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3 years ago

Type the correct answer in the box,

Mathematics

1 answer:

HACTEHA [7]3 years ago

3 0

We'll assume the quadratic equation has real coefficients

Answer:

The other solution is x=1-8i.

Step-by-step explanation:

The Complex Conjugate Root Theorem

if P(x) is a polynomial in x with real coefficients, and a + bi is a root of P(x) with a and b real numbers, then its complex conjugate a − bi is also a root of P(x).

The question does not specify if the quadratic equation has real coefficients, but we will assume that.

Given x=1+8i is one solution of the equation, the complex conjugate root theorem guarantees that the other solution must be x=1-8i.

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Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

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So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

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So the middle integral is

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