MARCO TEÓRICO
3.2. El Color como Parte del Código de los Cráneos
A one-variable, nth-order equation can be written in the following standard form: a1xn+a 2x n-1+a 3x n-2+. . .+ a n-2x 2+a n-1x +an=0
where a1, a2, . . .an are constants, x is the variable, and a1 ≠ 0. We won’t even think about trying to factor an equation like this in general, although specific cases might lend themselves to factorization. Solving nth-order equations, where n> 5, practically demands the use of a computer.
PROBLEM 6-9
What are the solutions to the following factored equation? (x−j3)(x+2)(x+j5)=0
SOLUTION 6-9
To solve this, simply identify the three quantities that make any one of the factors equal to 0. These quantities are j3 for the first factor, −2 for the second factor, and −j5 for the third factor. Mathematicians speak of the set containing the solutions of an equation like this as the solution set. If we call that set S in this case, then:
S={j3,−2,−j5}
PROBLEM 6-10
Write the equation from Problem 6-9 as a cubic equation in standard form. The coefficients will be complex. Be sure these coefficients are expressed in the form a+ jb, where aand bare real numbers and jis equal to the positive square root of −1.
SOLUTION 6-10
The conversion of a factored equation to standard form is easier than the reverse process, although it can be tedious if there are more than two factors. Let’s multiply the second two factors first:
(x+2)(x+j5)=x2+j5x+2x+j10
(x−j3)(x2+j5x+2x+j10)=x3+j5x2+2x2+j10x−j3x2−j215x−j6x −j230
=x3+j5x2+2x2+j10x−j3x2+15x−j6x+30 =x3+(j5+2−j3)x2+(j10+15−j6)x+30 =x3+(2+j2)x2+(15+j4)x+30
The cubic equation in standard form is therefore: x3+(2+j2)x2+(15+j4)x+30=0
Quick Practice
Here are some practice problems that cover the material presented in this chap- ter. Solutions follow the problems.
PROBLEMS
1. Factor the following quadratic equation: y2+7y+12=0
2. State the solution set S for the quadratic equation given in Practice Problem 1.
3. State the following quadratic equation in standard form: x(x−5)=0
4. What can be said about the solutions to this quadratic equation? 2z2+2z+5=0
5. State the solution set S for the quadratic equation given in Practice Problem 4. Use the quadratic formula to find the solutions.
SOLUTIONS
1. The process of factoring a quadratic equation involves some intuition. In this case, you might be able to quickly see the factors:
(y+3)(y+4)=y2+4y+3y +12 =y2+7y+12
so the factored form is:
(y+3)(y +4)=0
2. It’s easy to solve a quadratic or higher-order equation when we see it in factored form. The solutions are those values of the variable that make either of the factors equal to 0. In this case, these values are y= −3 and y
= −4. The solution set is therefore: S={−3,−4}
3. Multiply the factors. Here, the first factor is x, and the second factor is (x − 5). Therefore, the side of the equation to the left of the equality symbol is:
x(x −5)=x2−5x
and the quadratic equation in standard form is: x2−5x+0=0
It is not necessary to write down any addend in the standard form of a quadratic or higher-order equation when the coefficient of that addend is equal to 0. It is perfectly all right, in this case, to write:
x2−5x =0
4. Consider the discriminant d=b2−4ac, where a=2,b=2, and c=5: d=22−(4×2×5)=4−40= −36
In this example, d< 0, indicating that there are two distinct solutions, and that they are complex conjugates.
5. We already know that b2−4ac= −36. The first solution is: z=[−2+(−36)1/2] / (2 ×2)
=(−2+j6) / 4
= −2/4+(j6)/4
The second solution is:
z=[−2 −(−36)1/2] / (2 ×2) =(−2 −j6) / 4
= −2/4−(j6)/4
= −1/2−j(3/2) The solution set is therefore:
S={[−1/2+j(3/2)], [−1/2−j(3/2)]}
Quiz
This is an “open book” quiz. You may refer to the text in this chapter. A good score is 8 correct. Answers are in the back of the book.
1. Consider the following equation: z2+25=0
How many solutions does this equation have, and what is its, or their, nature?
(a) There are two distinct real-number solutions. (b) There is a single real-number solution. (c) There are two complex-conjugate solutions. (d) It is impossible to tell without more information. 2. What sort of equation is the following?
−21w2−17w+45=6w2+10w+3
(a) A linear equation. (b) A quadratic equation.
(c) An equation of unknown order. (d) An invalid equation.
3. Consider the following equation: x3−x=0
(a) x(x−1)(x+1)=0 (b) x2x3=0
(c) x+2x−3x=0
(d) It cannot be expressed in factored form. 4. Consider the following equation:
6y+3=8y+4
How many solutions does this equation have, and what is its, or their, nature?
(a) There are two distinct real-number solutions. (b) There is a single real-number solution. (c) There are two complex-conjugate solutions. (d) It is impossible to tell without more information. 5. Consider again the following equation:
x3−x=0
What is the real-number solution set Sfor this equation? (a) S={0}
(b) S={0, 1} (c) S={0,−1, 1}
(d) S= ∅, because there are no real-number solutions. 6. Which of the following is nota quadratic equation?
(a) (x +2)(x−3)=5 (b) x+3x−4=3 (c) (x +j5)(x−j4)=8
(d) None of the above are quadratic equations.
7. Consider the following quadratic equation in standard form: px2+qx+r=0
wherep,q, and rare real-number constants, and xis the unknown. This equation has a single real-number solution if and only if
(a) p=q=r. (b) q2=4pr.
(c) r2< 4pq.
8. Consider, yet again, the following equation: x3−x=0
It is tempting to suppose that we can divide this equation through on each side by x, getting a simpler equation, as follows:
x3−x=0
(x3−x) / x=0/x x3/x−x/x=0
x2−1=0
Unfortunately, we cannot legitimately do this because (a) this is not a cubic equation in standard form. (b) there are no real-number solutions.
(c) it cannot generally be done for higher-order equations.
(d) the solution set of the original equation contains the element 0. 9. In a quadratic equation with two distinct solutions, both of which are real
numbers,
(a) the discriminant is a positive real number. (b) the discriminant is a positive imaginary number. (c) the discriminant is a negative real number. (d) the discriminant is a negative imaginary number. 10. Consider the following equation:
x4=1
Which, if any, of the following numbers is notan element of the solution set of this equation?
(a) 1. (b) −1. (c) 0 +j1.
(d) All of three of the above are elements of the solution set of this equation.