Quadratic Equations and Discriminant

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# Quadratic Equations and Discriminant ## 🎯 Learning Objectives By the end of this topic, you will be able to: 1. **Solve** quadratic equations by factoring, completing the square, and the quadratic formula 2.

Quadratic Equations and Discriminant

🎯 Learning Objectives

By the end of this topic, you will be able to:

Solve quadratic equations by factoring, completing the square, and the quadratic formula
Compute the discriminant and determine the nature of roots
Classify roots as real and distinct, real and equal, or complex
Find the sum and product of roots without solving the equation
Form a quadratic equation given its roots

📋 Prerequisites

Quadratic Functions — standard form, vertex, parabola
Algebra — factoring, square roots, rationalizing denominators
Number systems — real vs. complex numbers

📖 Core Content

4.1 Intuition: Finding the Zeros

A quadratic equation is the equation

ax^2 + bx + c = 0

. Its solutions (roots) are the

x

-coordinates where the parabola

y = ax^2 + bx + c

crosses the

x

-axis. These are called the zeros of the quadratic function. Imagine a ball thrown in the air:

h(t) = -5t^2 + 20t + 2

. Solving

h(t) = 0

tells us when the ball hits the ground. Solving quadratic equations answers "when does something become zero?"

💡 Why this matters: Quadratic equations appear everywhere in science, engineering, and finance: projectile motion, optimization, break-even analysis, compound interest problems, and eigenvalues of $2\times 2$ matrices.

4.2 Methods of Solving

4.2.1 Method 1: Factoring (When Applicable)

Recipe:

Write in standard form: $ax^2 + bx + c = 0$
Factor the left side
Set each factor to zero
Solve for $x$ Example 1.1 (Easy): Solve $x^2 - 5x + 6 = 0$ . Step 1: Factor: $(x - 2)(x - 3) = 0$ Step 2: $x - 2 = 0$ or $x - 3 = 0$ Step 3: $x = 2$ or $x = 3$

\boxed{x = 2,\ 3}

Example 1.2: Solve

2x^2 + 7x + 3 = 0

. Step 1: Factor by grouping:

2x^2 + 7x + 3 = 2x^2 + 6x + x + 3

= 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

Step 2:

2x + 1 = 0

x + 3 = 0 \implies x = -1/2

x = -3

\boxed{x = -\frac{1}{2},\ -3}

4.2.2 Method 2: Completing the Square

Recipe:

Move $c$ to RHS: $ax^2 + bx = -c$
Divide by $a$ : $x^2 + \frac{b}{a}x = -\frac{c}{a}$
Add $\left(\frac{b}{2a}\right)^2$ to both sides
Factor LHS as perfect square
Take square root and solve Example 2.1: Solve $x^2 + 6x + 2 = 0$ by completing the square. Step 1: $x^2 + 6x = -2$ Step 2: Add $(6/2)^2 = 9$ : $x^2 + 6x + 9 = -2 + 9$ Step 3: $(x + 3)^2 = 7$ Step 4: $x + 3 = \pm\sqrt{7} \implies x = -3 \pm \sqrt{7}$

\boxed{x = -3 \pm \sqrt{7}}

4.2.3 Method 3: The Quadratic Formula (Universal)

For

ax^2 + bx + c = 0

, the quadratic formula:

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

This formula works for all quadratic equations — real or complex roots. Example 3.1: Solve

2x^2 - 3x - 5 = 0

. Step 1:

a = 2

b = -3

c = -5

Step 2:

x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4}

Step 3:

x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}

x = \frac{3 - 7}{4} = \frac{-4}{4} = -1

\boxed{x = \frac{5}{2},\ -1}

4.3 The Discriminant

4.3.1 Intuition

The discriminant

D = b^2 - 4ac

tells us what kind of roots the equation has — without actually solving it. It's like a medical test that tells you the nature of the disease before treatment.

