Algebra I Workbook For Dummies
Page 12
Q. Use the square root rule to solve: 7x2 = 28.
A. (which is the same as x = 2 or x = –2). First divide each side by 7 to get x2 = 4. Then find the square root of 4 to get the answer.
Q. Use the square root rule to solve: 3y2 – 75 = 0.
A. (which is the same as y = 5 or y = –5). Before using the square root rule, first add 75 to each side to get the number on the right. Then divide each side by 3 to get y2 = 25. Finally, find the square root of 25 to get the answer.
1. Use the square root rule to solve: x2 = 9.
Solve It
2. Use the square root rule to solve: 5y2 = 80.
Solve It
3. Use the square root rule to solve: z2 – 100 = 0.
Solve It
4. Use the square root rule to solve: 20w2 – 125 = 0.
Solve It
Solving by Factoring
The quickest and most efficient way of solving a quadratic equation that has three terms is to factor it (if it can be factored) and then use the multiplication property of zero (MPZ) to solve for the solutions. The MPZ says that, if the product ab = 0, then either a or b must be equal to 0. You use this property on the factored form of a quadratic equation to set the two linear factors equal to 0 and solve those simple equations for the value of the variable.
Sometimes when you factor the quadratic, you get two identical factors. That’s because the quadratic was a perfect square. You still get two answers, but they’re the same number, which is called a double root.
Q. Solve for x by factoring: 18x2 + 21x – 60 = 0.
A. . The factored form of the original equation is 3(6x2 + 7x – 20) = 3(3x – 4)(2x + 5) = 0. When the product of three factors is zero, then one or more of the factors must be equal to zero. The first factor here, the 3, is certainly not equal to zero, so you move on to the next factor, 3x – 4, and set it equal to zero to solve for the solution. You use the same process with the last factor, 2x + 5. Only the variables can take on a value to make the factor equal to 0. That’s why, when you set the last two factors equal to 0, you solve those equations and get the answers.
Q. Solve for x by factoring: 4x2 – 48x = 0.
A. x = 0 or x = 12. You don’t need a constant term (it’s technically 0 in this case) in order for this process to work. Just factor out the GCF 4x from each term in the equation.
The two factors, set equal to 0 give you
5. Solve for x by factoring: x2 – 2x – 15 = 0.
Solve It
6. Solve for x by factoring: 3x2 – 25x + 28 = 0.
Solve It
7. Solve for y by factoring: 4y2 – 9 = 0.
Solve It
8. Solve for z by factoring: z2 + 64 = 16z.
Solve It
9. Solve for y by factoring: y2 + 21y = 0.
Solve It
10. Solve for x by factoring: 12x2 = 24x.
Solve It
11. Solve for z by factoring: 15z2 + 14z = 0.
Solve It
12. Solve for y by factoring: .
Solve It
Using the Quadratic Formula
You can use the quadratic formula to solve a quadratic equation, regardless whether the terms can be factored. Factoring and using the MPZ is almost always easier, but you’ll find the quadratic formula most useful when the equation can’t be factored or the equation has numbers too large to factor in your head.
If a quadratic equation appears in its standard form, ax2 + bx + c = 0, then the quadratic formula gives you the solutions to that equation. You can find the solutions by using the following formula:
If you use the quadratic formula on a quadratic that can be factored, then the number under the radical (the square root symbol) always turns out to be a perfect square, and no radical appears in the solution. If the equation isn’t factorable, then your answer — a perfectly good one, mind you — has radicals in it. If the number under the radical turns out to be negative, then that equation doesn’t have a real solution. If you want to know more about this real versus imaginary situation, look at the “Dealing with Impossible Answers” section later in this chapter.
Q. Use the quadratic formula to solve: 2x2 + 11x – 21 = 0 with a = 2, b = 11, and c = –21.
A. . Fill in the formula and simplify to get
Because the 289 under the radical is a perfect square, you could have solved this problem by factoring.
