My Math Forum help with problem

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 24th, 2009, 04:55 PM #1 Member   Joined: May 2009 Posts: 37 Thanks: 0 help with problem the problem is this solve this problem by completing the square Code: 2x^2-12x+11 I keep getting 2(x-3)^2+11 they (the book) says the answer is 2(x-3)^2-7 could anyone offer a quick step by step so i can see where I am wrong or where they are wrong? thanks
 July 24th, 2009, 05:58 PM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: help with problem Expand $2(x-3)^2$ algebraically, then this obviously tells you the reason.
 July 24th, 2009, 06:04 PM #3 Member   Joined: May 2009 Posts: 37 Thanks: 0 Re: help with problem ya, no duh, but if i don't know the answer, how do i find it? what I am saying is this, I move the 11 over to the other side complete the square and end up with 2(x-3)^2 = -11 then i move the 11 back over and have 2(x-3)^2 + 11 and if i expand the problem then I have 2(x^2 -6x +9) + 11 which then becomes 2x^2 -12x +18 + 11 which becomes 2x^2 - 12x + 29 good luck factoring that am I saying 2x^2 - 12x +36 - 7 ??? i am getting lost here
 July 24th, 2009, 06:07 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: help with problem Be smart. Do what I tell you, and you will logically understand the meaning of my words.
 July 24th, 2009, 06:12 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: help with problem Sigh. Expanding the thing will become $2x^2-12x+18$, and this is what you've got for part of your answer, which was wrong. But, this obviously tells you how to solve it now.
 July 24th, 2009, 06:13 PM #6 Member   Joined: May 2009 Posts: 37 Thanks: 0 Re: help with problem I did that before I posted the question...
 July 24th, 2009, 06:14 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: help with problem I hope they've taught you subtraction back in the school days.
 July 24th, 2009, 06:15 PM #8 Member   Joined: May 2009 Posts: 37 Thanks: 0 Re: help with problem if i am not given the answer, how do i work it forwards............
July 24th, 2009, 06:20 PM   #9
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Re: help with problem

Quote:
 Originally Posted by SteveThePirate if i am not given the answer, how do i work it forwards............ tell me that or stop posting you arrogant guy(cant curse for fear of being banned)
This problem is not hard, as long you follow what I say dude, calm down pirate! LET ME CLARIFY.

The problem asks to CTS in order to find the values for $x$ when the equation equals to . Let , where $k$ is our value we are looking for. Expand the expression, and we will get $2x^2-12x+11= 2(x-3)^2+k = 2(x^2-6x+9)+k = 2x^2-12x+18+k$. $18+k= 11$, so the $k= 11-18 = -7$ will tell ya what you need. $7= 2(x-3)^2$, and so on.

 July 24th, 2009, 06:26 PM #10 Member   Joined: May 2009 Posts: 37 Thanks: 0 Re: help with problem that does make sense to me but my big problem is that, the answer is not a given, so you do not know that it will be - 7 at the end, and I realized how to get the answer once i had the answer, but, is there always going to be a ( + k at the end of a factor problem? how would i choose it to be -7 if i do not know that it is supposed to be -7?

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