F(x)=(-2/((x+y-2)^(1/2))-(x+y+2)^(1/2) the only irrational part of this expression is the (x+y-2)^(1/2) in the denominator, so, to rationalize this, you multiply the numerator and denominator by the denominator, as well as the other parts of the expression also, you must multiply the -sqrt(x+y+2) by sqrt(x+y-2)/sqrt(x+y-2) to form a common denominator (-2)/(x+y-2)^(1/2)-(x+y+2)^(1/2)(x+y-2)^(1/2)/(x+y-2)^(1/2) (common denominator) (-2-(x^2+xy+2x+xy+y^2+2y-2x-2y-4))/(x+y-2)^(1/2) (FOIL) (-2-x^2-y^2-2xy+4)/(x+y-2)^(1/2) (Distribute negative) (-x^2-y^2-2xy+2)/(x+y-2)^(1/2) (Simplify numerator) (-x^2-y^2-2xy+2)(x+y-2)^(1/2)/(x+y-2)^(1/2)(x+y-2)^(1/2) (Rationalize denominator by multiplying both top and bottom by sqrt) (-x^2-y^2-2xy+2)((x+y-2)^(1/2))/(x+y-2) (The function is now rational) =(-x^2-y^2-2xy+2)(sqrt(x+y-2))/(x+y-2)
I'm not sure if this dividing,multiplying,subtracting or adding but for adding i got 6 as my answer.I don't know if this the correct answer but its better than having no answer right?
D) construction of the angle bisector. So if you have a triangle that is not equilateral the perpendicular bisectors would not work so you need angle bisectors. I have drawn a diagram The green is the angle bisectors and the yellow the perpendicular bisectors. You can see the problem straight away