Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
2. Section 8.2
Continue to next subchapter

Exercise 73 Page 393

Substitute 5+2i into the given equation. Recall the definition of the imaginary unit i.

Yes, see solution.

Practice makes perfect

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
( 5+2i)^2-10( 5+2i)? =-29
â–¼
(a+b)^2=a^2+2ab+b^2
5^2+2(5)(2i)+(2i)^2-10(5+2i)? =-29
25+2(5)(2i)+(2i)^2-10(5+2i)? =-29
25+20i+(2i)^2-10(5+2i)? =-29
25+20i+2^2 i^2-10(5+2i)? =-29
25+20i+4i^2-10(5+2i)? =-29
25+20i+4(-1)-10(5+2i)? =-29
25+20i-4-10(5+2i)? =-29
21+20i-10(5+2i)? =-29
21+20i-50-20i? =-29
-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.