" (iii) "(x^(2)-2x)^(2)-23(x^(2)-2x)+120

Solve (x^(2) - 2x)^(2) - 23 (x^(2) - 2x) + 120 = 0

The equation (x^(2)+x+1)^(2)+1=(x^(2)+x+1)(x^(2)-x-5) for x in(-2,3) will have number of solutions.1 b.2 c.3d.0

Factorize each of the following algebraic expressions: x^(2)+14x+45 (2) x^(2)-22x+120x^(2)-11x-42(4)a^(2)+2a-3

int_( then a+b equals to )^( If )dx=((a+x^(2))^((3)/(2))*(x^(2)-b))/(120x^(5))+C

Divide the following polynomial p(x) by polynomial s(x) p(x)=2x^(3)-13x^(2)+23x-12,s(x)=2x-3

If the lines 2x+3y+1=0 and 3x-y-4=0 lie along diameters of a circle of circumference 10 pi ,then the equation of the circle is (A) x^(2)+y^(2)-2x+2y-23=0 (B) x^(2)+y^(2)+2x-2y-23=0 (C) x^(2)+y^(2)+2x+2y-23=0 (D) x^(2)+y^(2)-2x-2y-23=0

Solution of the equation 3^(2x^(2))-2.3^(x^(2)+x+6)+3^(2(x+6))=0 is

Solution of the equation 3^(2x^2)-2.3^(x^2+x+6)+3^(2(x+6)) =0 is

If the lines 2x + 3y + 1 = 0 and 3x - y-4 = 0 lie along two diameters of a circle of circumference 10 pi , then the equation of circle is (i) x^(2) + y^(2) + 2x + 2y + 23 = 0 (ii) x^(2) + y^(2) - 2x - 2y - 23 = 0 (iii) x^(2) + y^(2) - 2 x + 2y - 23 = 0 (iv) x^(2) + y^(2) + 2x - 2y + 23 = 0

Factorise x^(3)-23x^(2)+142x-120