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Hitman42 [59]
3 years ago
7

Please help!!! 25 points or I’ll mark brainliest!!

Mathematics
1 answer:
iogann1982 [59]3 years ago
4 0

Answer:

The origin is (0, 0) so let's say we have (0, 5) we find that this point is 5 units away from the origin. We find that by subtracting the origin's y coordinate from our (0, 5) point's y coordinate. This method can be used for the x-axis too.

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Which of the following are square roots of —8 + 8i/3? Check all that apply.
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Answer:

Options (2) and (3)

Step-by-step explanation:

Let, \sqrt{-8+8i\sqrt{3}}=(a+bi)

(\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2

-8 + 8i√3 = a² + b²i² + 2abi

-8 + 8i√3 = a² - b² + 2abi

By comparing both the sides of the equation,

a² - b² = -8 -------(1)

2ab = 8√3

ab = 4√3 ----------(2)

a = \frac{4\sqrt{3}}{b}

By substituting the value of a in equation (1),

(\frac{4\sqrt{3}}{b})^2-b^2=-8

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b⁴ - 8b² - 48 = 0

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b²(b² - 12) + 4(b² - 12) = 0

(b² + 4)(b² - 12) = 0

b² + 4 = 0 ⇒ b = ±√-4

                     b = ± 2i

b² - 12 = 0 ⇒ b = ±2√3

Since, a = \frac{4\sqrt{3}}{b}

For b = ±2i,

a = \frac{4\sqrt{3}}{\pm2i}

  = \pm\frac{2i\sqrt{3}}{(-1)}

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But a is real therefore, a ≠ ±2i√3.

For b = ±2√3

a = \frac{4\sqrt{3}}{\pm 2\sqrt{3}}

a = ±2

Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)

Options (2) and (3) are the correct options.

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