sasogeek
  • sasogeek
if i add (x to one) and subtract one from (x plus root of one squared), i get a number twice as big as (x minus itself), what is x?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AravindG
  • AravindG
First form the equation using the info given
sasogeek
  • sasogeek
go right ahead :)
AravindG
  • AravindG
lol why not you try first :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

sasogeek
  • sasogeek
alright... this is what i get :P lol uhmm... add x to one, so, x+1 and subtract one from x plus root of one squared, (x+1)+(x+1^2-1) that's equal to a number twice as big as x minus itself, so (x+1)+(x+1^2-1)=2*(x-x) correct?
ParthKohli
  • ParthKohli
(Hey, congrads!)
AravindG
  • AravindG
yes
sasogeek
  • sasogeek
nope! LOOOL, look at this line :P and subtract one from x plus root of one squared, (x+1)+(x+1^2-1) where's the root of one squared? it's only one squared, there's no root :P
AravindG
  • AravindG
\[\sqrt{1^2}=1\]
AravindG
  • AravindG
No difference :)
ParthKohli
  • ParthKohli
The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2 - 1\)?
sasogeek
  • sasogeek
that makes all the difference in the initial equation. i think it matters though result may be the same, lol
sasogeek
  • sasogeek
@ParthKohli the full equation is what i have written down, (x+1)+(x+1^2-1)=2*(x-x) but then this part is supposed to be \(\large (x+ \sqrt{1^2}-1) \) :)
ParthKohli
  • ParthKohli
Is the question this?\[(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)\]If so, you have \((x + 1) + (x + \sqrt{1})^2 - 1 = 0\)
sasogeek
  • sasogeek
tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P
AravindG
  • AravindG
actually When you frame the equation it should be important \[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]
ParthKohli
  • ParthKohli
Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...
sasogeek
  • sasogeek
root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)
ParthKohli
  • ParthKohli
No... this can't be
ParthKohli
  • ParthKohli
You will have two solutions then, I guess.
sasogeek
  • sasogeek
nope, just one
AravindG
  • AravindG
just one
AravindG
  • AravindG
\[\sqrt{1^2}=1\]
AravindG
  • AravindG
Okay so why not solve it straight away @sasogeek ?
ParthKohli
  • ParthKohli
Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x - 1)-1 = 0\]\[(x+1) + (x + 1) -1 =0\]
AravindG
  • AravindG
omg \[\sqrt{1} \neq \pm 1\]
ParthKohli
  • ParthKohli
@AravindG Why not?
sasogeek
  • sasogeek
\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)
AravindG
  • AravindG
Because square root is a function having range as positive real numbers
ParthKohli
  • ParthKohli
\[(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range
ParthKohli
  • ParthKohli
Or actually, the domain is nonnegative real numbers.
dumbsearch2
  • dumbsearch2
Wow.
AravindG
  • AravindG
@ParthKohli you are having a serious misconception on square roots .
sasogeek
  • sasogeek
you don't have to spam
ParthKohli
  • ParthKohli
@AravindG Wasn't my demonstration enough?
sasogeek
  • sasogeek
can we just solve my problem and take this to another thread, lol
anonymous
  • anonymous
so where was this problem from?
AravindG
  • AravindG
I can explain it to you separately not here
sasogeek
  • sasogeek
my head, @Peter14
AravindG
  • AravindG
Lets finish this problem first
anonymous
  • anonymous
you have an interesting head.
sasogeek
  • sasogeek
lol why thank you :)
AravindG
  • AravindG
ok @sasogeek proceed :)
ParthKohli
  • ParthKohli
If you want to neglect one root of \(1\), you may. Just consider \(1\) as the root of 1 if you want to. But seriously, this equation has two possible solutions.
dumbsearch2
  • dumbsearch2
LOL @Peter14
AravindG
  • AravindG
@ParthKohli I can make a separate thread for your query .
sasogeek
  • sasogeek
it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD
sasogeek
  • sasogeek
x=\(-\frac{1}{2} \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.