ROOTS AND RADICALS On the flip side of exponents and powers are roots and radicals. The square root of a number n is a number that you “square” (multiply by itself, or raise to the power of 2) to obtain n. 2 5 = 4 (the square root of 4) because 2 3 2(or2 2 ) 5 4 The cube root of a number n is a number that you raise to the power of 3 (multiply by itself twice) to obtain n. You determine greater roots (for example, the “fourth root”) in the same way. Except for square roots, the radical sign will indicate the root to be taken. 2 5 = 3 8 (the cube root of 8) because 2 3 2 3 2(or2 3 ) 5 8 2 5 = 4 16 (the fourth root of 16) because 2 3 2 3 2 3 2(or2 4 ) 5 16 For the GMAT, you should know the rules for simplifying and for combining radical expressions. Simplifying Radicals On the GMAT, always look for the possibility of simplifying radicals by moving what’s under the radical sign to the outside of the sign. Check inside your square-root radicals for perfect squares: factors that are squares of nice tidy numbers or other terms. The same advice applies to perfect cubes, and so on. = 4a 2 5 2|a| 4 and a 2 are both perfect squares; remove them from under the radical sign, and find each one’s square root. = 8a 3 5 = ~4!~2!a 3 5 2a = 2a 8 and a 3 are both perfect cubes, which contain perfect-square factors; remove the perfect squares from under the radical sign, and find each one’s square root. You can simplify radical expressions containing fractions in the same way. Just be sure that what’s in the denominator under the radical sign stays in the denominator when you remove it from under the radical sign. Î 20x x 3 5 Î ~4!~5! x 2 5 2 = 5 x Î 3 3 8 5 Î 3 3 2 3 5 1 2 = 3 3 TIP Whenever you see a non-prime number under a square-root radical sign, factor it to see whether it contains perfect- square factors you can move outside the radical sign. More than likely, you need to do so to solve the problem at hand. Chapter 10: Math Review: Number Theory and Algebra 243 www.petersons.com 10. Î 28a 6 b 4 36a 4 b 6 5 (A) a b Î a 2b (B) a 2b Î a b (C) |a| 3|b| = 7 (D) a 2 3b 2 = 2 (E) 2a 3b The correct answer is (C). Divide a 4 and b 4 from the numerator and denominator of the fraction. Also, factor out 4 from 28 and 36. Then, remove perfect squares from under the radical sign: Î 28a 6 b 4 36a 4 b 6 5 Î 7a 2 9b 2 5 |a| = 7 3|b| ,or |a| 3|b| = 7 In GMAT questions involving radical terms, you might want to remove a radical term from a fraction’s denominator to match the correct answer. To accomplish this, multiply both numerator and denominator by the radical value. (This process is called “rationalizing the denominator.”) Here’s an example of how to do it: 3 = 15 5 3 = 15 = 15 = 15 5 3 = 15 15 or 1 5 = 15 Combining Radical Terms The rules for combining terms that include radicals are quite similar to those for exponents. Keep the following two rules in mind; one applies to addition and subtraction, while the other applies to multiplication and division. Addition and subtraction: If a term under a radical is being added to or subtracted from a term under a different radical, you cannot combine the two terms under the same radical. = x 1 = y Þ = x 1 y = x 2 = y Þ = x 2 y = x 1 = x 5 2 = x, not = 2x On the GMAT, if you’re asked to combine radical terms by adding or subtracting, chances are you’ll also need to simplify radical expressions along the way. 244 PART IV: GMAT Quantitative Section www.petersons.com 11. = 24 2 = 16 2 = 6 5 (A) = 6 2 4 (B) 4 2 2 = 2 (C) 2 (D) = 6 E. 2 = 2 The correct answer is (A). Although the numbers under the three radicals combine to equal 2, you cannot combine terms this way. Instead, simplify the first two terms, then combine the first and third terms: = 24 2 = 16 2 = 6 5 2 = 6 2 4 2 = 6 5 = 6 2 4 Multiplication and Division: Terms under different radicals can be combined under a common radical if one term is multiplied or divided by the other, but only if the radical is the same. = x = x 5~ = x! 2 ,orx = x = y 5 = xy = x = y 5 Î x y = 3 x = x 5 ? (you cannot easily combine = 3 x = x 5 x 1 3 x 1 2 5 x 1 3 1 1 2 5 x 5 6 12. ~2 = 2a! 2 5 (A) 4a (B) 4a 2 (C) 8a (D) 8a 2 (E) 6a The correct answer is (C). Square each of the two terms, 2 and = 2a, separately. Then combine their squares by multiplication: ~2 = 2a! 2 5 2 2 3~ = 2a! 2 5 4 3 2a 5 8a. Roots You Should Know For the GMAT, memorize the roots in the following table. If you encounter one of these radical terms on the exam, chances