Question

Which values of p and q would make the value of the following expression equal to 29i?

(2 – 5i)(p + q)i

Answer 1

Ok, expand the (2-5i)(p+q)i
and what do you then get?

Answer 2

expand?

Answer 3

yes in other words multiply out the bracket

Answer 4

its still confusing

Answer 5

\(\large\color{black}{ \displaystyle (a+b)(c+d)=ab+ac+bc+bd }\)

have you seen anything like this before?

Answer 6

yes

Answer 7

i need to look into which property is which still get them confused

Answer 8

\(\large\color{black}{ \displaystyle (2 – 5i)(p + q)=2(p+q)-5i(p+q) }\)

same way as: \((a-b)(C)=a(C)-b(C)\)
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And therefore:
\(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5i^2(p+q) }\)

we know \(i^2=-1\)
because \(i^2=\left(\sqrt{-1}\right)^2=-1,\) like any \(\left(\sqrt{a}\right)^2=a\)

So,
\(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5(-1)(p+q) }\)
\(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)+5(p+q) }\)

Answer 9

Now, how would you get 29i in a "a+bi" form?
0+29i

and so here we have this form as well

5(p+q) + 2(p+q)i

so

a=0=5(p+q)
bi=29i=2(p+q)i

Answer 10

solve the system
0=5(p+q)
29=2(p+q) (---> this eq. results after canceling the i's on both sides.)

Answer 11

good luck