Find all real or imaginary Solutions to each equation. Use the method of your choice?

Question

3v^2 + 4v - 1 = 0

Answer 1

3v^2 + 4v - 1 = 0
3(v^2 + 4v/3) - 1 = 0
3(v^2 + 4v/3 + (2/3)^2 - (2/3)^2) - 1 = 0
3(v^2 + 2(2v/3) + (2/3)^2) - (2^2)/3 - 1 = 0
3(v + 2/3)^2 - 7/3 = 0
3(v + 2/3)^2 = 7/3
(v + 2/3)^2 = 7/9
v + 2/3 = +-Sqrt[7/9]
v = -2/3 +- Sqrt[7]/3
v = (-2 + Sqrt[7])/3 or (-2 - Sqrt[7])/3

Answer 2

when dealing with quadratics, using the quadratic formula is always the best way to solve them.

3v^2 + 4v - 1 = 0

v = (-b ± sqrt(b^2 - 4ac))/(2a)

v = (-4 ± sqrt(4^2 - 4(3)(-1)))/(2(3))
v = (-4 ± sqrt(16 + 12))/6
v = (-4 ± sqrt(28))/6
v = (-4 ± sqrt(4 * 7))/6
v = (-4 ± 2sqrt(7))/6
v = (1/3)(-2 ± sqrt(7))

Answer 3

Perhaps the simplest method of finding solutions is to recognize that the equation is a quadratic of the form:

Ax^2 + Bx + C = 0

Since you already know the quadratic formula:

x = [-B +/- sqrt(B^2 - 4AC)] / 2A

it becomes a simple plug and chug problem.

Answer 4

v = - 4 ± √ [16 + 12 ] / 6

v= [- 4 ± √28] / 6

v = [- 4 ± 2√7] / 6

V = - 2/3 ± (1/3 )√7

Answer 5

v=4+/-sq.rt.[16+12)]/6
=4+/-2sq.rt7/6
=(2/3)+/-(1/3)sq.rt.7