trinomial [sp]

[silverlining326]

please help:

√(X^2) - 4X + 4

and its square root of all that

thanks, i dont know how to do the actual symbols, but thank you for the help.

 

[tkhunny]

Sadly, you have provided neither sufficient information nor comprehensible notation. Give it another go and use more parentheses.

 

[stapel]

√(X^2) - 4X + 4 and its square root of all that

What's square root of all which? And what are the instructions?

Please reply with clarification, including the full and exact text of the exercise, the complete instructions, and a clear listing of what you have tried thus far.

For formatting information, please review the links in the "Forum Help" pull-down menu at the very top of the page.

Thank you.

Eliz.

 

[silverlining326]

I hope this is more understandable:
√((X^2)-4X+4)

solving for x

 

[stapel]

I hope this is more understandable: √((X^2)-4X+4)

So the expression is as follows...?

. . . . .\(\displaystyle \sqrt{x^2\,-\,4x\,+\,4}\)

solving for x

You have to have an equation (something with an "equals" sign in it) in order to "solve". All we have so far is an expression.

Eliz.

 

[silverlining326]

ok, it must not be to solve for X... i had asked a friend what the goal of the problem was, she must of written it down wrong...

if possible, can someone please help with simplifying it??

 

[Gene]

Factor x²-4x+4. It comes out as a nice square which you can take the square root of.

 

[silverlining326]

Factor x²-4x+4. It comes out as a nice square which you can take the square root of.

hmm, then maybe im totally lost..

because you can use "-2" right?

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

but -2 isnt a perfect square, or did i skip a part?

 

[Mrspi]

Factor x²-4x+4. It comes out as a nice square which you can take the square root of.

hmm, then maybe im totally lost..

because you can use "-2" right?

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

but -2 isnt a perfect square, or did i skip a part?

You may not have skipped a "part", but I think you have skipped over a "concept":

For any real number n, sqrt(n<SUP>2</SUP>) = | n |

So, sqrt(y<SUP>2</SUP>) = | y |
and sqrt[(k + 3)<SUP>2</SUP>] = | k + 3 |

Now, you have
sqrt[(x - 2)<SUP>2</SUP>]

Can you finish it?

 

[silverlining326]

l x - 2 l

if its not that then im really stupid. lol

 

[Gene]

That's it. (You do know that |x| is the absolute value of x?)
-----------------
Gene

 

[silverlining326]

yes yes i do =]

thank you again for the help, that did help clear up the concept for me..

i think the reason i didnt even remember that was because i didnt understand it, now i deffff understand it =] =] =]