purplemexican
  • purplemexican
Which values of p and q would make the value of the following expression equal to 29i? (2 – 5i)(p + q)i
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
Ok, expand the (2-5i)(p+q)i and what do you then get?
purplemexican
  • purplemexican
expand?
LynFran
  • LynFran
yes in other words multiply out the bracket

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purplemexican
  • purplemexican
its still confusing
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle (a+b)(c+d)=ab+ac+bc+bd }\) have you seen anything like this before?
purplemexican
  • purplemexican
yes
purplemexican
  • purplemexican
i need to look into which property is which still get them confused
SolomonZelman
  • 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, \(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5(-1)(p+q) }\) \(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)+5(p+q) }\)
SolomonZelman
  • 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
SolomonZelman
  • SolomonZelman
solve the system 0=5(p+q) 29=2(p+q) (---> this eq. results after canceling the i's on both sides.)
SolomonZelman
  • SolomonZelman
good luck

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