SOLUTION: Write a quadratic equation for which the following are solutions for x: ± 3i This one has me a bit confused. Here's what I get x = ±i sqrt 9 x = ± sqrt -9 x^2=-9 x^2+9=0 That

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Write a quadratic equation for which the following are solutions for x: ± 3i This one has me a bit confused. Here's what I get x = ±i sqrt 9 x = ± sqrt -9 x^2=-9 x^2+9=0 That      Log On


   

Question 135617: Write a quadratic equation for which the following are solutions for x: ± 3i
This one has me a bit confused. Here's what I get
x = ±i sqrt 9
x = ± sqrt -9
x^2=-9
x^2+9=0
That's not a complete quadratic equaiton, but I don't know what to do next or if I'm doing it right to begin with.
Found 2 solutions by Earlsdon, solver91311:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
I see nothing wrong with what you have done and your answer is correct.
When you say that it is not a "complete" quadratic equation, what you mean is that in the general form:ax%5E2%2Bbx%2Bc+=+0 , b happens to be zero but that doesn't mean that you don't have a quadratic equation.
The proof, of course, is to solve your answer to see if you get the given solutions of: x = 3i and x = -3i
x%5E2%2B9+=+0 Subtract 9 from both sides.
x%5E2+=+-9 Take the square root of both sides.
x+=+sqrt%28-9%29 or x+=+-sqrt%28-9%29 and sqrt%28-9%29+=+3sqrt%28-1%29 = 3i so...
x+=+3i or x+=+-3i

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
You sort of took a convoluted path to get there, but you did arrive at the right answer.

Here's how I would have done it:

A polynomial equation has a root x=a if and only if x-a is a factor of the polynomial. Furthermore, a polynomial equation of degree n has exactly n roots.

You are given two roots, so you know that you must have a quadratic (degree 2) polynomial equation.

The roots are 3i and -3i, therefore, the factors of the polynomial must be x-3i and x-%28-3i%29=x%2B3i.

Multiply the two factors together: %28x-3i%29%28x%2B3i%29=x%5E2%2B9 (factorization of the difference of two squares in reverse)

So your equation is x%5E2%2B9=0.

And, by the way, it most certainly is a 'complete' quadratic equation of the form ax%5E2%2Bbx%2Bc=0: a=1, b=0, and c=9. The only restriction on the possible values for the coefficients in ax%5E2%2Bbx%2Bc=0 is that a%3C%3E0. b and c can be any values, including zero.