Question

Solve \[\sqrt{(x-1)}=\sqrt{(x)}-1\]

Answer 1

square both sides

Answer 2

x-1=x+1?

Answer 3

No @Hollywood_chrissy

Answer 4

simply sqaure both sides and then taking the surds part on one side again square it to get the ans
....x-1=x+1-2Vx
or 2Vx=2
or Vx=1
or x=1

Answer 5

use (a+b)^2 on the right side and solve it

Answer 6

sorry I am comfused by the first step.

Answer 7

you square left term it becomes x-1. right.

Answer 8

yes

Answer 9

to solve the term on right use (a+b)^2.
so it becomes ((sqrt(x)-1)^2 = x+1-2sqrt(x)

Answer 10

huh why?

Answer 11

why not?

Answer 12

so
x-1=x+1-2sqrt(x)
2sqrt(x)=2
sqrt(x)=1
x=1

Answer 13

sorry but I do not understand, why can't we just sqaure the rhs. what is this (a+b)^2?????

Answer 14

your answer is correct, but I do understand the work.

Answer 15

*not

Answer 16

when you square the left term it becomes\[(\sqrt{x-1})^{2}=(x-1)^{\frac{ 1 }{ 2} \times 2}=x-1\]

Answer 17

you got this

Answer 18

yes

Answer 19

now square the right term
\[(\sqrt{x}-1)^{2}=(\sqrt{x}-1) \times (\sqrt{x}-1)\]
\[x-\sqrt{x} -\sqrt{x}+1\]=\[x-2\sqrt{x}+1\]
which can be computed using identity
\[(a+b)^{2}=(a+b) \times (a+b) = a^{2}+b^{2}+2ab \]

Answer 20

Is it clear?

Answer 21

oh I was doing \[\sqrt{x}^2-1^2\]

Answer 22

yes

Answer 23

thanks @pradipgr817

Answer 24

:)
my pleasure

Answer 25

well go for hit and trial,,1 is a solution CLEARLY !!

Answer 26

Great