Forgotten basics.

[bobisaka]

How did they come to simplify the part of the equation that is highlighted?

 

[Steven G]

The distributed law over subtraction says that a( b - c) = a*b - a*c
and a*b - a*c = a( b - c). This is called factoring.

In calculating 7(8-6) you can compute 7(2)=14 OR 7*8 - 7*6 = 56 -42 =14

 

[pka]

\(4(1-\sqrt3)=4-4\sqrt3\)

 

[bobisaka]

4(1−√3)=4−4√3

I still do not understand how to do the distributive property to get the 4(1−√3).

For example this is what i can come up with, with my current understanding:

4-4√3

2*2-2*2√3 / 2

2-4√3

-2√3


Even better, which chapter in the algebra 1 or 2 texbooks should i be referring to?
Currently cannot find examples similar to what I am looking for..

 

[skeeter]

[MATH]4[/MATH] and [MATH]4\sqrt{3}[/MATH] are not like terms and cannot be combined as you did in post #4

think about [MATH]4-4x[/MATH], where [MATH]x[/MATH] represents [MATH]\sqrt{3}[/MATH] ... both terms have [MATH]4[/MATH] as a common factor

[MATH]4 \cdot 1 - 4 \cdot x = 4(1 - x)[/MATH]

 

[bobisaka]

I still do not understand how to do the distributive property to get the 4(1−√3).

For example this is what i can come up with, with my current understanding:

4-4√3

2*2-2*2√3 / 2

2-4√3

-2√3

[MATH]4[/MATH] and [MATH]4\sqrt{3}[/MATH] are not like terms and cannot be combined as you did in post #4

think about [MATH]4-4x[/MATH], where [MATH]x[/MATH] represents [MATH]\sqrt{3}[/MATH] ... both terms have [MATH]4[/MATH] as a common factor

[MATH]4 \cdot 1 - 4 \cdot x = 4(1 - x)[/MATH]

Ah I see. The 'like terms' is what confused me, thanks for clarifying.
I will do some of the other questions to test my understanding.

 

[bobisaka]

Try this question:

10 -√75 / 5

10 - √25√3 / 5

10 - 5√3 / 5 ----> Where do i take it from here?

The answer should be 2 - √3

 

[JeffM]

Try this question:

10 -√75 / 5

10 - √25√3 / 5

10 - 5√3 / 5 ----> Where do i take it from here?

The answer should be 2 - √3

One of your problems is that you do not use parentheses properly

[MATH]\dfrac{4 - \sqrt{48}}{4} = (4 - \sqrt{48})/4[/MATH]
[MATH]\dfrac{4 - \sqrt{16 * 3}}{4} = (4 - \sqrt{16 * 3})/4[/MATH]
[MATH]\dfrac{4 - 4\sqrt{3}}{4} = (4 - 4 \sqrt{3})/4[/MATH]
[MATH]\dfrac{4(1 - \sqrt{3})}{4} = 4(1 - \sqrt{3})/4[/MATH]
[MATH]1 - \sqrt{3} = 1 - \sqrt{3}[/MATH]
You have the same problem with parentheses in your latest problem.

 

[bobisaka]

One of your problems is that you do not use parentheses properly

[MATH]\dfrac{4 - \sqrt{48}}{4} = (4 - \sqrt{48})/4[/MATH]
[MATH]\dfrac{4 - \sqrt{16 * 3}}{4} = (4 - \sqrt{16 * 3})/4[/MATH]
[MATH]\dfrac{4 - 4\sqrt{3}}{4} = (4 - 4 \sqrt{3})/4[/MATH]
[MATH]\dfrac{4(1 - \sqrt{3})}{4} = 4(1 - \sqrt{3})/4[/MATH]
[MATH]1 - \sqrt{3} = 1 - \sqrt{3}[/MATH]
You have the same problem with parentheses in your latest problem.

I have posted another thread, this time with my question in parenthesis.