The numbers i�ã5 and -i�ã5 are solutions to which equation?
a. x^2+�ã5=0
b. x^2+5=0
d. x^2-5=0
thank you do you know how to do it though?
of course. I got the answer
you say √5i and -√5i are the solutions, so
(x-√5i)(x+√5i) = 0
x^2 - (√5i)^2 = 0
x^2 - 5i^2 = 0
but i^2 = -1, so
x^2 + 5 = 0
omg thank you soooo much (:
To determine which equation has the solutions i�ã5 and -i�ã5, we need to evaluate each equation using these values.
a. x^2+�ã5=0
We substitute i�ã5 into the equation:
(i�ã5)^2+�ã5 = -5+�ã5 = -4+�ã5
This expression is not equal to 0, so i�ã5 is not a solution to this equation.
b. x^2+5=0
We substitute i�ã5 into the equation:
(i�ã5)^2+5 = -5+5 = 0
This expression is equal to 0, so i�ã5 is a solution to this equation.
c. x^2-5=0
We substitute i�ã5 into the equation:
(i�ã5)^2-5 = -5-5 = -10
This expression is not equal to 0, so i�ã5 is not a solution to this equation.
Thus, the equation that has the solutions i�ã5 and -i�ã5 is option b. x^2+5=0.