Question

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?

Answer 1

First form the equation using the info given

Answer 2

go right ahead :)

Answer 3

lol why not you try first :)

Answer 4

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?

Answer 5

(Hey, congrads!)

Answer 6

yes

Answer 7

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

Answer 8

\[\sqrt{1^2}=1\]

Answer 9

No difference :)

Answer 10

The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2 - 1\)?

Answer 11

that makes all the difference in the initial equation. i think it matters though result may be the same, lol

Answer 12

@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 \(\large (x+ \sqrt{1^2}-1) \)
:)

Answer 13

Is the question this?\[(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)\]If so, you have \((x + 1) + (x + \sqrt{1})^2 - 1 = 0\)

Answer 14

tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P

Answer 15

actually

When you frame the equation it should be important

\[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]

Answer 16

Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...

Answer 17

root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)

Answer 18

No... this can't be

Answer 19

You will have two solutions then, I guess.

Answer 20

nope, just one

Answer 21

just one

Answer 22

\[\sqrt{1^2}=1\]

Answer 23

Okay so why not solve it straight away @sasogeek ?

Answer 24

Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x - 1)-1 = 0\]\[(x+1) + (x + 1) -1 =0\]

Answer 25

omg \[\sqrt{1} \neq \pm 1\]

Answer 26

@AravindG Why not?

Answer 27

\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)

Answer 28

Because square root is a function having range as positive real numbers

Answer 29

\[(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range

Answer 30

Or actually, the domain is nonnegative real numbers.

Answer 31

Wow.

Answer 32

@ParthKohli you are having a serious misconception on square roots .

Answer 33

you don't have to spam

Answer 34

@AravindG Wasn't my demonstration enough?

Answer 35

can we just solve my problem and take this to another thread, lol

Answer 36

so where was this problem from?

Answer 37

I can explain it to you separately not here

Answer 38

my head, @Peter14

Answer 39

Lets finish this problem first

Answer 40

you have an interesting head.

Answer 41

lol why thank you :)

Answer 42

ok @sasogeek proceed :)

Answer 43

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.

Answer 44

LOL @Peter14

Answer 45

@ParthKohli I can make a separate thread for your query .

Answer 46

it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD

Answer 47

x=\(-\frac{1}{2} \)