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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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2 years ago
How do uou do this problem
Ronch [10]
The first step to solving an equation like this is to find the slope of a line that will be perpendicular to the line given. The slope of a line that's perpendicular to another line is the negative reciprocal. The negative reciprocal of -1/5 is 5. So, so far our equation is y = 5x + b. Now, to find what b is equal to, we should substitute the values of x and y from the point (1,2) since we know that our line goes through the point. Our equation becomes:

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3 years ago
Find the volume of this cylinder. Use 3 for T.
DIA [1.3K]

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3 0
3 years ago
The point V(4, –1) is rotated 90° clockwise around the origin. What are the coordinates of its image V’? V’(1, 4) V’(–1, 4) V’(–
tresset_1 [31]

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Step-by-step explanation:

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that means the right part of the x-axis turns into the bottom part of the y-axis, and the bottom part of the y-axis into the left part of the x-axis.

so, positive x turns into negative y, and the negative y turns into negative x.

therefore, (4 -1) turns into (-1, -4).

7 0
2 years ago
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