6x^(2)+kx-sqrt(6)=0;x=(1sqrt(3))/(2)

6x^(2)-sqrt(2)x-2=0

int(x+3sqrt(x^(2))+6sqrt(x))/(x(1+3sqrt(x)))dx

ln Delta ABC cosA= -(2)/(3) the equation with roots sin A ,tan A is 1) 6x^(2)+sqrt(5)x-5=0 2) 6x^(2)-sqrt(5)x+5=0 3) 6x^(2)-sqrt(5)x-5=0 4) 6x^(2)+sqrt(5)x+5=0

If the area of the triangle formed by the lines x=0,y=0,3x+4y-a(a>0) is 1, then a=(A)sqrt(6)(B)2sqrt(6)(C)4sqrt(6)(D)6sqrt(2)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

If f(x)=sqrt(1+cos^(2)(x^(2))), then f'((sqrt(pi))/(2)) is (sqrt(pi))/(6)(b)-sqrt(pi/6)1/sqrt(6)(d)pi/sqrt(6)

solve by factorisation: 3x^(2)-2sqrt(6)x+2=0 and (1)/(x-1)-(1)/(x+5)=(6)/(7)