Re: Simplying expressions

Quote:

Originally Posted by

**daigo** A problem in my book shows that:

$\displaystyle \sqrt{({x_2 - \frac{x_1 + x_2}{2})^{2} + (y_2 - \frac{y_1 + y_2}{2}})^{2}} = \frac{1}{2}\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\$

But I don't understand how they got this. This is what I have so far:

$\displaystyle \sqrt{({x_2 - \frac{x_1 + x_2}{2})^{2} + (y_2 - \frac{y_1 + y_2}{2}})^{2}} = \sqrt{({\frac{2x_2}{2} - \frac{x_1 + x_2}{2})^{2} + (\frac{2y_2}{2} - \frac{y_1 + y_2}{2}})^{2}} = \sqrt{({\frac{-x_1 + 3x_2}{2})^{2} + (\frac{-y_1 + 3y_2}{2}})^{2}}$

Do you understand that $\displaystyle \sqrt {{{\left( {\frac{x}{2}} \right)}^2} + {{\left( {\frac{y}{2}} \right)}^2}} = \frac{1}{2}\sqrt {{x^2} + {y^2}}~?$

Do you understand that $\displaystyle \left( {a - \frac{{a + b}}{2}} \right) = \left( {\frac{{a - b}}{2}} \right)~?$

Re: Simplying expressions

Quote:

Originally Posted by

**daigo** A problem in my book shows that:

$\displaystyle \sqrt{({x_2 - \frac{x_1 + x_2}{2})^{2} + (y_2 - \frac{y_1 + y_2}{2}})^{2}} = \frac{1}{2}\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\$

But I don't understand how they got this. This is what I have so far:

$\displaystyle \sqrt{({x_2 - \frac{x_1 + x_2}{2})^{2} + (y_2 - \frac{y_1 + y_2}{2}})^{2}} = \sqrt{({\frac{2x_2}{2} - \frac{x_1 + x_2}{2})^{2} + (\frac{2y_2}{2} - \frac{y_1 + y_2}{2}})^{2}} = \sqrt{({\frac{-x_1 + 3x_2}{2})^{2} + (\frac{-y_1 + 3y_2}{2}})^{2}}$

You are not distributing the negative correctly.

$\displaystyle x_2- \frac{x_1+ x_2}{2}= \frac{2x_2}{2}- \frac{x_1+ x_2}{2}= \frac{2x_2- x_1- x_2}{2}= \frac{x_2- x_1}{2}$

Re: Simplying expressions

Ah, okay, I understand what I did wrong now.

But I still don't understand how you can do this:

$\displaystyle \sqrt {{{\left( {\frac{x}{2}} \right)}^2} + {{\left( {\frac{y}{2}} \right)}^2}} = \frac{1}{2}\sqrt {{x^2} + {y^2}}~?$

How can you factor something out from inside a radical sign? Is that allowed?

Re: Simplying expressions

Quote:

Originally Posted by

**daigo** Ah, okay, I understand what I did wrong now.

But I still don't understand how you can do this:

$\displaystyle \sqrt {{{\left( {\frac{x}{2}} \right)}^2} + {{\left( {\frac{y}{2}} \right)}^2}} = \frac{1}{2}\sqrt {{x^2} + {y^2}}~?$

How can you factor something out from inside a radical sign? Is that allowed?

$\displaystyle \sqrt {{{\left( {\frac{x}{2}} \right)}^2} + {{\left( {\frac{y}{2}} \right)}^2}} = \sqrt {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{4}} = \sqrt {\frac{{{x^2} + {y^2}}}{4}} = \frac{{\sqrt {{x^2} + {y^2}} }}{{\sqrt 4 }} = \frac{1}{2}\sqrt {{x^2} + {y^2}} $

Re: Simplying expressions

Oh, that makes sense now..

How do you know when to group terms and when you don't?

i.e. $\displaystyle \frac{2x_2 - x_1+ x_2}{2} = \frac{2x_2 - (x_1+ x_2)}{2} = \frac{2x_2- x_1- x_2}{2}$

But when it's something like:

$\displaystyle x^{2} - 4x + 5$ It's not $\displaystyle x^{2} - (4x + 5)$

Re: Simplying expressions

Quote:

Originally Posted by

**daigo** Oh, that makes sense now..

How do you know when to group terms and when you don't?

i.e. $\displaystyle \frac{2x_2 - x_1+ x_2}{2} = \frac{2x_2 - (x_1+ x_2)}{2} = \frac{2x_2- x_1- x_2}{2}$

But when it's something like:

Before you continue go learn basic algebra. **We are not a tutorial service.**

Re: Simplying expressions

But I am taking an Algebra class now...what textbooks do you recommend to supplement my learning? School is not being very helpful, so I have to learn everything on my own..

Re: Simplying expressions

Quote:

Originally Posted by

**Plato** Before you continue go learn basic algebra. **We are not a tutorial service.**

There's no need to be so rude. The OP was just asking why the negative value is distributed over the entire fraction when it doesn't have brackets.

Basically, we treat the fraction as an entire quantity, so when you are subtracting a fraction, you are subtracting ALL terms in the fraction. This is why you need to distribute the negative across all terms in the fraction.

Re: Simplying expressions

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Originally Posted by

**Prove It** There's no need to be so rude.

I would rather be rude than not to understand the mathematics.

Re: Simplying expressions

Quote:

Originally Posted by

**Plato**

If you're referring to that particular question, it's not my fault the question is worded ambiguously.

Re: Simplying expressions

I don't mind if people from the upper-levels of a subject are rude (i.e. my maths professor who responds with "Oh, that's easy" to every question asked towards him and ignores the student entirely; though I don't expect much better academic quality from a community college) but I think it's just counter-productive when a student is willing to learn on his/her own and the only response to that is "go learn on your own" without pointing them towards a specific textbook and maybe even which books/resources to avoid (as to prevent the student from learning the material incorrectly, as I find many websites practice the spread of disinformation).

Re: Simplying expressions

Well Daigo, we've been exposed to "math teachers" too; there's a couple I would have loved to shoot!

But you seem like a reasonable student; you do "see" WHY it's difficult to teach "some stuff" from a

"type it out" site like this (like, no blackboard and stuff...); agree???

You showed in post#6 some weird work(!). I'll try to show the "how" (I'll use a and b !) :

2b - a + b = 2b + b - a = 3b - a

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

2b - (a - b) = 2b - a + b = 3b - a (same as the 1st one)

Make sure you "follow" that before going to something more involved...