Prove that. The equation whose roots are radii of escribed circles is `x^3 - 2x^2 (r+R) + s^2x-s^2r =0`

A: The equation whose roots are the reciprocals of the roots of 2x^(3) + 7x^(2) - 6x + 1 = 0 is x^(3) - 6x^(2) + 7x + 2 =0 . R: the equation whose roots are the reciprocals of those of f(x) = 0 is f(1/x) = 0.

A: the quadratic equation whose roots are the reciprocals of the roots of 3x^2 - 7x +2=0 is 3x^2 +7x +2=0 R : the quadratic equation whose roots are the reciprocals of the roots of the quadratic equation f(x) is f(1/x)=0

Show that the radii of the three escribed circles of a triangle are roots of the equation, x^3 - x^2 (4R +r)+ x s^2 - r s^2 =0 .

If alpha,beta are the roots of the equation px^(2)-qx+r=0, then the equation whose roots are alpha^(2)+(r)/(p) and beta^(2)+(r)/(p) is (i) p^(3)x^(2)+pq^(2)x+r=0 (ii) px^(2)-qx+r=0 (iii) p^(3)x^(2)-pq^(2)x+q^(2)r=0 (iv) px^(2)+qx-r=0

Show that that the radii of the three escribed circles of a triangle are the roots of the equation x^3-x^2(4R+r)+xs^2-rs^2=0

In Delta ABC the roots of the equation x^3-(r+4R)x^2+s^2x-s^2r=0 are

If s_1,s_2,s_3.........s_r are the sum of the products of the roots taken 'r' at a time then for x^5 - x^2 + 4x - 9 = 0 => s_3 +s_4 - s_5 =

If r_1 , r_2 , r_3 are the radii of the escribed circles of a Delta ABC and if r is the in radius then r_1r_2r_3-r( r_1r_2+r_2r_3+r_3r_1)

The abscissa of two A and B are the roots of the equation x ^(2) - 4x - 6=0 and their ordinates are the roots of x ^(2) + 5x - 4=0. If r is the radius of circle with AB as diameter, then r is equal to