To solve by completing the square, what needs to be moved in this equation?
1 answer:
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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:
And if we count the number of zeros before the number 7, we can rewrite the number like this:
We cansolve this problem also counting the number of positions that we need to move the decimal point to the right in order to obtain the first number (7)
And the best option would be:
B. 7 x 10-7
Step-by-step explanation:
For this case we have the following number given:
And if we count the number of zeros before the number 7, we can rewrite the number like this:
We cansolve this problem also counting the number of positions that we need to move the decimal point to the right in order to obtain the first number (7)
And the best option would be:
B. 7 x 10-7