Algebra Know-It-ALL
To figure out the value of j−^3 using the difference of powers law, note that j−^3 =j^1 −^4 =j/j^4 We have determined that j^4 ...
666 Worked-Out Solutions to Exercises: Chapters 21 to 29 (b) To find the difference (4 +j5)− (3 −j8), we multiply the second com ...
If we let a= 4, b= 5, c= 3, and d=−8, then we have c^2 +d^2 = 32 + (−8)^2 = 9 + 64 = 73 and therefore (4+j5) / (3 −j8) = [4 × 3 ...
668 Worked-Out Solutions to Exercises: Chapters 21 to 29 Now we can make these substitutions: Leta from the formula equal a in ...
absolute values equal to k. These are shown as points in Fig. C-1. There are infinitely many complex numbers with absolute value ...
670 Worked-Out Solutions to Exercises: Chapters 21 to 29 We’ve been told to factor the following quadratic, and we’ve been assu ...
Multiplying to confirm, we get (x+ 5)(2x− 2) = 0 2 x^2 − 2 x+ 10 x− 10 = 0 2 x^2 + 8 x− 10 = 0 The roots are found by solving th ...
672 Worked-Out Solutions to Exercises: Chapters 21 to 29 and 3 x− 2 = 0 The solution to the first of these equations is derived ...
This equation is equivalent to the other one. To show this, we can derive one squared binomial from the other: (4x− 5)^2 = (4x− ...
674 Worked-Out Solutions to Exercises: Chapters 21 to 29 Let’s take the square root of both sides of the binomial factor equati ...
The fact that the discriminant is positive tells us that this quadratic has two distinct real roots. To find the roots, we can u ...
676 Worked-Out Solutions to Exercises: Chapters 21 to 29 and x+j 7 = 0 In the top equation, we can add j7 to each side, getting ...
Here are the roots again, for reference: x=j7 or x=−j 3 Both of these statements are equations. In the first one, we can subtr ...
678 Worked-Out Solutions to Exercises: Chapters 21 to 29 Now let’s substitute back in where we left off. That gives us an expres ...
In the bottom equation, we can add the quantity (2 +j3) to each side, getting x= 2 +j 3 The roots can be formally expressed this ...
680 Worked-Out Solutions to Exercises: Chapters 21 to 29 Plugging in the first root and converting the subtractions to additions ...
Let’s use the commutative law for addition to rearrange the terms according to powers of x with each power of x on its own line: ...
682 Worked-Out Solutions to Exercises: Chapters 21 to 29 Because the parabola opens upward, we know that its vertex is an absol ...
has no real roots, so the quadratic function has no real zeros. That means the graph does not cross the x axis anywhere. If a pa ...
684 Worked-Out Solutions to Exercises: Chapters 21 to 29 Here’s the quadratic function again, for reference: y=− 2 x^2 + 2 x− ...
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