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LenaWriter [7]
3 years ago
10

X^4+2x^2-48=0 what is the solution set

Mathematics
1 answer:
jolli1 [7]3 years ago
6 0

Answer:

x. = sqrt 6 and x = - sqrt 6

Step-by-step explanation:

x^4 + 2x^2-48=0

(x^2 - 6) (x^2 + 8) = 0

x^2 - 6 = 0

x = + sqrt 6 and x = - sqrt 6

x^2 + 8 = 0

x^2 = sqrt -8 (a negative number doesn’t have a real square root)

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Step-by-step explanation:

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Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.
lesya692 [45]

Answer:

\displaystyle f(x) = -2(x-2)(x^2+4)

Step-by-step explanation:

We want to find a third degree polynomial with zeros <em>x </em>= 2 and <em>x</em> = 2i and f(-1) = 30.

First, note that by the Complex Root Theorem, since <em>x</em> = 2i is a root, <em>x</em> = -2i must also be a root.

Hence, we will have the three factors:

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Where <em>a</em> is the leading coefficient.

Expand and simplify the second and third factors:

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Hence:

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Since f(-1) = 30:

\displaystyle \begin{aligned}  f(x) &= a(x-2)(x^2+4) \\ \\ (30) &= a((-1)-2)((-1)^2+4) \\ \\ 30 &= -15a \\ \\ a&= -2\end{aligned}

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