are you’ll need to know its equivalent integer to answer the question. In the table on the following page, notice that the cube root of a negative number is negative and the cube root of a positive number is positive. Chapter 10: Math Review: Number Theory and Algebra 245 www.petersons.com Square roots of “perfect square” integers Cube roots of “perfect cube” integers (positive and negative) = 121 5 11 = 144 5 12 = 169 5 13 = 196 5 14 = 225 5 15 = 625 5 25 = 3 ~2!8 5~2!2 = 3 ~2!27 5~2!3 = 3 ~2!64 5~2!4 = 3 ~2!125 5~2!5 = 3 ~2!216 5~2!6 = 3 ~2!343 5~2!7 = 3 ~2!512 5~2!8 = 3 ~2!729 5~2!9 = 3 ~2!1000 5~2!10 ROOTS AND THE REAL NUMBER LINE As with exponents, the root of a number can bear a surprising relationship to the magnitude and/or sign (negative vs. positive) of the number (another of the test makers’ favorite areas). Here are three rules you should remember: If n . 1, then 1 , = 3 n , = n , n (the greater the root, the lesser the value). However, if n lies between 0 and 1, then n , = n , = 3 n , 1 (the greater the root, the greater the value). n 5 64 1 , = 3 64 , = 64 ,64 1 , 4 , 8 , 64 n 5 1 64 1 64 , Î 1 64 , Î 3 1 64 , 1 1 64 , 1 8 , 1 4 , 1 246 PART IV: GMAT Quantitative Section www.petersons.com Every negative number has exactly one cube root, and that root is a negative number. The same holds true for all other odd-numbered roots of negative numbers. = 3 2 27 523 (23)(23)(23) 5227 = 5 232 522 (22)(22)(22)(22)(22) 5232 Every positive number has only one cube root, and that root is always a positive number. The same holds true for all other odd-numbered roots of positive numbers. 13. Which of the following inequalities, if true, is sufficient alone to show that = 3 x , = 5 x? (A) 21 , x , 0 (B) x . 1 (C) |x| ,21 (D) |x| . 1 (E) x ,21 The correct answer is (E). If x ,21, then applying a greater root yields a lesser negative value—further to the left on the real number line. LINEAR EQUATIONS WITH ONE VARIABLE Algebraic expressions are usually used to form equations, which set two expressions equal to each other. Most equations you’ll see on the GMAT are linear equations, in which the variables don’t come with exponents. To solve any linear equation containing one variable, your goal is always the same: Isolate the unknown (variable) on one side of the equation. To accomplish this, you may need to perform one or more of the following four operations on both sides, depending on the equation: Add or subtract the same term from both sides Multiply or divide by the same term on both sides Clear fractions by cross-multiplication Clear radicals by raising both sides to the same power (exponent) Performing any of these operations on both sides does not change the equality; it merely restates the equation in a different form. NOTE The square root (or other even- number root) of any negative number is an imaginary number, not a real number. That’s why the preceding rules don’t cover these roots. ALERT! The operation you perform on one side of an equation must also be performed on the other side; otherwise, the two sides won’t be equal. Chapter 10: Math Review: Number Theory and Algebra 247 www.petersons.com Let’s take a look at each of these four operations to see when and how to use each one. Add or subtract the same term from both sides of the equation. To solve for x, you may need to either add or subtract a term from both sides of an equation—or do both. 14. If 2x 2 6 5 x 2 9, then x 5 (A) 29 (B) 26 (C) 23 (D) 2 (E) 6 The correct answer is (C). First, put both x-terms on the left side of the equation by subtracting x from both sides; then combine x-terms: 2x 2 6 2 x 5 x 2 9 2 x x 2 6 529 Next, isolate x by adding 6 to both sides: x 2 6 1 6 529 1 6 x 523 Multiply or divide both sides of the equation by the same non-zero term. To solve for x, you may need to either multiply or divide a term from both sides of an equation. Or, you may need to multiply and divide. 15. If 12 5 11 x 2 3 x , then x 5 (A) 3 11 (B) 1 2 (C) 2 3 (D) 11 12 (E) 11 3 The correct answer is (C). First, combine the x-terms: 12 5 11 2 3 x 248 PART IV: GMAT Quantitative Section www.petersons.com Next, clear the fraction by multiplying both sides by x: 12x 5 11 2 3 12x 5 8 Finally, isolate x by dividing both sides by 12: x 5 8 12 ,or 2 3 If each side of the equation is a fraction, your best bet is to cross-multiply. Where the original equation equates two fractions, use cross-multiplication to eliminate the fractions. Multiply the numerator from one side of the equation by the denominator from the other side. Set the product equal to the product of the other numerator and denominator. (In effect, cross-multiplication is a shortcut method of multiplying both sides of the equation by both denominators.) 