What is all of the real and imaginary zeros of #y= (x^2-9 ... - Socratic
Dec 12, 2017 · We have #4# zeros, #3# with multiplicity #3# and #-3,3i# and #-3i# with multiplicity of #1#.
How do you evaluate (3a -9i +2ai +6)/ (a^2+9) + (3-9i+3i+9 ... - Socratic
Aug 14, 2017 · Explanation: The first thing we notice with the two expression here is that the denominators are the same since #a^2+9=9+a^2#.
How do you evaluate # (19- 3i ) ( - 6- 14i )#? - Socratic
-156-248i (19-3i)* (-6-14i)=-114+18i-266i-42=-156-248i
Question #d629b - Socratic
b) I multiply the 2 brackets: #4*7-4*3i+7*3i-3*3i^2=28-12i+21i+9=37+9i# Answer link
How do you find \frac { - 3+ \sqrt { - 9} } { 6}? | Socratic
Explanation: #"note that "sqrt (-9)=3i# #rArr (-3+3i)/6# #= (-3)/6+ (3i)/6=-1/2+1/2i# Answer link
How do you combine like terms in #6- ( 4- 3i ) - ( - 2- 10i )#?
Apr 6, 2017 · See the entire solution process below: First, remove all of the terms from parenthesis. Be careful to handle the signs of each individual term correctly: 6 - 4 + 3i + 2 + 10i Next, group like …
How do you solve 11x ^ { 2} + 10= 2x ^ { 2} - 15? | Socratic
x = +-5/3i Given: 11x^2+10 = 2x^2-15 Subtract the right hand side from the left to get: 9x^2+25 = 0 Now x^2 >= 0 for all real values of x, so this has no real solutions.
How do you divide (5i)/(6+8i)? | Socratic
The answer is =0.4+0.3i The division of complex numbers is z_1/z_2 You multiply the numerator and denominator by the conjugate of the denominator (z_1*barz_2)/ (z_2*barz_2) If z=a+ib Then, barz=a …
How do you write x^4 + 13x^2 + 36 as a product of linear factors ...
Therefor; Y = (X + 4) (X + 9) y = (x^2 + 4) (x^2 + 9) To transform y to linear factors, we can use complex numbers, with i^2 = -1. y = (x - 2i) (x + 2i) (x - 3i) (x + 3i)
Factor 9a^5 -4a^3 -81a^2 +36 completely over the a ... - Socratic
Jul 26, 2017 · Factor #9a^5 -4a^3 -81a^2 +36# completely over the a) intergers c) reals b) rationals d) complex numbers ??