SOLUTION: Can you please help me understand the method of arriving at the answer of this problem? Thank you. In my textbook this was the exercise question Write the product in standard

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Can you please help me understand the method of arriving at the answer of this problem? Thank you. In my textbook this was the exercise question Write the product in standard       Log On


   

Question 121335This question is from textbook Demana, Waits, Foley, Kennedy Precalculus Graphical, Numerical, Algebraic
: Can you please help me understand the method of arriving at the answer of this problem? Thank you.
In my textbook this was the exercise question
Write the product in standard form:
(2+3i)(2-i)
The answer in the back of the book was X= -4 +or- the sqrt of the fraction 8/3.
I don't understand how they arrived at this answer. This question is from textbook Demana, Waits, Foley, Kennedy Precalculus Graphical, Numerical, Algebraic

Found 3 solutions by Earlsdon, checkley71, MathLover1:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the product and write the answer in standard form: %28a%2Bbi%29. Using FOIL:
%282%2B3i%29%282-i%29+=+4-2i%2B6i-3i%5E2 Simplifying this we get:
4%2B4i-3%28-1%29+=+4%2B4i%2B3 = 7%2B4i
The answer that you got from the back of the book doesn't appear to be correct! Did you read it correctly?
On the other hand, I have made the assumption that the i in your problem is i+=+sqrt%28-1%29.
Let's assume that the i is really a variable, so...
%282%2B3i%29%282-i%29+=+4%2B4i-3i%5E2 and we'll rearrange this for ease of computation.
-3i%5E2%2B4i%2B4+=+0 Multiply through by -1 to remove the negative coefficient of i%5E2
3i%5E2-4i-4+=+0 Now we'll using the quadratic formula:i+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where: a = 3, b = -4 and c = -4.
i+=%28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%283%29%28-4%29%29%29%2F2%283%29
i+=+%284%2B-sqrt%2816-%28-48%29%29%29%2F6
i+=+%284%2B-sqrt%2864%29%29%2F6
i+=+%284%2B8%29%2F6 or i+=+4-8%29%2F6
i+=+2 or i+=+-2%2F3 and this doesn't look like your answer either!

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
2+3i
2-i
----------------------------------------
4+6i-2i-3i^2 (i^2)=-1
4-4i-3*-1
4-4i+3
7-4i ANSWER.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
You can use the formula for multiplying two complex numbers:
%28a+%2B+bi%29+%2A+%28c+%2B+di%29+=+%28ac+-+bd%29+%2B+%28ad+%2B+bc%29i

and write the product %282%2B3i%29%282-i%29 in standard form like this:
%28a+%2B+bi%29+%2A+%28c+%2B+di%29+=+%28ac+-+bd%29+%2B+%28ad+%2B+bc%29i
= %282%2A2+%2B+3%29+%2B+%282%28-1%29+%2B+3+%2A2%29i
= %284+%2B+3%29+%2B+%28-2+%2B+6+%29i
= 7+-2i+%2B+6i
= 7+%2B+4i

or, just use the rule for multiplication of two binomials:
%282%2B3i%29%282-i%29
= 4+%96+2i+%2B6i+%96+3%2Ai%5E2……..i%5E2=-1
= 4+%2B+4i+%96+3%2A%28-1%29
= 4+%2B+4i+%2B+3
= 7+%2B+4i+
you said:
"the answer in the back of the book was X=+-4 + or - +sqrt+%288%2F3%29"
I think you saw wrong answer, here you have no x; so, try one more time.