If one root `x^2-x-k=0` is square of the other, then `k=` a.`2+-sqrt(5)` b. `2+-sqrt(3)` c. `3+-sqrt(2)` d. `5+-sqrt(2)`

Solve sqrt(3)x^(2) - sqrt(2)x + 3sqrt(3)=0

The length of the chord of the parabola y^2=x which is bisected at the point (2, 1) is (a) 2sqrt(3) (b) 4sqrt(3) (c) 3sqrt(2) (d) 2sqrt(5)

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)

Solve : sqrt(3)x^(2)+5x+2sqrt(3)=0

Solve sqrt(5)x^(2) + x + sqrt(5)=0

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

If x=1+1/(3+1/(2+1/(3+1/(2oo)))) a sqrt(5/2) b. sqrt(3/2) c. sqrt(7/3) d. sqrt(5/3)

If the vertices of the parabola be at (-2,0) and (2,0) and one of the foci be at (-3,0) then which one of the following points does not lie on the hyperbola? (a) (-6, 2sqrt(10)) (b) (2sqrt6,5) (c) (4, sqrt(15)) (d) (6, 5sqrt2)

Find the square root of 5+2sqrt6

If x ,y in R satify the equation x^2+y^2-4x-2y+5=0, then the value of the expression [(sqrt(x)-sqrt(y))^2+4sqrt(x y)]//(x+sqrt(x y)) is a. sqrt(2)+1 b. (sqrt(2)+1)/2 c. (sqrt(2)-1)/2 d. (sqrt(2)+1)/(sqrt(2))