`x^2+sqrt2x-4=0 ; x=sqrt3,x=-2sqrt2`

if f(x)=sqrt(x+2sqrt(2x-4))+sqrt(x-2sqrt(2x-4)) then f'(3)+f'(6)= (A) 0 (B) 1 (C) 2 (D) 1/2

If f(x)=sqrt(x+2sqrt(2x-4))+sqrt(x-2sqrt(2x-4)) then the value of 10 f'(102^(+)) , is

Solve the following quadratic equations : (i) 9x^(2) - 8x+2=0 (ii) sqrt7 x^(2) +x+ sqrt7=0 (iii) 2sqrt3x^(2) -sqrt2 x + 3sqrt3 =0

For x- (4sqrt6)/(sqrt2+sqrt3) , what is the value of (x+2sqrt2)/(x-2sqrt2)+(x+2sqrt3)/(x-2sqrt3) ?

If Delta(x)=|{:(4x-4,(x-2)^2,x^3),(8x-4sqrt2,(x-2sqrt2)^2,(x+1)^3),(12x-4sqrt3,(x-2sqrt3)^2,(x-1)^3):}| . Then the coefficient of x in Delta(x) is

Solve for x:sqrt(2x-1)+sqrt(3x-2)=sqrt(4x-3)+sqrt(5x-4)