Determine the sum of imaginary roots of the equation (2x + x - 1) ( 4x2 + 2x - 3) = 6
Given, (2x + x - 1) ( 4x2 + 2x - 3) = 6
Put 2x2 + x = y
Discriminants,
D1 = 1 + 4 x 3 = 13 > 0, real roots
D2 = 4 - 16 = - 12 < 0, imaginary roots
Sum of roots of 4x + 2x + 1 = 0 is
Let a, b, c be three real numbers, such that a + 2b + 4c = 0, Then, the equation ax2 + bx + c = 0
has both the roots complex
has its roots lying within - 1 < x < 0
has one of roots equal to
has its roots lying within 2 < x < 6
If are the roots of the equation x2 + x + 1 = 0, then the equation whose roots are is
x2 - x - 1 = 0
x2 - x + 1 = 0
x2 + x - 1 = 0
x2 + x + 1 = 0
For the real parameter t, the locus of the complex number in the complex plane is
an ellipse
a parabola
a circle
a hyperbola
If be the roots of the quadratic equation x2 + x + 1 = 0, then the equation whose roots are f is
x2 - x + 1 = 0
x2 - x - 1 = 0
x2 + x - 1 = 0
x2 + x + 1 = 0
The roots of the quadratic equation are
imaginary
real, rational and equal
real, irrational and unequal
real, rational and unequal
The quadratic equation x2 + 15 + 14 = 0 has
only positive solutions
only negative solutions
no solution
both positive and negative solution