## purplemexican one year ago Which values of p and q would make the value of the following expression equal to 29i? (2 – 5i)(p + q)i

1. SolomonZelman

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

2. Purplemexican

expand?

3. LynFran

yes in other words multiply out the bracket

4. Purplemexican

its still confusing

5. SolomonZelman

$$\large\color{black}{ \displaystyle (a+b)(c+d)=ab+ac+bc+bd }$$ have you seen anything like this before?

6. Purplemexican

yes

7. Purplemexican

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

8. SolomonZelman

$$\large\color{black}{ \displaystyle (2 – 5i)(p + q)=2(p+q)-5i(p+q) }$$ same way as: $$(a-b)(C)=a(C)-b(C)$$ ---------------------------------------- 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, $$\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5(-1)(p+q) }$$ $$\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)+5(p+q) }$$

9. SolomonZelman

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

10. SolomonZelman

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

11. SolomonZelman

good luck