# What is sqrt(1-x^2)/2 - (x^2-x/(2sqrt(1-x^2))=?Please show your steps

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### 1 Answer

You need to bring the fractions to a common denominator, hence, you need to multiply the first fraction by `sqrt(1-x^2)` such that:

`(sqrt(1-x^2)*sqrt(1-x^2) - (x^2 - x))/(2sqrt(1-x^2)) `

You need to open the brackets and to use a single square root such that:

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

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

You should remember that `sqrt(a^2) =|a| = +-a` , hence, reasoning by analogy, yields

`sqrt((1-x^2)^2) = |1 - x^2|`

Considering `|1 - x^2| = 1 - x^2` yields:

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

You need to find the roots of numerator -`2x^2+x+1 = 0 ` such that:

`x_(1,2) = (-1+-sqrt(1+8))/(-4) => x_(1,2) = (-1+-3)/(-4)`

`x_1 = -1/2, x_2 = 1`

You may write the factored form of quadratic `-2x^2+x+1` such that:

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

Substituting `-2(x + 1/2)(x - 1)` for `-2x^2+x+1` in the given expression yields:

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

You need to convert the difference of squares `1 - x^2` into a product such that:

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

`((x + 1/2)(1-x))/(sqrt((1-x)(1+x)))`

Reducing by `sqrt(1-x)` yields:

`((x + 1/2)sqrt(1-x))/(sqrt(1+x))`

You need to multiplicate by `sqrt(1+x)` to remove the square root from denominator such that:

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

**Hence, using `|1 - x^2| = 1 - x^2` and performing the simplification yields `((x + 1/2)sqrt(1-x^2))/(1+x).` **

Considering `|1 - x^2| = -(1 - x^2) = x^2 - 1` yields:

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

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

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

**Hence, using `|1 - x^2| =x^2 - 1` and performing the simplification yields `-sqrt(1-x^2)/(2(1+x)).` **