All notes Mathematics · Quadratic Equations · SSS1

Quadratic Equations II

18 views

General Form of Quadratic Equations Leading to Formula Method

Mathematics Lesson Notes

General Form of Quadratic Equations Leading to the Formula Method

Derivation of the Roots of the General Form of Quadratic Equations.
Using the Formula Method to solve Quadratic Equations.
Calculating the Sum and Product of quadratic roots directly.

1. The Anatomy of a Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable. The standard (or general) form is written as shown below:

ax² + bx + c = 0

ax²

Quadratic Term
a is the leading coefficient (a ≠ 0)

Linear Term
b is the linear coefficient

Constant Term
Pure number without variables

2. Derivation of the Quadratic Formula

We can find a universal formula for the roots of any quadratic equation by applying the Method of Completing the Square to the standard equation.

Start with the general form: $$\mathbf{ax^2 + bx + c = 0}$$

Divide the entire equation by the coefficient of $x^2$ (which is $a$): $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

Move the constant term to the right-hand side of the equal sign: $$x^2 + \frac{b}{a}x = -\frac{c}{a}$$

Find the square of half the coefficient of $x$: $$\left(\frac{1}{2} \cdot \frac{b}{a}\right)^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$$

Add this squared term to both sides of the equation: $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$$

Factor the left side as a perfect square binomial, and find a common denominator ($4a^2$) for the right side: $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$

Take the square root of both sides (remembering the $\pm$ symbol): $$x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$$

Isolate $x$ to get the final formula: $$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$

★ Core Mathematical Formula

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

Note: The term inside the square root ($b^2 - 4ac$) is called the Discriminant. It tells us the nature of the roots (real, repeated, or complex).

Evaluation 1

Suppose a general quadratic equation is given as: $Dy^2 + Ey + F = 0$.

Using the step-by-step method of completing the square above, derive the general roots for $y$.

3. Using the Formula Method to Solve Equations

Follow these three simple steps when using the quadratic formula:

Ensure the equation is arranged in standard form ($ax^2 + bx + c = 0$).
Identify your $a$, $b$, and $c$ coefficients (including their negative/positive signs).
Substitute the values carefully into the formula using brackets.

Example 1

Solve the equation: $3x^2 - 5x - 3 = 0$ (Give roots correct to 2 decimal places).

Solution:
Comparing with $ax^2 + bx + c = 0$, we find: $a = 3$, $b = -5$, $c = -3$.

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-3)}}{2(3)}$$ $$x = \frac{5 \pm \sqrt{25 + 36}}{6} \implies x = \frac{5 \pm \sqrt{61}}{6}$$

Since $\sqrt{61} \approx 7.8102$:

$$x = \frac{5 + 7.8102}{6} \quad\text{or}\quad x = \frac{5 - 7.8102}{6}$$ $$x = \frac{12.8102}{6} \quad\text{or}\quad x = \frac{-2.8102}{6}$$

Answer: x = 2.14 or x = -0.47

Example 2

Solve the equation: $6x^2 + 13x + 6 = 0$.

Solution:
Here, $a = 6$, $b = 13$, $c = 6$.

$$x = \frac{-13 \pm \sqrt{(13)^2 - 4(6)(6)}}{2(6)}$$ $$x = \frac{-13 \pm \sqrt{169 - 144}}{12} \implies x = \frac{-13 \pm \sqrt{25}}{12}$$ $$x = \frac{-13 \pm 5}{12}$$

Calculating the two separate pathways:

$$x = \frac{-13 + 5}{12} = \frac{-8}{12} = -\frac{2}{3} \approx -0.67$$ $$x = \frac{-13 - 5}{12} = \frac{-18}{12} = -\frac{3}{2} = -1.50$$

Answer: x = -0.67 or x = -1.50

Example 3

Solve the equation: $3x^2 - 12x + 10 = 0$ (Correct to 2 decimal places).

