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## If The Complex Number X = 3 + Bi And |X|2 = 13, Which Is A Possible Value Of B?

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## Answers ( 4 )

## If The Complex Number X = 3 + Bi And |X|2 = 13, Which Is A Possible Value Of B?

If you’ve ever studied complex numbers, you know that they are made up of two parts: a real part and an imaginary part. While the real part is easy to compute, understanding the imaginary part may require some extra effort. Let’s say we have a complex number x = 3 + bi, and we are asked to find the value of b if |x|2 = 13. In this article, we will show you how to do just that by using algebraic manipulation and basic math operations. So if you’re ready, let’s dive right in!

## What are complex numbers?

Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit. The number a is called the real part of the complex number, and b is called the imaginary part.

The absolute value of a complex number is the distance from the origin to the point represented by the complex number. In other words, it is the magnitude of the complex number. The absolute value of a complex number x = a + bi is given by |x| = √(a^2 + b^2).

One possible value of b is √(|x|^2 – a^2).

## The absolute value of a complex number

When working with complex numbers, the absolute value is defined as the distance of a complex number from the origin in the complex plane. In other words, it is the magnitude of a complex number. The absolute value of a complex number x = + bi can be written as |x| = .

There are two possible values for b that satisfy this equation: b = – and b = .## The equation |X|2 = 13

The answer to this question depends on what value you take for |x|. If you take |x| = 3, then a possible value for b is 4. However, if you take |x| = 5, then a possible value for b is -2.

## Possible values of B

There are two possible values of B:1) If X is a complex number, then |X|=sqrt(x^2+y^2). If we set |X|= and solve for y, we get y=sqrt(-x^2). Since we know that B must be a real number, this means that the only possible value of B is sqrt(-x^2).

2) The other possible value of B is when x=0. In this case, |X|=sqrt(y^2)=y. So the only possible value of B is y.

## Conclusion

In this article, we have discussed the problem of determining which is a possible value of B if the complex number X = 3 + Bi and |X|2 = 13. After solving for B in terms of X, we found that any value between -4 and 4 will satisfy the given equation. Therefore, it is evident that any real number within this range could be considered a possible value of B. This concludes our discussion on finding a suitable answer to the given question.

If the complex number x + 3bi is given, and the value of x2 equals 13, then it can be concluded that b is a real number with a certain value. To determine what this value might be requires further examination of the equation. By solving for b in the equation, it can be determined that b = -4/3 or 2.

This can be seen when examining the equation more closely. First, isolate for “b” by subtracting 13 from both sides of the equation which yields x2 – 13 = 3bi and then divide both sides by 3b to get (x2 – 13)/3b = i. Therefore, b must equal either -4/3 or 2 since its denominator cannot equal 0.

Have you ever encountered a complex number like X = 3 + Bi before? If so, you know that in mathematics, a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit.

But, what if you’re asked to solve a problem that involves a complex number like X = 3 + Bi and you’re asked to determine which is a possible value of B?

The answer lies in understanding the concept of a complex number’s absolute value, also known as its modulus. The absolute value of a complex number X = a + bi is defined as |X| = √(a2 + b2).

Therefore, if the problem states that |X|2 = 13, then that means that √(a2 + b2) = √13, or a2 + b2 = 13.

In this case, a is equal to 3, so 3² + b² = 13, or b² = 10. Thus, the possible value of B is ±√10, or ±3.14.

So, if the complex number X = 3 + Bi and |X|2 = 13, then one possible value of B is 3.14.

Hopefully this example has helped you understand the concept of complex numbers a little better and how to solve problems related to them.

Have you ever wondered what the answer to this complex equation might be? Well, let’s break it down and see what we can find!

First off, we’re told that the complex number X is equal to 3 + Bi. This means that the real part of the number is 3, while the imaginary part is B.

Next, we’re told that |X|2 = 13. In other words, we need to find the absolute value of X, and then square it, which will give us the value of 13.

Now, we can use some simple algebra to solve for B. We can start by subtracting 3 from both sides of the equation, which will leave us with |X|2 – 3 = 13 – 3.

Next, we can take the square root of both sides. This will give us |X| = √10. Now, we can substitute this back into our original equation and solve for B.

If we plug in |X| = √10, we get 3 + Bi = √10. We can then subtract 3 from both sides, leaving us with Bi = √10 – 3.

Finally, we can take the square root of both sides to solve for B. This will give us B = √7.

So, there you have it! If the complex number X = 3 + Bi and |X|2 = 13, then the possible value of B is √7.