4.3.2 Definition and Classification

D = b^2 - 4ac

Discriminant	Nature of Roots	Graph
$D > 0$	Two distinct real roots	Parabola crosses x-axis at two points
$D = 0$	One real root (double/repeated)	Parabola touches x-axis at one point
$D < 0$	Two complex conjugate roots	Parabola does NOT cross x-axis

(Diagram)

4.3.3 Worked Examples

Example 4.1 (Easy): Determine the nature of roots of

x^2 - 5x + 6 = 0

. Step 1:

a = 1

b = -5

c = 6

Step 2:

D = 25 - 24 = 1 > 0

Conclusion: Two distinct real roots.

\boxed{D = 1 > 0 \rightarrow \text{Two real roots}}

Example 4.2 (Medium): For what

k

does

x^2 + kx + 9 = 0

have a double root? Step 1:

a = 1

b = k

c = 9

Step 2:

D = k^2 - 36

Step 3: For double root,

D = 0 \implies k^2 = 36 \implies k = \pm 6

\boxed{k = 6 \text{ or } k = -6}

Example 4.3 (Hard): Show that

f(x) = x^2 + x + 1

has no real zeros. Step 1:

a = 1

b = 1

c = 1

Step 2:

D = 1 - 4 = -3 < 0

Conclusion: No real zeros (complex roots). The parabola lies entirely above the x-axis.

\boxed{D = -3 < 0 \rightarrow \text{No real roots}}

4.4 Sum and Product of Roots

4.4.1 Vieta's Formulas

For

ax^2 + bx + c = 0

with roots

\alpha

and

\beta

\alpha + \beta = -\frac{b}{a}

\alpha\beta = \frac{c}{a}

Derivation: If

\alpha

and

\beta

are roots, then

ax^2 + bx + c = a(x - \alpha)(x - \beta)

Expanding:

a(x^2 - (\alpha + \beta)x + \alpha\beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta

Comparing:

-a(\alpha + \beta) = b \implies \alpha + \beta = -b/a

And:

a\alpha\beta = c \implies \alpha\beta = c/a

4.4.2 Worked Examples

Example 5.1 (Easy): Find the sum and product of roots of

x^2 - 7x + 12 = 0

. Step 1:

a = 1

b = -7

c = 12

Step 2: Sum

= -(-7)/1 = 7

Step 3: Product

= 12/1 = 12

\boxed{\text{Sum} = 7,\ \text{Product} = 12}

Example 5.2 (Medium): If

\alpha

and

\beta

are roots of

2x^2 - 5x + 1 = 0

, find

\frac{1}{\alpha} + \frac{1}{\beta}

. Step 1:

\alpha + \beta = \frac{5}{2}

\alpha\beta = \frac{1}{2}

Step 2:

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{1/2} = 5

\boxed{5}

Example 5.3 (Hard): Form a quadratic equation whose roots are

2 + \sqrt{3}

and

2 - \sqrt{3}

. Step 1: Sum

= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4

Step 2: Product

= (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1

Step 3: Equation:

x^2 - (\text{sum})x + (\text{product}) = 0

x^2 - 4x + 1 = 0

\boxed{x^2 - 4x + 1 = 0}

4.5 Complex Roots

When

D < 0

, the roots are complex conjugates:

\alpha = p + qi

\beta = p - qi

, where

i = \sqrt{-1}

. Example 6.1: Solve

x^2 + 4x + 5 = 0

. Step 1:

a = 1

b = 4

c = 5

Step 2:

D = 16 - 20 = -4 < 0

Step 3:

x = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i

\boxed{x = -2 + i,\ -2 - i}

4.6 Why This Matters

The discriminant tells you everything about the behavior of a quadratic without solving it. In optimization:

$D > 0$ : Two break-even points
$D = 0$ : Exactly one break-even (touching)
$D < 0$ : No real break-even (always profitable or never) In machine learning, many loss functions are quadratic (or approximated as quadratic near the minimum). The convexity ( $a > 0$ ) guarantees a unique minimum.

