quiz 6A sol - Solutions to Quiz 6A...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Quiz 6A www.math.ufl.edu/˜harringt October 5, 2007 1. Reduce and simplify (a) (3+2i)5 (1 7i + i2 ) = (3+2i)5 (1 7i − 1) = (3+2i)5 7i ∗0=0 (b) i21 = (i4 )5 ∗ i = (1)5 ∗ i = i 2. Solve for x: (a) 3 + 2|x + 4| = 0 2|x + 4| = −3 −3 |x + 4| = 2 Since |x| ≥ 0 for all x ∈ R then |x + 4| = the solution set is ∅. −3 2 has no solution. So (b) √ 3x + 1 √ 3x + 1 √ ( 3x + 1)2 3x + 1 0 0 3+ = = = = = = x x−3 (x − 3)2 x2 − 6x + 9 x2 − 9x + 8 (x − 8)(x − 1) (1) (2) (3) (4) (5) (6) So x = 8 and x = 1. However in step (3) we could have created extranenous solutions. Thus, we must check our solutions. 1 Check x = 8: 3+ √ 3∗8+1=3+ √ 25 = 8. So 8 works. Check x = 1: 3+ √ 3∗1+1=3+ √ 4 = 5 = 1. So 1 does NOT work. So the solution set is {8}. 3. Determine whether the statement is true or false. (a) If 3 + 2i is a root to a quadratic equation with real coefficients then the other root is 3 − 2i. This statement is TRUE. The roots are complex conjugates of each other. √ √√ (b) −3 −2 = 6 √√ √ √√ This statement is FALSE. −3 −2 = i2 3 2 = − 6. 2 ...
View Full Document

This note was uploaded on 07/12/2011 for the course MAC 1140 taught by Professor Williamson during the Fall '08 term at University of Florida.

Page1 / 2

quiz 6A sol - Solutions to Quiz 6A...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online