Find all real or imaginary solutions to each equation. Use the method of your choice.
3y² + 4v – 1 = 0
No equation.
3y² +4v - 1 =0
To solve the equation 3y² + 4v - 1 = 0, we can use the quadratic formula. The quadratic formula gives the solutions to any quadratic equation of the form ax² + bx + c = 0.
The quadratic formula is given by:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, y corresponds to x, and we have a = 3, b = 4, and c = -1. Plugging these values into the quadratic formula, we get:
y = (-4 ± √(4² - 4(3)(-1))) / (2(3))
Simplifying further:
y = (-4 ± √(16 + 12)) / 6
= (-4 ± √(28)) / 6
= (-4 ± 2√(7)) / 6
The solutions to the equation 3y² + 4v - 1 = 0 are:
y = (-4 + 2√(7)) / 6
y = (-4 - 2√(7)) / 6
These solutions can be either real or imaginary, depending on the value of √(7).