16. If, 7a 8 5 a 1 1 3 , then a 5 (A) 8 13 (B) 7 8 (C) 2 (D) 7 3 (E) 15 The correct answer is (A). First, cross-multiply as we’ve described: (3)(7a) 5 (8)(a 1 1) Next, combine terms (distribute 8 to both a and 1): 21a 5 8a 1 8 Next, isolate a-terms on one side by subtracting 8a from both sides; then combine the a-terms: 21a 2 8a 5 8a 1 8 2 8a 13a 5 8 Finally, isolate a by dividing both sides by its coefficient 13: 13a 13 5 8 13 a 5 8 13 Chapter 10: Math Review: Number Theory and Algebra 249 www.petersons.com Square both sides of the equation to eliminate radical signs. Where the variable is under a square-root radical sign, remove the radical sign by squaring both sides of the equa- tion. (Use a similar technique for cube roots and other roots.) 17. If 3 = 2x 5 2, then x 5 (A) 1 18 (B) 2 9 (C) 1 3 (D) 5 4 (E) 3 The correct answer is (B). First, clear the radical sign by squaring all terms: ~3 2 !~ = 2x! 2 5 2 2 ~9!~2x!54 18x 5 4 Next, isolate x by dividing both sides by 18: x 5 4 18 ,or 2 9 LINEAR EQUATIONS WITH TWO VARIABLES What we’ve covered up to this point is pretty basic stuff. If you haven’t quite caught on, you should probably stop here and consult a basic algebra workbook for more practice. On the other hand, if you’re with us so far, let’s forge ahead and add another variable. Here’s a simple example: x 1 3 5 y 1 1 Quick . . . what’s the value of x? It depends on the value of y, doesn’t it? Similarly, the value of y depends on the value of x. Without more information about either x or y, you’re stuck—but not completely. You can express x in terms of y, and you can express y in terms of x: x 5 y 2 2 y 5 x 1 2 Let’s look at one more: 4x 2 9 5 3 2 y 250 PART IV: GMAT Quantitative Section www.petersons.com Solve for x in terms of y: 4x 5 3 2 y 1 9 x 5 3 8 y 1 9 4 Solve for y in terms of x: 4x 2 9 3 2 5 y 2 3 ~4x 2 9!5y 8 3 x 2 6 5 y To determine numerical values of x and y, you need a system of two linear equations with the same two variables. Given this system, there are two different methods for finding the values of the two variables: The substitution method The addition-subtraction method Next, we’ll apply each method to determine the values of two variables in a two-equation system. The Substitution Method To solve a system of two equations using the substitution method, follow these four steps (we’ll use x and y here): In either equation, isolate one variable (x) on one side. Substitute the expression that equals x in place of x in the other equation. Solve that equation for y. Now that you know the value of y,plugitintoeither equation to find the value of x. ALERT! You can’t solve an equation if it contains two unknowns (variables). You either need to know the value of one of the variables or you need a second equation. Chapter 10: Math Review: Number Theory and Algebra 251 www.petersons.com 18. If 2 5 p 1 q 5 3q 2 10, and if q 5 10 2 p, then p q 5 (A) 5 7 (B) 3 2 (C) 5 3 (D) 25 6 (E) 36 6 The correct answer is (A). Don’t let the fact that the question asks for p q (rather than simply p or q) throw you. Because you’re given two linear equations with two unknowns, you know that you can first solve for p and q, then divide p by q. First things first: Combine the q-terms in the first equation: 2 5 p 5 2q 2 10 Next, substitute (10 2 p)forq (from the second equation) in the first equation: 2 5 p 5 2~10 2 p!210 2 5 p 5 20 2 2p 2 10 2 5 p 5 10 2 2p Move the p-terms to the same side, then isolate p: 2 5 p 1 2p 5 10 12 5 p 5 10 p 5 S 5 12 D ~10! p 5 25 6 252 PART IV: GMAT Quantitative Section www.petersons.com . not = 2x On the GMAT, if you’re asked to combine radical terms by adding or subtracting, chances are you’ll also need to simplify radical expressions along the way. 244 PART IV: GMAT Quantitative. because 2 3 2 3 2 3 2(or2 4 ) 5 16 For the GMAT, you should know the rules for simplifying and for combining radical expressions. Simplifying Radicals On the GMAT, always look for the possibility. , = 3 64 , = 64 ,64 1 , 4 , 8 , 64 n 5 1 64 1 64 , Î 1 64 , Î 3 1 64 , 1 1 64 , 1 8 , 1 4 , 1 246 PART IV: GMAT Quantitative Section www.petersons.com Every negative number has exactly one cube