If , then the values of x are
- 2, 2, - 4
- 2, 2, 4
3, 2, - 2
4, 4, 3
B.
- 2, 2, 4
, then
In this case, the equation becomes
x2 - x - 6 = - x - 2
or x2 - 4 = 0
x =
Clearly, x = 2 satisfies the domain of the equation in this case. So, x = 2 is a solution.
Then, equation reduces to x2 - x - 6 = 0 = x + 2 i.e., x2 - 2x - 8 = 0 or x = - 2, 4
Both these values lies in the domain of the equation in this case, so x = - 2, 4 are the roots. Hence, roots are x = - 2, 2, 4
If are the roots of x2 - ax + b = 0 and if = Vn, then
Vn + 1 = aVn + bVn - 1
Vn + 1 = aVn + aVn - 1
Vn + 1 = aVn - bVn - 1
Vn + 1 = aVn - 1 + bVn
If the complex numbers z1, z2 and z3 are in AP, then they lie on a
circle
parabola
line
ellipse
If one root is square of the other root of the equation x2 + px + q = 0 , then the relations between p and q is
p3 - (3p - 1)q + q2 = 0
p3 - q(3p + 1) + q2 = 0
p3 + (3p - 1)q + q2 = 0
p3 + (3p + 1)q + q2 = 0