Question

Which values of x and y would make the following expression represent a real number? (4+5i)(x+ yi)
Ox=4,y= 5
Ox=-4, y = 0
Ox=4, y=-5
Ox=0, y= 5​

Answer 1

Answer:

C

Step-by-step explanation:

We have the expression $(4+5i)(x+yi)$ and we want to find the values of $x$ and $y$ such that the expression will evaluate to a real number.

So, let's first expand the expression. Distribute:

$=4(x+yi)+5i(x+yi)$

Distribute:

$=4x+4yi+5xi+5yi^2$

Simplify:

$=(4x-5y)+(4yi+5xi)$

So, we want to make the second part within the real numbers.

Notice that we only have two ways of doing this: 1) Either both $x$ and $y$ are imaginary numbers themselves canceling out the $i$, or 2), the entire expression equals 0.

Since our answer choices consists of only real numbers, this means that the imaginary part must be equal to 0. So:

$4yi+5xi=0$

We can divide everything by $i$:

$4y+5x=0$

Now, we can use our answer choices.

Running down the list, we can see that the choice that works is C. If we substitute the values of C into the equation, we get:

$4(-5)+5(4)=0\\\Rightarrow -20+20=0\stackrel{\checkmark}{=}0$

Therefore, our answer is C.

Answer 2

Answer:

c

Step-by-step explanation: