For what real number k does the product (25 + ki)*(3+2i) equal a real number?

it's -50/3

(25 + ki)*(3+2i) = 75+3ki+50i-2k

= 75-2k + (3k+50)i

So, you need 3k+50 = 0, since real numbers have zero imaginary part.

well, geez. You cant solve 3k+50 = 0?

k = -50/3

(25 - 50/3 i)*(3+2i) = 325/3

which is a real number

Um, Lelface the answer isnt 325/3. It's -50/3, like steve said earlier. He was just substituting the answer in to show it worked.

To find the real number k that makes the product (25 + ki) * (3 + 2i) equal a real number, we need to multiply the two complex numbers and set the imaginary part equal to zero.

Let's start by expanding the product using the distributive property:

(25 + ki) * (3 + 2i) = 25 * 3 + 25 * 2i + ki * 3 + ki * 2i

Now, simplify each term:

75 + 50i + 3ki + 2ki^2

Since i^2 is defined as -1, we can substitute -1 for i^2:

75 + 50i + 3ki + 2(-1)

Simplifying further:

75 + 50i + 3ki - 2

Now, to make the product a real number, the imaginary part must be zero. So, we need to ignore the terms with i and equate the remaining real numbers to zero:

75 + 3ki - 2 = 0

Combine like terms:

73 + 3ki = 0

To solve for k, divide both sides of the equation by 3:

(73/3) + ki = 0

This equation can only hold true if the imaginary part, ki, is also zero. Therefore, k = 0.

So, the real number k that makes the product (25 + ki) * (3 + 2i) equal a real number is k = 0.

are you sure? I need the actual answer.

Uhh 325/3 isn't right