Question

can anyone sove this?? challenging...

Answer 1

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here is your answer :




√x+1 - √x-1 = -1
(√x+1) = (√x-1 - 1)
(x+1 ) = (√x-1 - 1) power 2
(x+1) = (x-1)+1 - 2√x-1
x-x+1-1+1 = -2√x-1


1 = -2√x-1 -----------> EQ (1)

according to equation (1)......

1 = -2√x-1
-1/2 = √x-1
(-1/2) power 2 = x-1
1/4 = x-1
x= 1/4+1
x= 1+4/4

x = 5/4

Answer 2

$Given : \sqrt{x + 1} + \sqrt{x - 1} = -1$

On squaring both sides, we get

$= \ \textgreater \ ( \sqrt{x + 1} )^2 = (-1 - \sqrt{x - 1})^2$

$= \ \textgreater \ x + 1 = 2 \sqrt{x - 1} + x$

$= \ \textgreater \ 1 = 2 \sqrt{x - 1}$

On squaring both sides, we get

$= \ \textgreater \ 1^2 = (2 \sqrt{x - 1})^2$

$= \ \textgreater \ 1 = 2^2(x - 1)$

$= \ \textgreater \ 1 = 4(x - 1)$

$= \ \textgreater \ 1 = 4x - 4$

$= \ \textgreater \ 4x = 5$

$= \ \textgreater \ x = \frac{5}{4}$


Therefore, the final answer is : $\boxed{x = \frac{5}{4} }$


Verification:

$= \ \textgreater \ \sqrt{ \frac{5}{4} + 1 } + \sqrt{ \frac{5}{4} - 1 }$

$= \ \textgreater \ \sqrt{ \frac{9}{4} } + \sqrt{ \frac{1}{4} }$

$= \ \textgreater \ \frac{3}{2} + \frac{1}{2}$

$= \ \textgreater \ \frac{4}{2}$

$= \ \textgreater \ 2 \neq -1$



Therefore the given equation has No solution...