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Mathematics 8 Online
OpenStudy (anonymous):

Can someone please help me quickly? I only have about 20 minutes left so can someone try and help ASAP? Here is question: Find the product of (x + (3+5i))2.

OpenStudy (anonymous):

Find the product of (x + (3+5i))2.

OpenStudy (anonymous):

@mathstudent55 Can you help me out here?

OpenStudy (anonymous):

@ganeshie8 Can you help me out here?

OpenStudy (anonymous):

Also if you can show or describe the steps that would be useful...

OpenStudy (zzr0ck3r):

\(a(b+(c+d))=ab+a(c+d) = ab+ac+ad\)

OpenStudy (larseighner):

let a = x and b = 3 + 5i find \((a + b)^2 \) and then substitute back

OpenStudy (anonymous):

@zzr0ck3r Can you elaborate? and @LarsEighner I am going to try it but what do you mean substitute back?

OpenStudy (larseighner):

if you expanded \((a + b)^2\), replace b with 2 + 5i in the expansion.

OpenStudy (anonymous):

@LarsEighner I got x^2+6x+40ix-16 Is that right?

OpenStudy (larseighner):

No. I think you have add terms in ix to terms in i. I could be wrong.

OpenStudy (zzr0ck3r):

\((x + (3+5i))^2=(x + (3+5i))(x + (3+5i))=x^2+2x(3+5i)+(3+5i)^2=\\x^2+6x+10xi+(9+30i+25i^2)=x^2+6x+10xi+9+30i-25=\\x^2+6x+10xi+30i+16\)

OpenStudy (anonymous):

@LarsEighner what do you mean?

OpenStudy (zzr0ck3r):

you combined \(10xi\) with \(30i\)

OpenStudy (anonymous):

@zzr0ck3r thank you :D

OpenStudy (larseighner):

@zzr0ck3r the square of 5i is -25

OpenStudy (zzr0ck3r):

right, hence why I have -25

OpenStudy (anonymous):

Something weird is going on though... Whenever I check for the answer on mathway I get that the answer is supposed to be x^2+6x+9

OpenStudy (zzr0ck3r):

thats wrong

OpenStudy (larseighner):

Then how come your unit term comes out +16?

OpenStudy (zzr0ck3r):

-25+9 = 16

OpenStudy (larseighner):

If it is telling you -25 + 9 = 16 it is wrong.

OpenStudy (zzr0ck3r):

err lol -16, I wrote minus down and put + here

OpenStudy (zzr0ck3r):

im dyslexic

OpenStudy (anonymous):

Thank you guys! You both helped me out a lot!

OpenStudy (zzr0ck3r):

\((x + (3+5i))^2=(x + (3+5i))(x + (3+5i))=x^2+2x(3+5i)+(3+5i)^2=\\x^2+6x+10xi+(9+30i+25i^2)=x^2+6x+10xi+9+30i-25=\\x^2+6x+10xi+30i-16\)

OpenStudy (larseighner):

Anyway what @zzr0ck3r wrote in the big equation is right except it should by -16 at the end.

OpenStudy (zzr0ck3r):

edit* \(\huge \ \ \ \ \uparrow\)

OpenStudy (larseighner):

Now it is fixed

OpenStudy (zzr0ck3r):

that much more important stuff is everything else, so I hope you understand @silca if not ask....

OpenStudy (anonymous):

@zzr0ck3r Yeah I understand now it's just mathway was confusing me... I would get the answer you put in the first 3 trials but for some reason I got confused when I uploaded this question and wrote the wrong answer.

OpenStudy (larseighner):

This stuff is much easier if you do not try to work it out horizonatally, but set it up like a regular multiplication.

OpenStudy (larseighner):

\[ \large \begin{align} x +3 &+ 5i \cr x +3 &+ 5i \cr \hline 5ix + 15i &+ 25i^2 \cr 3x \quad + 9 \quad \quad \quad \quad + 15i& \cr x^2 + 3x \quad\quad \quad \quad + 5ix \quad \quad \quad & \cr\hline x^2 + 6x \;\;\; + 9 + 10ix +30i &+ -25 \end{align}\] Then remember to gather your units terms.

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