📐 Key Formulas — Summary Table

Concept	Formula	When to Use
Quadratic formula	$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$	Solving ANY quadratic
Discriminant	$D = b^2 - 4ac$	Nature of roots
$D > 0$	Two distinct real roots	Parabola crosses x-axis twice
$D = 0$	One double real root	Parabola touches x-axis
$D < 0$	Two complex conjugate roots	Parabola doesn't cross x-axis
Sum of roots	$\alpha + \beta = -b/a$	Vieta's formula
Product of roots	$\alpha\beta = c/a$	Vieta's formula
Forming equation	$x^2 - (\text{sum})x + (\text{product}) = 0$	Given roots
Completing square	$a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}$	Vertex form

⚠️ Common Pitfalls

Pitfall 1: Sign Error in the Quadratic Formula

Mistake: Writing

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

correctly but evaluating

-b

as negative even when

b

is already negative. Example: For

x^2 - 5x + 6 = 0

b = -5

, so

-b = 5

, not

-5

x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = 3, 2

. ✓

Pitfall 2: Forgetting the $\pm$ Sign

Mistake: Writing

x = \frac{-b + \sqrt{D}}{2a}

and forgetting

\pm

, thus finding only one root. Correct: The

\pm

gives both roots. Every quadratic (in the complex plane) has exactly two roots.

Pitfall 3: Discriminant Confusion

Mistake: Thinking

D > 0

means "one real root" or

D = 0

means "no real root." Correct:

D > 0

→ TWO real roots;

D = 0

→ ONE (double) real root;

D < 0

→ NO real roots (complex).

Pitfall 4: Forgetting to Write in Standard Form First

Mistake: Trying to apply the quadratic formula to

3x^2 + 5 = 7x

without rearranging. Correct: First write

3x^2 - 7x + 5 = 0

, then identify

a = 3

b = -7

c = 5

📝 Practice Questions

Q1: Solve $x^2 - 9 = 0$ .

Strategy Hint: Difference of squares.

$(x-3)(x+3) = 0 \implies x = 3 \text{ or } x = -3$

$\boxed{x = \pm 3}$ Q2: Solve $2x^2 - 5x - 3 = 0$ using the quadratic formula.

Strategy Hint: $a=2$ , $b=-5$ , $c=-3$ .

$x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$ $x = 3 \text{ or } x = -\frac{1}{2}$

$\boxed{x = 3,\ -\frac{1}{2}}$ Q3: Find the discriminant and nature of roots of $x^2 - 6x + 9 = 0$ .

Strategy Hint: Compute $D = b^2 - 4ac$ .

$D = 36 - 36 = 0$ → one double real root.

$\boxed{D = 0,\ \text{double root } x = 3}$ Q4: For what $k$ does $x^2 - 4x + k = 0$ have real and distinct roots?

Strategy Hint: Need $D > 0$ .

$D = 16 - 4k > 0 \implies k < 4$

$\boxed{k < 4}$ Q5: Find the sum and product of roots of $3x^2 + 7x - 2 = 0$ .

Strategy Hint: Sum $= -b/a$ , Product $= c/a$ .

Sum $= -\frac{7}{3}$ , Product $= -\frac{2}{3}$

$\boxed{\text{Sum} = -\frac{7}{3},\ \text{Product} = -\frac{2}{3}}$ Q6: Form a quadratic equation with roots $3$ and $-4$ .

Strategy Hint: $x^2 - (\text{sum})x + \text{product} = 0$ .

Sum $= 3 + (-4) = -1$ , Product $= 3(-4) = -12$ $x^2 - (-1)x + (-12) = 0 \implies x^2 + x - 12 = 0$

$\boxed{x^2 + x - 12 = 0}$ Q7: Solve $x^2 - 4x + 13 = 0$ (complex roots).

Strategy Hint: $D < 0$ , use $i = \sqrt{-1}$ .

$x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$

$\boxed{x = 2 + 3i,\ 2 - 3i}$ Q8: If $\alpha, \beta$ are roots of $x^2 - 5x + 3 = 0$ , find $\alpha^2 + \beta^2$ .

Strategy Hint: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ .

$\alpha + \beta = 5$ , $\alpha\beta = 3$ $\alpha^2 + \beta^2 = 25 - 6 = 19$

$\boxed{19}$ Q9: Solve $2x^2 + 3x - 2 = 0$ by completing the square.

Strategy Hint: Divide by 2 first, then complete.

$x^2 + \frac{3}{2}x = 1$ $x^2 + \frac{3}{2}x + \frac{9}{16} = 1 + \frac{9}{16} = \frac{25}{16}$ $(x + \frac{3}{4})^2 = \frac{25}{16}$ $x + \frac{3}{4} = \pm \frac{5}{4}$ $x = \frac{-3 \pm 5}{4} \implies x = \frac{1}{2} \text{ or } x = -2$

$\boxed{x = \frac{1}{2},\ -2}$ Q10: A rectangular garden has area 96 m². The length is 4 m more than the width. Find dimensions.

Strategy Hint: $l = w + 4$ , $A = w(w+4) = 96$ .

$w^2 + 4w - 96 = 0$ $D = 16 + 384 = 400$ $w = \frac{-4 \pm 20}{2} = 8 \text{ or } -12$ (discard negative) $w = 8$ , $l = 12$

$\boxed{8\text{m} \times 12\text{m}}$ Q11: If one root of $x^2 - 7x + k = 0$ is $2$ , find $k$ and the other root.

Strategy Hint: If $2$ is a root, $f(2) = 0$ .

$4 - 14 + k = 0 \implies k = 10$ Other root: product $= k = 10$ , so other root $= 10/2 = 5$ Check: $2 + 5 = 7 = -b/a$ ✓

$\boxed{k = 10,\ \text{other root} = 5}$ Q12: Prove that $x^2 + px + p^2 = 0$ has no real roots for any non-zero real $p$ .

Strategy Hint: Show $D < 0$ for $p \neq 0$ .

$D = p^2 - 4p^2 = -3p^2$ For $p \neq 0$ , $-3p^2 < 0$ → no real roots.

$\boxed{D = -3p^2 < 0 \text{ for } p \neq 0}$

🔗 Cross-References

Previous topic: Quadratic Functions
Next topic: Polynomials — quadratics generalize to higher-degree polynomials
Related: Complex numbers — needed when $D < 0$
Across courses: BSMA1003 Maths 2 (eigenvalues use characteristic polynomial); BSMA1002 Stats 1 (quadratic regression) Join Discord Previous3.1 Quadratic FunctionsNext4.0 Polynomial Operations

What this note is really saying

Quadratic Equations and Discriminant

🎯 Learning Objectives

📋 Prerequisites

📖 Core Content

4.1 Intuition: Finding the Zeros

4.2 Methods of Solving

4.2.1 Method 1: Factoring (When Applicable)

4.2.2 Method 2: Completing the Square

4.2.3 Method 3: The Quadratic Formula (Universal)

4.3 The Discriminant

4.3.1 Intuition

4.3.2 Definition and Classification

4.3.3 Worked Examples

4.4 Sum and Product of Roots

4.4.1 Vieta's Formulas

4.4.2 Worked Examples

4.5 Complex Roots

4.6 Why This Matters

📐 Key Formulas — Summary Table

⚠️ Common Pitfalls

Pitfall 1: Sign Error in the Quadratic Formula

Pitfall 2: Forgetting the ±\pm± Sign

Pitfall 3: Discriminant Confusion

Pitfall 4: Forgetting to Write in Standard Form First

📝 Practice Questions

🔗 Cross-References

Pitfall 2: Forgetting the $\pm$ Sign