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lys-0071 [83]
3 years ago
11

Find a quadratic equation with roots -1, 4i and -1 - 4i

Mathematics
1 answer:
vazorg [7]3 years ago
3 0
\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=-1+4i\implies &x+1-4i=0\\
x=-1-4i\implies &x+1+4i=0
\end{cases}
\\\\\\
(x+1-4i)(x+1+4i)=\stackrel{y}{0}
\\\\\\\
[(x+1)~~-~~(4i)][(x+1)~~+~~(4i)]=y
\\\\\\\
[(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y
\\\\\\
(x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y
\\\\\\
x^2+2x+1~~+16=y\implies x^2+2x+17=y
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nasty-shy [4]

Answer:

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Step-by-step explanation:

All values rounded to the hundredth

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Answer:

D) 3x^2 - 12

Step-by-step explanation:

Using PEMDAS;

There is no need to evaluate the part of the equation (x^2 - 8) because is no need to, as it is already in its simplest form.

We must evaluate the part of the equation continuing with, "- (-2x^2+4)," as it is not in its simplest form.

Evaluating "- (-2x^2+4)":

Step 1: Distributing the negative

Once distributing the negative symbol amongst the values within the parenthesis according to PEMDAS, we get "2x^2 - 4" as the product.

Step 2: Consider the rest of the equation to evaluate

Since the part of the equation is still in play here as it is a part of the original equation to be solved, we must evaluate it as a whole to get the final answer.

Thus,

x^2 -8 + 2x^2 - 4 = ___

*we can remove the parenthesis as it has no purpose, since it makes no difference.

Evaluating for the answer, we get,

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Step-by-step explanation:

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