# Math First take the tangent of both sides to get

`tan(arctan(x)+arctan(2x)) = tan(pi/4)`

`tan(pi/4) = 1` and `tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b))`

so

`(tan(arctan(x))+tan(arctan(2x)))/(1-tan(arctan(x))tan(arctan(2x))) = 1`

` tan(arctan(a))=a`

so

`(x+2x)/(1-(x)(2x))=1`    Simplifying and cross multiplying we get

`3x = 1-2x^2`    Put in standard form:

`2x^2 + 3x - 1 = 0`   ...

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First take the tangent of both sides to get

`tan(arctan(x)+arctan(2x)) = tan(pi/4)`

`tan(pi/4) = 1` and `tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b))`

so

`(tan(arctan(x))+tan(arctan(2x)))/(1-tan(arctan(x))tan(arctan(2x))) = 1`

` tan(arctan(a))=a`

so

`(x+2x)/(1-(x)(2x))=1`    Simplifying and cross multiplying we get

`3x = 1-2x^2`    Put in standard form:

`2x^2 + 3x - 1 = 0`    Now use the quadradic formula to get

` x = (-3+-sqrt(9-(4)(-1)(2)))/(2(2)) = (-3+-sqrt(9+8))/2 = (-3+-sqrt(17))/4`

So the anwers are `x = (-3+sqrt(17))/4` and `x=(-3-sqrt(17))/4`

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