If x + sqrt {({x^2} + 1)} = a, then x = | vedclass

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Question

(b) $x + \sqrt {{x^2} + 1} = a \Rightarrow \sqrt {{x^2} + 1} = a - x$

$ \Rightarrow $${x^2} + 1 = {(a - x)^2} = {x^2} - 2ax + {a^2}$

$ \Rightarr

Vedclass · Accepted Answer

${1 \over 2}(a - 1/a)$

Vedclass · Answer

(b) $x + \sqrt {{x^2} + 1} = a \Rightarrow \sqrt {{x^2} + 1} = a - x$

$ \Rightarrow $${x^2} + 1 = {(a - x)^2} = {x^2} - 2ax + {a^2}$

$ \Rightarrow $$x = {{1 - {a^2}} \over { - 2a}} = {{{a^2} - 1} \over {2a}} = {1 \over 2}\left( {a - {1 \over a}} \right)$.

If $x + \sqrt {({x^2} + 1)} = a,$ then $x =$

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