your original thinking that
sqrt(1+x) -sqrt(1-x) = sqrt (1-x-sqrt(1+x))
is incorrect
test it with a value of x
e.g. suppose x = .5
then you would have
√1.5 - √.5 = √(1 - .5 + √1.5 )
or
appr .5176 = appr .455 , not even close
What was the rest of the question?
sqrt(1+x) -sqrt(1-x) = sqrt (1-x-sqrt(1+x))
Please help
This question is.....
.
.
√(1+x) - √(1-x) = whole root under OR
√1-x-√(1+x)
Please help me
3 answers
But how did you get +√1.5 it's-√1.5
But still according to you the question is incomplete.
However if x=0 then you would get a true equation. But how do I solve it to get even at least 0
But still according to you the question is incomplete.
However if x=0 then you would get a true equation. But how do I solve it to get even at least 0
If you're trying to solve the equation,
√(1+x) - √(1-x) = √(1-x-√(1+x))
(√(1+x) - √(1-x)) = √(1-x-√(1+x))^2
(1+x) - 2√(1-x^2) + (1-x) = 1-x-√(1+x)
2 - 2√(1-x^2) = 1-x-√(1+x)
1+x = 2√(1-x^2)-√(1+x)
(1+x)^2 = 4(1-x^2) - 4√((1-x^2)(1+x)) + (1+x)
5x^2+x-4 = -4(x+1)√(1-x)
(5x^2+x-4)^2 = 16(x+1)(1-x)
25x^4+26x^3-23x^2-24x = 0
x(x+1)^2(25x-24) = 0
x = 1, 0, 24/25
Check for extraneous roots which might not fit the original equation.
√(1+x) - √(1-x) = √(1-x-√(1+x))
(√(1+x) - √(1-x)) = √(1-x-√(1+x))^2
(1+x) - 2√(1-x^2) + (1-x) = 1-x-√(1+x)
2 - 2√(1-x^2) = 1-x-√(1+x)
1+x = 2√(1-x^2)-√(1+x)
(1+x)^2 = 4(1-x^2) - 4√((1-x^2)(1+x)) + (1+x)
5x^2+x-4 = -4(x+1)√(1-x)
(5x^2+x-4)^2 = 16(x+1)(1-x)
25x^4+26x^3-23x^2-24x = 0
x(x+1)^2(25x-24) = 0
x = 1, 0, 24/25
Check for extraneous roots which might not fit the original equation.