`sqrt((x-y)/(x+y)) -sqrt((x+y)/(x-y))` `` `sqrt(xy^-1+x^-1y)` `` `(x^-1 + 2x^-2)^-0.5` ``

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to simplify the expression `sqrt((x-y)/(x+y)) - sqrt((x+y)/(x-y))` , using the properties of radicals, such that:

`sqrt((x-y)/(x+y)) - sqrt((x+y)/(x-y)) = (sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y))`

You need to rationalize the denominators of fractions, such that:

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (sqrt(x-y)sqrt(x+y))/(x+y) - (sqrt(x+y)sqrt(x-y))/(x-y)`

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (sqrt((x-y)(x+y)))/(x+y) - (sqrt((x+y)(x-y)))/(x-y)`

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (sqrt(x^2 - y^2))/(x+y) - (sqrt(x^2 - y^2))/(x-y)`

You need to bring the fractions to a common denominators, such that:

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (xsqrt(x^2 - y^2) - ysqrt(x^2 - y^2) - xsqrt(x^2 - y^2) - ysqrt(x^2 - y^2))/((x+y)(x-y))`

Reducing duplicate terms yields:

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (-2ysqrt(x^2 - y^2))/(x^2 - y^2)`

Hence, reducing the expression `sqrt((x-y)/(x+y)) - sqrt((x+y)/(x-y))` to a simpler form, yields:

`(sqrt(x-y))/(sqrt(x+y)) - (sqrt(x+y))/(sqrt(x-y)) = (-2ysqrt(x^2 - y^2))/(x^2 - y^2)`

You need to reduce the expression `sqrt(xy^(-1) + x^(-1)y)` to a simpler form, such that:

`sqrt(xy^(-1) + x^(-1)y) = sqrt(x/y+ y/x)`

Bringing the fractions under the square root to a common denominator, yields:

`sqrt(xy^(-1) + x^(-1)y) = sqrt((x^2+y^2)/(xy))`

Rationalizing the expression, yields:

`sqrt(xy^(-1) + x^(-1)y) = (sqrt(xy(x^2+y^2)))/(xy)`

Hence, reducing the expression `sqrt(xy^(-1) + x^(-1)y)`   to a simpler form, yields:

`sqrt(xy^(-1) + x^(-1)y) = (sqrt(xy(x^2+y^2)))/(xy)`

You need to reduce the expression `(x^(-1) + 2x^(-2))^(-0.5)` to a simpler form, hence, you need to use the negative power property first, such that:

`(x^(-1) + 2x^(-2))^(-0.5) = 1/((1/x + 2/(x^2))^(1/2))`

Converting the rational power into radical yields:

`(x^(-1) + 2x^(-2))^(-0.5) = 1/(sqrt(1/x + 2/(x^2)))`

`1/(sqrt(1/x + 2/(x^2))) = 1/(sqrt((x + 2)/(x^2)))`

`1/(sqrt((x + 2)/(x^2))) = 1/((sqrt(x + 2))/|x|)`

`1/((sqrt(x + 2))/|x|) = |x|/(sqrt(x + 2))`

Rationalizing the fraction yields:

`|x|/(sqrt(x + 2)) = (|x|(sqrt(x + 2)))/(x+2)`

Hence, reducing the expression `(x^(-1) + 2x^(-2))^(-0.5)` to a simpler form, yields:

`(x^(-1) + 2x^(-2))^(-0.5) = (|x|(sqrt(x + 2)))/(x+2).`

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oldnick | (Level 1) Valedictorian

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`sqrt((x-y)/(x+y))-sqrt((x+y)/(x-y))=(sqrt((x-y)^2)-sqrt((x+y)^2))/(sqrt((x+y)(x-y)))=`

`=((x-y)-(x+y))/(sqrt(x^2+y^2))=-2y/sqrt(x^2-y^2)`

`sqrt(xy^(-1)+ x^(-1)y)=sqrt(x/y+y/x)=sqrt((x^2+y^2)/(xy))=1/sqrt(xy)sqrt(x^2+y^2)`

`sqrt(x^(-1)+2x^(-2))=sqrt(1/x+2/x^2)=sqrt((x+2)/x^2)=1/xsqrt(x+2)`

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