Maths Help - Yes, Its homework.... (Algebra)?

Question

Hello, I'm doing Quadratics I think it is. I have this question here...
Now, this equation has a surd. Im going to use & for the surd sign.

(x + &3)(x - &3)

Now, Can I get the working for it. Now, in my exercise book I have the answer thats why i want the working for it.

- Cheers, Daniel

Answer 1

(x + &3)(x - &3)
= x^2 - 3

Answer 2

Dear Daniel,

1. First Method (Difference of Two Squares)
===================================

The algebraic expression given is of the form :-

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

Therefore, by using the theorem above :-

(x + √3)(x - √3) = x^2 - (√3)^2
(x + √3)(x - √3) = x^2 - 3

ANSWER : x^2 - 3

2. Second Method (Direct Expansion)
===============================

Expand the algebraic expression directly :-

(x + √3)(x - √3) = x^2 - x√3 + x√3 - (√3)^2
(x + √3)(x - √3) = x^2 - (√3)^2
(x + √3)(x - √3) = x^2 - 3

ANSWER : x^2 - 3

3. Third Method (Long Expansion)
===========================

This method is quite long :-

(x + √3)(x - √3) = x(x - √3) + (√3)(x - √3)
(x + √3)(x - √3) = x^2 - x√3 + x√3 - (√3)^2
(x + √3)(x - √3) = x^2 - (√3)^2
(x + √3)(x - √3) = x^2 - 3

ANSWER : x^2 - 3

4. Fourth Method (Long Expansion-Another Way)
======================================

This method is quite long :-

(x + √3)(x - √3) = x(x + √3) - (√3)(x + √3)
(x + √3)(x - √3) = x^2 + x√3 - x√3 - (√3)^2
(x + √3)(x - √3) = x^2 - (√3)^2
(x + √3)(x - √3) = x^2 - 3

ANSWER : x^2 - 3

==FOOTNOTE==

The theorem (Difference of Two Squares)

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

can be proven as follows :-

Expand the algebraic expression

(a + b)(a - b) = a^2 - ab + ab - b^2
(a + b)(a - b) = a^2 - b^2

Thus,

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

======================================
Thank you for your patient.

From tarmeasy.
6th February 2007 (Tuesday) 4:26 PM Malaysia

Answer 3

(x + √3)(x - √3) has a difference-of-squares expansion:
x^2 - x√3 + x√3 -3 =
x^2 - 3

Answer 4

Try using the foil method
F-first x's first
O- outer x's outer
I- Inner x's Inner
L- last x's last
example below:

http://www.mathwords.com/f/foil_method.htm