Exercise 73 Page 393

We want to determine whether 5+2i is a solution to the given equation. To do so, we can substitute x=5+2i into the equation. If we receive a true statement, then it is indeed a solution. Before we do that, recall that the imaginary unit i is the complex number whose square is - 1. i^2=-1 With this in mind, let's substitute 5+2i to the given equation and simplify.

x^2-10x=-29

Substitute

x= 5+2i

( 5+2i)^2-10( 5+2i)? =-29

▼

(a+b)^2=a^2+2ab+b^2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

5^2+2(5)(2i)+(2i)^2-10(5+2i)? =-29

CalcPow

Calculate power

25+2(5)(2i)+(2i)^2-10(5+2i)? =-29

Multiply

Multiply

25+20i+(2i)^2-10(5+2i)? =-29

PowProdII

(a b)^m=a^m b^m

25+20i+2^2 i^2-10(5+2i)? =-29

CalcPow

Calculate power

25+20i+4i^2-10(5+2i)? =-29

IDefPow

i^2=- 1

25+20i+4(-1)-10(5+2i)? =-29

MultPosNeg

a(- b)=- a * b

25+20i-4-10(5+2i)? =-29

SubTerm

Subtract term

21+20i-10(5+2i)? =-29

Distr

Distribute -10

21+20i-50-20i? =-29

SubTerms

Subtract terms

-29=-29 ✓

Since the substitution of 5+2i resulted in a true statement, we know that it is indeed a solution to the given equation.