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Naily [24]
4 years ago
10

If a polynomial function f(x) has roots 3 and radical 7 what must also be a root of f(x)?

Mathematics
2 answers:
Zepler [3.9K]4 years ago
8 0

Answer:

The answer is -√7

Step-by-step explanation:

∵ The polynomial has roots 3 and √7

∴ It must have root -√7 the conjugate of √7

∴ The roots of f(x) are 3 , √7 , -√7

   (one rational and two irrational conjugate to each other)

finlep [7]4 years ago
4 0

Answer:   -√7 must also be a root of f(x).

Step-by-step explanation:  Given that a polynomial function f(x) has roots 3 and radical 7.

We are to find the other number that must be a root of f(x).

We know that

irrational roots of a polynomial function always occur in conjugate pairs.

That is, if (a + √b) is a root of a polynomial function, then its conjugate pair (a - √b) is also a root of the polynomial function.

For the given polynomial f(x), the given roots are 3 and √7.

Now, \sqrt7=0+\sqrt 7.

Then, the other root will be

0-\sqrt7=-\sqrt7.  

Thus, -√7 must also be a root of f(x).

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The probability that x is lower or equal to a is given by:

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P(X \leq 5) = 1 - e^{-5\lambda}

(a) lambda=.5

P(X \leq 5) = 1 - e^{-5*0.5} = 0.9179

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(b) lambda=0.9

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P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9959 = 0.0041

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9514 1404 393

Answer:

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