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liq [111]
4 years ago
15

A polynomial function P(x) with rational coefficients has the given roots find two additional roots of P(x)=0 i and 7+8i

Mathematics
2 answers:
leonid [27]4 years ago
8 0

Answer:

As, the given polynomial is P(x)=0,

It is given that two of it's roots are i and 7+ 8 i.

If this quadratic equation has four roots, there is no effect whether the powers of x has rational or any real coefficients.Why i have written this because imaginary roots or non real roots occur in pairs.

It is not necessary that this polynomial has only four roots, but number of non real root or imaginary root will be even.

So, if P(x)=0, has two roots → i and 7+ 8 i, other two roots will be -i and 7- 8 i.




den301095 [7]4 years ago
3 0
Note that if a + bi is a root of P(x) = 0, then a – bi is also a root of P(x) = 0.

In this case, i and 7 + 8i are two roots of P(x) = 0. So –i and 7 – 8i are two additional roots of P(x) = 0.
 
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The sum of 2 numbers is 100. 50 x 2 = 100

Twice the first number plus twice the second number is 200.

50+50=100 so therfore 100+100 makes 200

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Iliana was part of a group that was working on changing 0.4 repeated to a fraction. Each member of the group had a different ans
amm1812
\bf 0.444444444\overline{4}\impliedby \textit{and keeps on going}\\\\
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\textit{let's say }\boxed{x=0.444444444\overline{4}}\quad \textit{ thus }10\cdot x=4.44444444\overline{4}
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\textit{wait a minute! }4.44444444\overline{4}\textit{ is really just }4+0.444444444\overline{4}

\bf \textit{but we know }x=0.444444444\overline{4} \textit{ so then }4+0.444444444\overline{4}=\boxed{4+x}
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thus\qquad 10x=4+x\implies 10x-x=4\implies 9x=4\implies \boxed{x=\cfrac{4}{9}}

you can check in your calculator.

anyhow, to get the "recurring decimal to fraction", you start by setting to some variable, "x" in this case, then move the repeating part to the left of the point by multiplying it by some power of 10, and then do the equating.
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Lina20 [59]

Answer:

(0 , 3)

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y = \frac{3}{2}x + 3

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