Q. Use the quadratic formula to solve: x2 – 8x + 2 = 0.
A. . The quadratic equation is already in the standard form, with a = 1, b = –8, and c = 2. Fill in the formula with those values and simplify:
13. Use the quadratic formula to solve: x2 – 5x – 6 = 0.
Solve It
14. Use the quadratic formula to solve: 6x2 + 13x = –6.
Solve It
15. Use the quadratic formula to solve: x2 – 4x – 6 = 0.
Solve It
16. Use the quadratic formula to solve: 2x2 + 9x = 2.
Solve It
17. Use the quadratic formula to solve: 3x2 – 5x = 0. Hint: When there’s no constant term, let c = 0.
Solve It
18. Use the quadratic formula to solve: 4x2 – 25 = 0. Hint: Let b = 0.
Solve It
Completing the Square
An alternative to using factoring or the quadratic formula to solve a quadratic equation is a method called completing the square. Admittedly, this is seldom the method of choice, but the technique is used when dealing with conic sections in higher mathematics courses. I include it here to give you some practice using it on some nice quadratic equations.
Here are the basic steps you follow to complete the square:
1. Rewrite the quadratic equation in the form .
2. Divide every term by a (if a isn’t equal to 1).
3. Add to each side of the equation.
This is essentially just half the x term’s newest coefficient squared.
4. Factor on the left (it’s now a perfect square trinomial).
5. Take the square root of each side of the equation.
6. Solve for x.
Q. Use completing the square to solve .
A. x = 5, x = –1.Working through the steps:
1. Rewrite the equation as .
2. No division is necessary.
3. Add to each side to get
.
4. Factor on the left,
5. Find the square root of each side:
You only need use the ± in front of the radical on the right. Technically, it belongs on both sides, but because you end up with two pairs of the same answer, just one ± is enough.
6. Solve for x by adding 2 to each side of the equation: , giving you x = 5 or x = –1.
Q. Use completing the square to solve .
A. .Working through the steps:
1. Rewrite the equation as .
2. Divide each term by 2: .
3. Add to each side to get
.
4. Factor on the left, . Notice
that the constant in the binomial is the number that got squared in Step 3.
5. Find the square root of each side:
6. Solve for x:
19. Use completing the square to solve for x in .
Solve It
20. Use completing the square to solve for x in .
Solve It
Dealing with Impossible Answers
The square root of a negative number doesn’t exist — as far as real numbers are concerned. When you encounter the square root of a negative number as you’re solving quadratic equations, it means that the equation doesn’t have a real solution.
You can report your answers when you end up with negatives under the radical by using imaginary numbers (or numbers that are indicated with an i to show that they aren’t real). With imaginary numbers, you let . Then you can simplify the radical using the imaginary number as a factor. Don’t worry if you’re slightly confused about imaginary numbers. This concept is more advanced than most of the algebra prob
lems you’ll encounter in this book and during Algebra I. For now, just remember that using i allows you to make answers more complete; imaginary numbers let you to finish the problem.
Q. Rewrite , using imaginary numbers.
A. 3i. Split up the value under the radical into two factors:
Q. Rewrite , using imaginary numbers.
A. . You can write the value under the radical as the product of three factors:
The last two lines of the equation look almost identical. Putting the i after the 4 is just a mathematical convention, a preferred format. It doesn’t change the value at all.
21. Rewrite as a product with a factor of i.
Solve It
22. Rewrite as an expression with a factor of i.
Solve It
Answers to Problems on Solving Quadratic Equations
This section provides the answers (in bold) to the practice problems in this chapter.
1. Use the square root rule to solve: . The answer is .
2. Use the square root rule to solve: . The answer is .
3. Use the square root rule to solve: . The answer is .
4. Use the square root rule to solve: . The answer is .
5. Solve for x by factoring: . The answer is .
6. Solve for x by factoring: . The answer is .
7. Solve for y by factoring: . The answer is .
You could also have used the square root rule on this problem.
