Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 151573: Perform the indicated operation..
 (9-8i)+(7+2i) .
 (3+4i)(2-6i) .
 (6+3i)/(2+2i) .
: Perform the indicated operation..
 (9-8i)+(7+2i) .
 (3+4i)(2-6i) .
 (6+3i)/(2+2i) .

Answer by jim_thompson5910(9217) About Me  (Show Source):
You can put this solution on YOUR website!
 (9-8i)+(7+2i) Start with the given expression.


 (9+7)+(-8i+2i) Group like terms


 16-6i Combine like terms

So (9-8i)+(7+2i)=16-6i

So the expression is in standard form a+bi where a=16 and b=-6





(3+4i)(2-6i) Start with the given expression.


Now let's FOIL the expression.


Remember, when you FOIL an expression, you follow this procedure:


(highlight(3)+4i)(highlight(2)-6i) Multiply the First terms:(3)*(2)=6.


(highlight(3)+4i)(2+highlight(-6i)) Multiply the Outer terms:(3)*(-6*i)=-18*i.


(3+highlight(4i))(highlight(2)-6i) Multiply the Inner terms:(4*i)*(2)=8*i.


(3+highlight(4i))(2+highlight(-6i)) Multiply the Last terms:(4*i)*(-6*i)=-24*i^2=-24(-1)=24.


6-18*i+8*i+24 Now collect every term to make a single expression.


30-10*i Now combine like terms.


So (3+4i)(2-6i) FOILS to 30-10*i.


In other words, (3+4i)(2-6i)=30-10*i.

So the expression is in standard form a+bi where a=30 and b=-10








(6+3i)/(2+2i) Start with the given expression


((6+3i)/(2+2i))((2-2i)/(2-2i)) Multiply by the fraction by (2-2i)/(2-2i). Note 2-2i is the conjugate of 2+2i.


((6+3i)(2-2i))/((2+2i)(2-2i)) Combine the fractions


((6+3i)(2-2i))/(4+4i-4i-4i^2) Foil the denominator


((6+3i)(2-2i))/(4-4i^2) Combine like terms


((6+3i)(2-2i))/(4-4(-1)) Rewrite i^2 as -1


((6+3i)(2-2i))/(4+4) Multiply


((6+3i)(2-2i))/(8) Add


(12-12i+6i-6i^2)/(8) FOIL the numerator


(12-12i+6i-6(-1))/(8) Rewrite i^2 as -1


(12-12i+6i+6)/(8) Multiply


(18-6i)/(8) Combine like terms.


(18)/(8)-(6i)/(8) Break up the fraction


(9)/(4)-(3/4)i Reduce


So (6+3i)/(2+2i)=(9)/(4)-(3/4)i.


So the expression is in standard form a+bi where a=9/4 and b=3/4