a(b + c) = ab + ac

Example 1
= 2x(3x + 4) - 3(3x + 4)

= 6x^{2} + 8x - 9x -12

= 6x^{2} - x - 12

Earlier, you read that:

a(b + c) = ab + ac

Example 1
Expand (2x - 3)(3x + 4)

= 2x(3x + 4) - 3(3x + 4)

= 6x^{2} + 8x - 9x -12

= 6x^{2} - x - 12

They are represented as:

Example 3

1. (a + b)(a + b)

2. (a - b)(a - b)

3. (a + b)(a - b)

Consider the third identity Expand (a + b)(a - b)

= a(a - b) + b(a - b)

= a^{2} - ab + ba - b^{2}

= a^{2} - b^{2}

This is referred to as the difference of two squares.

Example 3

Using Quadratic identity Expand (3x + 4)(3x - 4)

From difference of two squares:

(a + b)(a - b) = a^{2} - b^{2}

= (3x)^{2} - 4^{2}

= 9x^{2} - 16

- This is the opposite of expansion (explained above).

Example 5

While a(b + c) = ab + ac is expansion

ab + ac = ab + ac is factorization

Factorize 3x^{2} - 9x

3x is common hence:

= 3x(x - 3)

Example 5

Factorize x^{2} + 7x + 12

Use the rule: ax^{2} + bx + c

* p x q = c* and * p + q = b*

p x q = 12 and p + q = 7

The numbers are 4 and 3 since 4 x 3 = 12 and 4 + 3 = 7

x^{2} + 3x + 4x + 12

= x(x + 3) + 4(x + 3)

= (x + 4)(x + 3)

- If a x b = 0, either a = 0 or b = 0.

Example 6

Example 6

Solve the equation 2x^{2} + 7x + 6 = 0

2x^{2} + 7x + 6 = 0

2x^{2} + 3x + 4x + 6 = 0

x(2x+3) + 2(2x + 3) = 0

(2x + 3)(x + 2) = 0

__First Part:__

2x + 3 = 0

2x = -3

x = -3/2

__Second Part:__

x + 2 = 0

x = -2

Hence x = -2 and x = -3/2

The solution obtained in solving quadratic equations may also be termed as roots or a quadratic equation.

Note:

Note:

Hence:

Example 7

Given that the roots of the equation 3x^{2} - 4x - 4 = 0
are a and b find 1/a + 1/b

Given the roots an equation, a quadratic equation can be formed.

Note:

Example 8

Note:

Example 8

Form the quadratic equation whose roots are:

1/2 or 2/3

1/2 or 2/3

x = ½ or x = 2/3

2x = 1 or 3x = 2

2x - 1 = 0 or 3x - 2 = 0

(2x - 1)(3x - 2) = 0

2x(3x - 2) - 1(3x - 2) = 0

6x^{2} - 4x - 3x + 2 = 0

6x^{2} - 7x + 2 = 0

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