Solve (x+3)2+(x+3)-2=0
2 answers:
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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:
y = - 2x + 13
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Calculate m using the slope formula
m = (y₂ - y₁ ) / (x₂ - x₁ )
with (x₁, y₁ ) = (5, 3) and (x₂, y₂ ) = (4, 5)
m = =
= - 2, thus
y = - 2x + c ← is the partial equation
To find c substitute either of the 2 points into the partial equation
Using (4, 5), then
5 = - 8 + c ⇒ c = 5 + 8 = 13
y = - 2x + 13 ← equation of line