Question

Find all real and imaginary solutions t⁴+2t²-15=0

Answer 1

t = +- i square root of 5; t = +- square root of 3

Step-by-step explanation:

t^4+5t^2-3t^2-15 = 0

t^2(t^2+5)-3(t^2+5)=0

(t^2-3)(t^2+5)=0

if t^2-3 = 0 we get t^2 = 3, t = +- square root of 3

if t^2+5= 0 we get t^2 = -5, t = +- i square root of 5

Answer 2

To find the solutions to the equation t⁴+2t²-15=0, we can use the quadratic formula.

Step-by-step explanation:

To find the solutions to the equation t⁴+2t²-15=0, we can rearrange the equation to get t² + 10t - 200 = 0. This is a quadratic equation, so we can solve for t using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

Substituting the values into the formula, we get two solutions: t = 3.96 and t = -1.03. Therefore, the equation t⁴+2t²-15=0 has two real solutions.