8. Solve for z by factoring: . The answer is , which is a double root.
9. Solve for y by factoring: . The answer is .
10. Solve for x by factoring: . The answer is .
11. Solve for z by factoring: . The answer is .
12. Solve for y by factoring: . The answer is .
has 12 as a common denominator, so
13. Use the quadratic formula to solve: . The answer is .
14. Use the quadratic formula to solve: . The answer is .
First, rewrite in the standard form:
15. Use the quadratic formula to solve: . The answer is .
16. Use the quadratic formula to solve . The answer is . First rewrite the equation in standard form.
17. Use the quadratic formula to solve: . The answer is .
Note: You could have factored the quadratic: .
18. Use the quadratic formula to solve: . The answer is .
19. Use completing the square to solve for x in . The answer is x = or x = –4.
20. Use completing the square to solve for x in . The answer is
.
21. Rewrite as a product with a factor of i. The answer is 2i.
22. Rewrite as an expression with a factor of i. The answer is .
Chapter 14
Yielding to Higher Powers
In This Chapter
Tallying up the possible number of roots
Making educated guesses about solutions
Finding solutions by factoring
Recognizing patterns to make factoring and solutions easier
Polynomial equations are equations involving the sum (or difference) of terms where the variables have exponents that are whole numbers. (Trinomials are special types of polynomials; I thoroughly cover solving trinomial equations in Chapter 13.) Solving polynomial equations involves setting polynomials equal to zero and then figuring out which values create true statements. You can use the solutions to polynomial equations to solve problems in calculus, algebra, and other mathematical areas. When you’re graphing polynomials, the solutions show you where the curve intersects with the x-axis — either crossing it or just touching it at that point. Rather than just taking some wild guesses as to what the solutions might be, you can utilize some of the available techniques that help you make more reasonable guesses as to what the solutions are and then confirm your guess with good algebra.
This chapter provides several examples of these techniques and gives you ample opportunities to try them out.
Determining How Many Possible Roots
Mathematician René Descartes came up with his rule of signs, which allows you to determine the number of real roots that a polynomial equation may have. The real roots are the real numbers that make the equation a true statement. This rule doesn’t tell you for sure how many roots there are; it just tells you the maximum number there could be. (If this number is less than the maximum number of roots, then it’s less than that by two or four or six, and so on.)
To use Descartes’s rule, first write the polynomial in decreasing powers of the variable; then do the following:
To determine the maximum possible positive roots, count how many times the signs of the terms change from positive to negative or vice versa.
To determine the possible number of negative roots, replace all the x’s with negative x’s. Simplify the terms and count how many times the signs change.
Q. How many possible real roots are there in 3x5 + 5x4 – x3 + 2x2 – x + 4 = 0?
A. At most four positive and one negative. The sign changes from positive to negative to positive to negative to positive. That’s four changes in sign, so you have a maximum of four positive real roots. If it doesn’t have four, then it could have two. If it doesn’t have two, then it has none. You step down by twos. Now count the number of possible negative real roots in that same polynomial by replacing all the x’s with negative x’s and counting the number of sign changes:
This version only has one sign change — from negative to positive, which means that it has one negative real root. You can’t go down by two from that, so one negative real root is the only choice.
Q. How many possible real roots are there in 6x4 + 5x3 + 3x2 + 2x – 1 = 0?
A. One positive and three or one negative. Count the number of sign changes in the original equation. It has only one sign change, so there’s exactly one positive real root. Change the function by replacing all the x’s with negative x’s and count the changes in sign:
It has three sign changes, which means that it has three or one negative real roots.
1. Count the number of possible positive and negative real roots in x5 – x3 + 8x2 – 8 = 0.
Solve It
2. Count the number of possible positive and negative real roots in 8x5 – 25x4 – x2 + 25 = 0.
Solve It
Applying the Rational Root Theorem
In the preceding section, you discover that Descartes’s rule of signs counts the possible number of real roots. Now you see a rule that helps you figure out just what those real roots are, when they’re rational numbers.
Real numbers can be either rational or irrational. Rational numbers are numbers that have fractional equivalents; that is, they can be written as fractions. Irrational numbers can’t be written as fractions; they have decimal values that never repeat and never end.