Create an equation that results in at least one extraneous solution. Work through your equation, justify each step, and explain how the solution is extraneous.
I AM WILLING TO WORK. PLZ HELP
there are three parts and i have no clue of what to do!!!plz
@Ghostgate @Bixslo !!!!
I'm kind of lost on what to do... I'm trying to figure out what an extraneous solution is, because of this I may not be much help to you. Sorry...
This will take some time for me to figure but when that time comes someone else would of helped you. But there are lots of good people here that can help you with math.
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.
Alright, I think I can begin to help you, you'll just have to give me awhile to write out my next reply, hopefully this helps you. ;)
That sword art online picture though ;)
ok that's fine. I just need someone to attempt with me.
\[\sqrt{x+4=x-2}\]
I don't know if that could be my answer or am i wrong
Alright, first you have this equation: (To be honest, I'm not good with making example equations, so I kind of stole various bits for this one.) x^2/x - 4 = 16/x - 4 - Alright, first you view the equation, knowing that x can't equal 4 otherwise the equation would become undefinable due to how you will have 0's as your denominators. So...: x^2/x - 4 = 16/x - 4 - Multiply both equations by (x - 4). This would cause you to have 1 as your denominator(I'm hoping you can understand why, because I don't fully understand it myself.). x^2 = 16 - Now that you are left with this you won't simply find the square root of 16, instead you shall take an alternative measure which is completely abnormal, in my opinion at least. x^2 - 16 = 0 - Now, we shall solve by splitting apart this. (x + 4)(x - 4) = 0 - Having this means that: (x + 4) = 0 [or] (x - 4) = 0 - You can solve these normally now. x = 4 [or] x = -4 - But in the beginning we know that x can't equal 4, so...: x = 4 is our extraneous solution. However this is a basic interpretation of it, I don't think I've learned about extraneous solutions or I just don't remember them, whatever the case, I hope this helps, I'm not entirely sure if your answer is correct, let me try to review it. Again, hopefully this helps in someway! ;) {---Ghostgate---} @Austin1617 Lol - I found the picture while looking for pictures of Kirito and Asuna together. I'm going to take it you like SAO?
@zasharra For an equation that leads to an extraneous solution, you got it! I managed to recheck and your answers fit personally, nice job! ;) - I'm pretty sure your solution works out, have a great day! {---Ghostgate---}
@Ghostgate wait i still need you
Alright, what is it?
So for number 1 i just put my equation down and that's the entire answer? AND THERE IS ANOTHER PART
my bad one the caps.
Well it says to simply create an equation that results in at least one extraneous solution, so yes your equation should work. So... what's this 'other part'? It's completely fine, don't worry about the caps.
Using the properties of exponents and radicals, design at least three different equivalent forms of x². You must show how each one can be simplified back to x² in two or more steps. Stretch your mind and get creative! Keep in mind that something too simple, like x • x would not be acceptable since it takes only one step to convert it to x².
Sword Art Online is a good anime @Ghostgate
With this part i have no clue.
@Austin1617 Yeah it is. I'm glad to see there's another fan of it. ;) {---} Alright, as for this, we need to create three different ways to end up with x^2 that takes at least 2 steps to do so. This shouldn't be too difficult.
Thanks for helping me by the way!!! I'm at a dead line and NO ONE would help.
i even tried to cheat on this assignment because it was so hard. (but my teacher is letting me do it over)
Alright, here's one at least: 1. √(x^6 / x^2) = x^2 - Since you divide you get x^4 and square root of that = x^2, so at least two steps here.
@zasharra I'm glad I can help, I just hope you do well this time around. ;)
Second one: x(x) + 5 - 5 = x^2 - This is a bit unusual, but it gives you the answer, you multiply to get x^2 and then you subtract 5 from the positive 5 to get 0, leaving you with x^2.
Third and final one: √x^2(x) = x^2 - I believe you would find the square root, resulting in x and then multiplying it against x to get x^2. @zasharra Do I get a medal for this? I know I sound pretty stupid requesting it, but I figured you would have given me one by now. I don't care about the medals as long as I help people and it sounds as if all I do is care about it, but... well I like to know my help was appreciated, I guess.
How do i give medals? im new to this site
Click the 'best response' button and it will give someone a medal. ;) - It's as simple as that. Also, welcome to OpenStudy then!
Do you have any other questions?
did you get it? so sorry about that
yes i dod
Yes, I did, thank you and it's no problem! ;)
Substitute your birthday into the equation Square root of the quantity x minus y + m = d, where y is the last two digits of your birth year, m is the month, and d is the day. If you were born on 7/10/1856 like Nikola Tesla, your equation would be Square root of the quantity x minus 56 + 7 = 10. Solve for x and identify if it is an extraneous solution.
i did this one but i dont know if its right
@Ghostgate Square root of the quantity x - 99 + 5 = 14.
My b-day is May 14, 1999. I don't know how to solve for x. and did you make up those equations your self in question two?
Alright, so for this, let me quickly go over these steps because I have to go eat really soon: Step 1: Add -99 and 5 together to get -94. Step 2: Add 94 to both sides of the equation giving you: √x = 108 Step 3: Find the square root of 108. Step 4: Once you plug it back into the equation and multiply you will get: 108 - 99 + 5 = 14 Step 5: 108 - 99 = 9 Step 6: 9 + 5 = 14 So your answer should be correct, if I did my math right, anyhow I gtg. See ya!
Thank you so much i wish i could give you more medals!! god bless
@zasharra You're welcome! I'm glad I could help. ;) - Also to answer your previous question, yes I made up, however stole some bits and pieces, the equations for question 2, as I stated before.
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