How would I write a quadratic equation having the given numbers as solutions
4i, -4i
factors would be
(x-4i)(x+4i)
so x^2 - 16i^2 = 0
x^2 + 16 = 0
or x = ±4i
square both sides
x^2 = 16i^2
x^2 = -16
x^2 + 16 = 0
i liked YAY MATH! the best. it was funny but it also taught me alot, of ourcse i doubt i'll remember any of it in the mornin, but i watched the rest of the magician video and learned how to like multiply square routes or whatever, partially. i didn't really like the like math playground place, i got sorta confused there, but i like the like cool math one. i played this game though called like snorzies, i have no idea how i ended up on it, but that wasn't exactly related to math, but i saw some other things that were useful on the cool math page.
To write a quadratic equation having 4i and -4i as solutions, you can use the fact that complex solutions come in conjugate pairs.
Since the solutions 4i and -4i are conjugates, we can use the fact that conjugate pairs have the same real part and opposite imaginary parts.
Let's denote the unknown quadratic equation as ax^2 + bx + c = 0.
The real part of the solutions is 0, so the sum of the solutions is also 0.
Therefore, according to the sum of solutions formula for a quadratic equation, we have:
(-b / a) = 0
This implies that b = 0.
Since the only term in the quadratic equation that would give rise to imaginary solutions is the term involving the square root of -1 (i), we need to have i^2 in the equation.
i^2 = -1.
So, our quadratic equation becomes:
ax^2 + c = 0
Now, let's substitute the known solutions in the equation:
For the solution 4i: when x = 4i, we have a(4i)^2 + c = 0. Simplifying this, we get:
16ai^2 + c = 0
16a(-1) + c = 0
-16a + c = 0 (equation 1)
For the solution -4i: when x = -4i, we have a(-4i)^2 + c = 0. Simplifying this, we get:
16ai^2 + c = 0
16a(-1) + c = 0
-16a + c = 0 (equation 2)
Now, let's solve equations 1 and 2 simultaneously:
From equation 1, we have c = 16a.
Substituting this into equation 2, we get:
-16a + 16a = 0
0 = 0
This means that the equations are dependent, and there are infinitely many quadratic equations that have 4i and -4i as solutions.
However, an example of a quadratic equation that satisfies these conditions is:
ax^2 + 16a = 0
where a represents any nonzero real number.
To write a quadratic equation with the given solutions, follow these steps:
Step 1: Identify the given solutions
The given solutions are 4i and -4i.
Step 2: Use the fact that complex solutions come in conjugate pairs
Since the given solutions are complex conjugates, the quadratic equation will have the form: (x - 4i)(x + 4i) = 0
Step 3: Simplify the equation
To simplify the equation, use the difference of squares formula:
(x - 4i)(x + 4i) = (x)^2 - (4i)^2 = x^2 - 16i^2
Step 4: Recall the definition of i^2
The definition of i^2 is -1.
Step 5: Replace i^2 with -1 in the equation
Substitute -1 for i^2 in the equation:
x^2 - 16(-1) = x^2 + 16 = 0
This is the final quadratic equation.
So, the quadratic equation with solutions 4i and -4i is:
x^2 + 16 = 0