SOLUTION: how do I solve - x2 - 6x + 16 = 0 It says there are no real answers, but that there are complex answers. What does this mean please? Thank you.

Question 128762: how do I solve -
x2 - 6x + 16 = 0
It says there are no real answers, but that there are complex answers. What does this mean please?
Thank you.
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=-6, and c=16

Negate -6 to get 6

Square -6 to get 36 (note: remember when you square -6, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

After simplifying, the quadratic has roots of

or

Notice if we graph the quadratic , we get

graph of

And we can see that there are no real roots

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how do I solve -
x^2 - 6x + 16 = 0
It says there are no real answers, but that there are complex answers. What does this mean please?
--------------------
The Real Numbers include the Rational Numbers and the Irrational Numbers.
Complex Numbers are a set of numbers that look like a+bi and where the
a and the b are Real numbers and "i" is the square root of negative-one.
-------------------
Your Problem:
x^2 - 6x + 16 = 0
Use the Quadratic formula to get:
x = [6 +- sqrt(36 - 4*16)]/2
x = [6 +- sqrt(-28)]/2
x = [6 +- 2isqrt(7)]/2
x = [3 +- isqrt(7)]
x = 3+isqrt(7) or x = 3-isqrt(7)
=========================
Notice that the "3" is rational, and the sqrt(7) is irrational
and the answers are in the form a+bi
=========================================
Cheers,
Stan H.
Cheers,
Stan H.

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