Solution:
Here, $a = 3$, $b = -12$, $c = 10$.

$$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(10)}}{2(3)}$$ $$x = \frac{12 \pm \sqrt{144 - 120}}{6} \implies x = \frac{12 \pm \sqrt{24}}{6}$$

Since $\sqrt{24} \approx 4.8990$:

$$x = \frac{12 + 4.8990}{6} = 2.8165 \quad\text{or}\quad x = \frac{12 - 4.8990}{6} = 1.1835$$

Answer: x = 2.82 or x = 1.18

Evaluation 2

Use the formula method to solve the following quadratic equations:

$t^2 - 8t + 2 = 0$
$t^2 + 3t + 1 = 0$

4. Sum and Product of Quadratic Roots

We can find the sum and product of the roots directly from the coefficients in the equation without solving it. If an equation has roots $\alpha$ (alpha) and $\beta$ (beta), it can be factored as:

$$(x - \alpha)(x - \beta) = 0 \implies \mathbf{x^2 - (\alpha + \beta)x + \alpha\beta = 0} \quad\text{--- (Eq. 1)}$$

Comparing this directly with the standard equation divided by $a$:

$$\mathbf{x^2 + \left(\frac{b}{a}\right)x + \left(\frac{c}{a}\right) = 0} \quad\text{--- (Eq. 2)}$$

Sum of Roots

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

Product of Roots

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

★ Important Algebraic Root Identities

To solve complex root problems, memorize these expansions:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \implies \alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta}$

Examples on Sum & Product

1. If the roots of $3x^2 - 4x - 1 = 0$ are $\alpha$ and $\beta$, find $\alpha + \beta$ and $\alpha\beta$.

Solution: $a = 3, b = -4, c = -1$.

$$\text{Sum } (\alpha + \beta) = -\frac{b}{a} = -\left(\frac{-4}{3}\right) = \mathbf{\frac{4}{3}}$$ $$\text{Product } (\alpha\beta) = \frac{c}{a} = \mathbf{-\frac{1}{3}}$$

2. Using the same equation, calculate the value of: (a) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ and (b) $\alpha - \beta$.

Solution (a): First, find a common denominator:

$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$$ $$= \frac{\left(\frac{4}{3}\right)^2 - 2\left(-\frac{1}{3}\right)}{-\frac{1}{3}} = \frac{\frac{16}{9} + \frac{6}{9}}{-\frac{1}{3}} = \frac{\frac{22}{9}}{-\frac{1}{3}} = \frac{22}{9} \times (-3) = \mathbf{-7\frac{1}{3}}$$

Solution (b): Using the difference identity:

$$(\alpha - \beta) = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{\left(\frac{4}{3}\right)^2 - 4\left(-\frac{1}{3}\right)}$$ $$= \sqrt{\frac{16}{9} + \frac{12}{9}} = \sqrt{\frac{28}{9}} = \mathbf{\frac{\sqrt{28}}{3}}$$

Evaluation 3

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 11x + 5 = 0$, find the value of:

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

General Practice & Assignments

General Evaluation

Solve by rearranging into standard form first:

$63z = 49 + 18z^2$
$8s^2 + 14s = 15$

Solve using the formula method:

$12y^2 + y - 35 = 0$
$h^2 - 15h + 54 = 0$

Reading Assignment

New General Mathematics SS Bk2
Pages 41–42.

Solve Exercise 3e, Nos. 19 and 20 (page 42).

Weekend Assignment (Multiple Choice)

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 7x - 3 = 0$, find the value of:

$\alpha + \beta$
(a) 2/3 (b) 7/2 (c) 2/5 (d) 5/3
$\alpha\beta$
(a) -3/2 (b) 2/3 (c) 3/2 (d) -2/3
$\alpha\beta^2 + \alpha^2\beta$
(a) 21/4 (b) 4/21 (c) -4/21 (d) -21/4

Solve using the formula method:

$6p^2 - 2p - 7 = 0$
$3 = 8q - 2q^2$

Theory Examination Practice

Solve the equation $2x^2 + 6x + 1 = 0$ using the formula method. (Leave answer in surd form).
If $\alpha$ and $\beta$ are the roots of the equation $3x^2 - 9x + 2 = 0$, find the exact values of:
- (i) $\alpha\beta^2 + \alpha^2\beta$
- (ii) $\alpha^2 - \alpha\beta + \beta^2$

Test yourself on Mathematics

Trigonometry Quiz

4 questions 10 min SSS2

Start quiz

Algebraic Expressions Quiz

3 questions 10 min JSS2

Start quiz

Sets and Venn Diagrams Practice Test

50 questions 15 min SSS1

Start quiz

Simultaneous Linear Equations Practice Test

50 questions 15 min JSS3

Start quiz

Whole Numbers and Basic Operations Practice Test

50 questions 15 min JSS1

Start quiz

Trigonometry Practice Test

50 questions 15 min SSS2

Start quiz

Track your reading & take quizzes

Create free account