3 years ago

List 8 numbers that can go into six without leaving a remainder

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer:

they are

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3. n = $2$ +/- $i\sqrt{2}$

Step-by-step explanation:

1. Divide both sides by 2: (x + 7)^2 = 8

Square root both sides: x + 7 = +/- 2 $\sqrt{2}$

Subtract 7 from both sides: x = +/- 2 $\sqrt{2}$ - 7

2. Square root both sides: x - 3 = $\sqrt{-12}$

Since there is a negative inside the radical, we need to have an imaginary number: $i=\sqrt{-1}$ . So, $\sqrt{-12} =i\sqrt{12} =2i\sqrt{3}$

Add 3 to both sides: x = $3$ +/- $2i\sqrt{3}$

3. Divide by -5 from both sides: (n - 2)^2 = -2

Square root both sides: n - 2 = $\sqrt{-2}$

Again, we have to use i: $n-2=\sqrt{-2} =i\sqrt{2}$

Add 2 to both sides: n = $2$ +/- $i\sqrt{2}$

Hope this helps!

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