Didn't you mean "Find the possible value or values of z? y is not at all present in this equation.
z^2 - 4z + 4 = 0 can be factored into (z-2)(z-2). Thus, the given equation has two equal, real solutions: z=2 and z=-2.
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:
$22.68
Step-by-step explanation:
Answer:
About 62.1371
Step-by-step explanation:
Answer:
So do you already have the answer? Also 1/20 looks right.
Step-by-step explanation: