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3 years ago

Find a third-degree polynomial equation with rational coefficients that has the given numbers as roots. 1 and 3i

Mathematics

1 answer:

andre [41]3 years ago

4 0

Answer:

x³ − x² + 9x − 9 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs. So if 3i is a root, then -3i is also a root.

(x − 1) (x − 3i) (x − (-3i)) = 0

(x − 1) (x − 3i) (x + 3i) = 0

(x − 1) (x² − 9i²) = 0

(x − 1) (x² + 9) = 0

x (x² + 9) − (x² + 9) = 0

x³ + 9x − x² − 9 = 0

x³ − x² + 9x − 9 = 0

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Read below.

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2 years ago

Which equation has a graph with a vertex at (-1,-1)?

11111nata11111 [884]

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3 years ago

Write each complex number in rectangular form. a. 2(cos(135)+i sin(135)) b. 3(cos(120))+i sin(120)) c. 5(cos(5pi/4)+i sin(5pi/4)

bagirrra123 [75]

$\bf r\left[ cos\left( \theta \right)+i\ sin\left( \theta \right) \right]\quad 
\begin{cases}
x=rcos(\theta )\\
y=rsin(\theta )
\end{cases}\implies 
\begin{array}{llll}
x&,&y\\
a&&b
\end{array}\implies a+bi\\\\
-------------------------------\\\\
2\left[ cos\left( 135^o\right)+i\ sin\left( 135^o\right) \right]\impliedby r=2\qquad \theta =135^o
\\\\\\
2\left( -\frac{\sqrt{2}}{2} \right)+i\ 2\left( \frac{\sqrt{2}}{2}\right)\implies -\sqrt{2}+\sqrt{2}\ i$

$\bf -------------------------------\\\\
3\left[ cos\left( 120^o\right)+i\ sin\left( 120^o\right) \right]\impliedby r=3\qquad \theta =120^o
\\\\\\
3\left( -\frac{1}{2} \right)+i\ 3\left( \frac{\sqrt{3}}{2}\right)\implies -\frac{3}{2}+\frac{3\sqrt{3}}{2}\ i$

$\bf \\\\
-------------------------------\\\\
5\left[ cos\left( \frac{5\pi }{4}\right)+i\ sin\left( \frac{5\pi }{4}\right) \right]\impliedby r=5\qquad \theta =\frac{5\pi }{4}
\\\\\\
5\left( -\frac{\sqrt{2}}{2} \right)+i\ 5\left( -\frac{\sqrt{2}}{2}\right)\implies -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\ i$

$\bf -------------------------------\\\\
4\left[ cos\left( \frac{5\pi }{3}\right)+i\ sin\left( \frac{5\pi }{3}\right) \right]\impliedby r=4\qquad \theta =\frac{5\pi }{3}
\\\\\\
4\left( \frac{1}{2} \right)+i\ 4\left( -\frac{\sqrt{3}}{2}\right)\implies \frac{1}{2}-\frac{\sqrt{3}}{2}\ i$

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3 years ago