Math, asked by shivanshusharma14022, 5 months ago

find the value of √x²+a² + √x²-a² / √x²+a² - √x²-a² when x²=a⁴+1 / 2
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Answered by Shivampanwar2020
0

Multiply the denominator term with both numerator and denominator

</p><p></p><p>\frac { ( \sqrt { x ^ { 2 } + a ^ { 2 } } + \sqrt { x ^ { 2 } - a ^ { 2 } } ) ^ { 2 } } { ( x ^ { 2 } + a ^ { 2 } ) - ( x ^ { 2 } - a ^ { 2 } ) }

which becomes

\frac { x ^ { 2 } + a ^ { 2 } + x ^ { 2 } - a ^ { 2 } + 2 \sqrt { x ^ { 2 } + a ^ { 2 } } \sqrt { x ^ { 2 } - a ^ { 2 } } } { ( x ^ { 2 } + a ^ { 2 } ) - ( x ^ { 2 } - a ^ { 2 } ) }

\frac { 2 x ^ { 2 } + 2 \sqrt { ( x ^ { 2 } + a ^ { 2 } ) ( x ^ { 2 } - a ^ { 2 } ) } } { 2 a ^ { 2 } }

\frac { x ^ { 2 } + \sqrt { x ^ { 4 } - a ^ { 4 } } } { a ^ { 2 } }

\frac { x ^ { 2 } + \sqrt { x ^ { 4 } ( 1 - \frac { a ^ { 4 } } { x ^ { 4 } } ) } } { a ^ { 2 } }

\frac { x ^ { 2 } ( 1 + \sqrt { 1 - \frac { a ^ { 4 } } { x ^ { 4 } } } ) } { a ^ { 2 } }

\frac { a ^ { 4 } + 1 ( 1 + \sqrt { 1 - \frac { 4 a ^ { 4 } } { ( a ^ { 4 } + 1 ) ^ { 2 } } } } { 2 a ^ { 2 } }

solving what's inside the root

</strong></p><p><strong>[tex]\frac { ( a ^ { 4 } ) ^ { 2 } + 1 - 4 a ^ { 4 } } { ( a ^ { 4 } + 1 ) ^ { 2 } }

\sqrt { \frac { ( a ^ { 4 } - 1 ) ^ { 2 } } { ( a ^ { 2 } + 1 ) ^ { 2 } } }

now it can come out of the root in two ways. either this

\frac { a ^ { 4 } + 1 ( 1 + \frac { a ^ { 4 } - 1 } { a ^ { 4 } + 1 } ) } { 2 a ^ { 2 } }

or

\frac { a ^ { 4 } + 1 ( 1 - \frac { a ^ { 4 } - 1 } { a ^ { 4 } + 1 } ) } { 2 a ^ { 2 } }

From 1st way

\left. { a ^ { 2 } ( a ^ { 4 } + 1 ) \text { or } \frac { a ^ { 4 } + 1 } { a ^ { 2 } } } \\ { = a ^ { 6 } + a ^ { 2 } \quad \text { or } \quad a ^ { 2 } + \frac { 1 } { a ^ { 2 } } } \right.

It depends on the value of a which will determine how that fraction will come out of the square root. since no value is given we must take account all possibilities.

Hope you have understood the solution.

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