition 4.26 (Rational equation). A rational expression is a quotient of two polynomials, and a rational equation is an equation containing at least one such expression. The domain of the equation — the allowed values of the variable — excludes every root of every denominator.

Intuition. You cannot divide by zero, so any number that would put a zero in a denominator is forbidden from the start.

The equation only "exists" for the other numbers.

Example 4.27. Find the domain of $\frac{2}{x - 3} = \frac{1}{x}$ before solving anything.

Solution.

Set each denominator equal to zero and solve: $x - 3 = 0 \Rightarrow x = 3$ , and $x = 0$ .
Those are the forbidden values, so the domain is all real numbers except $x = 3$ and $x = 0$ . Any candidate solution equal to $3$ or $0$ must be rejected no matter what the algebra later produces.

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Example 4.28 ((factored denominator):). Find the excluded values of $\frac{5}{x ^{2} - x - 6} = 1$ .

Solution.

Factor the denominator: $x^{2} - x - 6 = (x - 3) (x + 2)$ .
Set each factor to zero: $x = 3$ or $x = - 2$ . The domain excludes both.

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Exercises for §4.11 →

4.12Clearing Denominators

Method 4.29 (Clearing denominators).

List the excluded values: every root of every denominator.
Multiply every term on both sides by the least common multiple of all the denominators appearing in the equation.
Solve the resulting fraction-free equation, which is equivalent to the original for values in the domain.

Intuition. Multiplying by the "magic number" that is a multiple of every bottom turns every fraction into a whole number or polynomial.

You are allowed to do this only because you already excluded the zeros.

Example 4.30. Solve $\frac{1}{x} + \frac{1}{3} = \frac{1}{2}$ . Denominators are $x$ , $3$ , $2$ ; excluded value: $x \neq = 0$ . The LCM of $x$ , $3$ , and $2$ is $6 x$ . Multiply every term by $6 x$ : $6 x \cdot \frac{1}{x} + 6 x \cdot \frac{1}{3} = 6 x \cdot \frac{1}{2}$ , which simplifies to $6 + 2 x = 3 x$ . Solve the linear equation: $6 = 3 x - 2 x = x$ , so $x = 6$ . Check: $x = 6 \neq = 0$ is allowed, and $\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ . ✓

Exercises for §4.12 →

4.13Extraneous Roots

Definition 4.31 (Extraneous root). A root of the cleared polynomial equation that makes a denominator zero in the original equation is extraneous, and must be discarded.

Even when the multiplication step itself does not introduce them, substitution is the only way to be sure a candidate is valid.

Intuition. The multiplication step can "hide" the fact that we divided by zero in disguise.

Checking is the honesty step that keeps only the answers that were legal all along.

Example 4.32. Solve $x / (x - 2) = 2/ (x - 2) + 1$ .

Solution.

Multiply by $(x - 2)$ : $x = 2 + (x - 2) \to 0 = 0$ , identity.
But original is undefined at $x = 2$ , so solution is all reals except 2.

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Exercises for §4.13 →

4.14Examples

Practice includes proportions, equations with several fractions, and equations that become quadratic after clearing.

Always state excluded values first, solve, then filter.

Intuition. Write the "forbidden list" before you start so you remember to reject them at the end.

The algebra gives candidates; the original equation is the final judge.

Example 4.33 ((becomes quadratic after clearing):). Solve $\frac{1}{x + 1} - \frac{1}{x - 1} = \frac{1}{x}$ . Excluded values (denominators zero): $x \neq = - 1$ , $x \neq = 1$ , $x \neq = 0$ . The LCM is $x (x + 1) (x - 1)$ . Multiply every term by it: $x (x - 1) - x (x + 1) = (x + 1) (x - 1)$ . Expand each piece: $x (x - 1) = x^{2} - x$ , $x (x + 1) = x^{2} + x$ , and $(x + 1) (x - 1) = x^{2} - 1$ . So $(x^{2} - x) - (x^{2} + x) = x^{2} - 1$ , which simplifies to $- 2 x = x^{2} - 1$ . Move everything to one side: $x^{2} + 2 x - 1 = 0$ . This is quadratic, so use the formula with $a = 1$ , $b = 2$ , $c = - 1$ : $D = 4 - 4 (1) (- 1) = 4 + 4 = 8$ , giving $x = \frac{- 2 \pm 8}{2} = - 1 \pm 2$ . Filter against the excluded list: neither $- 1 + 2 \approx 0.414$ nor $- 1 - 2 \approx - 2.414$ equals $- 1$ , $1$ , or $0$ , so both are valid solutions. Sanity check of $x = - 1 + 2$ : it satisfies $x^{2} + 2 x - 1 = 0$ by construction, and it is in the domain, so it is a genuine solution. ✓

Exercises for §4.14 →

Common Mistakes

Forgetting to list excluded values before multiplying: You will not know which roots to throw away later.
Multiplying only one side or only some terms: Every term on both sides must be multiplied by the LCM.
Keeping a root that zeros a denominator: It never was a solution; the original equation was not defined there.
Canceling a variable factor without considering it could be zero: That is exactly when extraneous roots appear.
Assuming the cleared equation has the same domain as the original: The polynomial equation is defined everywhere; the rational one is not.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter