For what real number k does the product (25 + ki)(3+2i) equal a real number?
I tried dividing by i. So I got: ((25/i)+k)((3/i+2) which would equal (-25i+k)*(-3i+2)
which means k would=25i. I don't think that's right but can you guys check it?
when you FOIL the binomials, the i terms have to cancel
(25 * 2i) + (3 * ki) = 0
Thank you!
meow
To find the real number value of k that makes the product (25 + ki)(3 + 2i) equal a real number, we can follow these steps:
Step 1: Expand the expression:
(25 + ki)(3 + 2i) = 25(3) + 25(2i) + ki(3) + ki(2i)
Step 2: Simplify the expression:
75 + 50i + 3ki + 2ki²
Step 3: Remember that i² is equal to -1:
75 + 50i + 3ki - 2k
Step 4: Collect the real and imaginary terms separately:
(75 - 2k) + (50 + 3k)i
For the product to be a real number, the imaginary term (50 + 3k)i must be equal to zero, which means 50 + 3k = 0.
Step 5: Solve for k:
50 + 3k = 0
3k = -50
k = -50/3 ≈ -16.67 (rounded to two decimal places)
So, the real number value of k that makes the product (25 + ki)(3 + 2i) equal a real number is approximately -16.67.