Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Answer:
x = 4 + sqrt(73) or x = 4 - sqrt(73)
Step-by-step explanation by completing the square:
Solve for x:
(x - 12) (x + 4) = 9
Expand out terms of the left hand side:
x^2 - 8 x - 48 = 9
Add 48 to both sides:
x^2 - 8 x = 57
Add 16 to both sides:
x^2 - 8 x + 16 = 73
Write the left hand side as a square:
(x - 4)^2 = 73
Take the square root of both sides:
x - 4 = sqrt(73) or x - 4 = -sqrt(73)
Add 4 to both sides:
x = 4 + sqrt(73) or x - 4 = -sqrt(73)
Add 4 to both sides:
Answer: x = 4 + sqrt(73) or x = 4 - sqrt(73)
Answer:
Quadrant III
Step-by-step explanation: