square root of (2x-5) + square root of (x+1)=3 Any one?? all the math websites blow ass and its 2am, class early tomorrow morning

Damn, I wish you had asked this earlier - Eyebette teaches this kind of stuff daily as part of her job, but alas, she's gone to bed. Sorry man and good luck!

its cool player, thanks though damn me and my procrastinating!! teach lets us take tests home and its due tomorrow morning, waited till now to finish up last 2 questions, o well

I am sure you can find a website to just plug the equations into? On another note, I was always good at math.. but looking at that I have no clue what to do, or what your even looking for.. I remember doing something like it.. but ya. dunno why they teach that... You only use that type of math in a select fields.. no real practical use.

i'm with ya there player.... all the websites i tried are bunk. thought i'd throw a hail mary up before i crashed, its cool

I'll try, to get rid of the square root you must square the square root and get. 2x - 5 + x +1 =9 Add the -5 and the +1 and get -4. Then add the 2x and the x . 3x -4 = 9 Add the -4 to both sides. 3x = 13 Divide both sides by 3 and get X = 13/3 * Doesn't seem right ,but not sure if I understand the question, sorry.

This question really doesn't make sense as they are all to the power of something different. So they cant be added together. So if this makes sense to do. Although the variables are the same (x), the powers are not the same (1, 2,and 5). You can't simplify these terms because only the variables are the same, and both the variables and the powers need to be the sam

Lets see of how much help I can be as it's also 2am for me. First off darklide that's a good attempt but your starting set up was wrong. You have to square BOTH ENTIRE sides. So, the easiest method is to put one square root on one side, and the other on the other side. (makes for simpler math also known as a better solution.) sqrt(2x-5) + sqrt(x+1)=3 1. Move one square root to the other side. sqrt(2x-5)=3-sqrt(x+1) 2. Square both sides. The ENTIRE sides. (sqrt(2x-5))^2=(3-sqrt(x+1))^2 ( 3. Simplify it The square of a binomial. Perfect square trinomials - A complete course in algebra Check this link if you have problems with taking the square of a binomial. It takes the form of (a-b)^2=(a^2-2ab+b^2) which is what you get when you times out (a-b)(a-b) 2x-5=9-6sqrt(x+1)+(x+1) 9. Move over terms. x-15=-6sqrt(x+1) 10. Square both sides again to remove sqrt. (x-15)^2=(-6sqrt(x+1)^2 11. Simplify using (a-b)^2 formula again for the left side this time. x^2-30x+225=36(x+1) 12. Move everything to one side making the equation = to 0 and now it's a quadratic equation which you can solve either by completing the square, factoring, or by the quadratic formula. x^2-66x+189=0 13. Since this is able to factor, I will use that method. And the factor form is (x-63)(x-3)=0 14. Solve for x X=63, x=3 15. NOW IS THE MOST IMPORTANT PART. YOu have to check each value of x to make sure that when plugged back into the original equation to make sure we don't have a negative number in the root, and that the statement is equal to 3. 16. The answer is x=3 since 63 gives us 19 as the answer, while when you plug in 3 for the value of x you get 3 the equation is true.

1. I believe this problem is solved by factoring by grouping. So try grouping it by what you can factor out. x^5-9x^3+8x^2-72=0 2. Factor what you can on each side. What I put in parenthesis is what I see can be factored. (x^5-9x^3)+(8x^2-72)=0 3. For the left pair you can take out x^3 and for the right you can take out 8. x^3(x^2-9)+8(x^2-9)=0 4. Notice how what we're left with is the same, that's a GOOD sign, that what we have is right. So what you do with that is bring what is the same in the parenthesis and that's one factor, and then what's left on the outside as another factor. So: (x^3+8)(x^2-9)=0 5. Set each factor as equal to 0 and solve. 6a Solving for the left factor gives. (x^3+8)=0 Factor it more using the formula (a^3+b^3)=(a+b)(a^2-ab+b^2) So; (x+2)(x^2-2x+4)=0 For the real number part (x+2)=0 so x=-2 And then solve that quadratic equation using either completing the square or the quadratic formula which will give you x = 1-i sqrt(3) and x = 1+i sqrt(3) as the complex answer. Back to the other part of the factor from earlier: (x^2-9)=0 1. Move the number to the right side and square root, or factor it. Either method. x^2=9 x=+or-3 so x=3 x=-3 So all the answers to that problem are: x=3 x=-3 x=-2 x=1-isqrt(3) x=1+isqrt(3)

That was where i was going with that also. and it worked for me. until the end... looks good to me!! thanks for the help player!!

You're welcome, just maybe next time procrastinate a little less so you can get an answer sooner? Also if you don't have a dedicated program for solving equations like this you can try using Wolfram|Alpha—Computational Knowledge Engine It does a pretty good job, that and a combo of google works wonders for checking problems.

Yea procrastination is a disease that I suffer from. Not sure what all sites I tried last night but have up on them late last night. Didn't even think to just Google the actual equations. /facepalm..... Good looking out though, that was just what I needed player

Thanks for everyone's help. Wolframalpha is a great site but it doesn't show the work, i.e. how they got to the answer... still... didn't know about that site at all and will use it in the future if/when i get stuck. Again, thanks guys

Amazing website when you're stuck on a math problem. Also, I wish math was still this easy. Differential Equations and Multivariable Calculus FTL.

What? its a take home test, plus my man ^^ up there worked it out so well i was able to see what he did, work it out myself and remember wtf i was supposed to do Damn!! I know who i'm pm'ing next time