147. Division of Surds. To divide one surd, Va, by another, Vb, we have the principle This principle is simply what Formula XI gives when n=2. EXERCISES Express each of the following quotients as a fraction under one radical sign, and reduce to simplest form. 13 Solution. V15 = V15 = V5. Ans. 6. Vrs 7. Va? – b?. Va-6 8. V3(x2 — y?). V12(x+y) 3 148. Rationalizing the Denominator of a Fraction. If the denominator of a fraction consists of a single surd, or is a binomial containing surds, the fraction may be changed into one which has surds only in its numerator. The process of doing this is called rationalizing the denominator, and is illustrated below. EXAMPLE 1. Rationalize the denominator in the fraction V3 SOLUTION. Here the denominator consists of the single surd V5. To get this surd out of the denominator it is merely necessary to multiply both numerator and denominator of the fraction by V5, giving V15/5. Ans. EXAMPLE 2. Rationalize the denominator in the fraction V3+V2 SOLUTION. Multiplying both numerator and denominator by V3–V2, the given fraction takes the form V3-V2 V3-V2 V3-V2 V3-V2 (V3_V2)(V3+V2) (V3)2-(V2)2 3-2 = 1 =V3-V2. Ans. EXAMPLE 3. Rationalize the denominator in the fraction 3 V5+2V2 V5-V2 Solution. Multiplying both numerator and denominator by V5+V2, we obtain (V5+V2)(3V5+2V2) _ 3.5+3V10+2V10+2.2 _ 19+5V10. Ane (V5)2–(V2)2 5-2 3 A careful examination of the examples above shows that if the denominator is of the form Va; that is, consists of a single surd, the fraction may be rationalized by simply multiplying both numerator and denominator by Va. If, on the other hand, the denominator has either of the binomial forms Va+Vb, or Vā- Vb, the fraction should have its numerator and denominator multiplied by Va-Vī in the first case and by Va+ Vī in the second. EXERCISES Write each of the following expressions with rationalized denominators. 3. Va. 6. V3+V2. V3- V2 V 6. 8. 3+ V5. 7. – 3 Va-4Vī Væ+1+3. " Væ+1+2 * 149. Finding the Value of Fractions Containing Surds. Suppose we wish to find the value of the expression 1 V3+V2 We begin by rationalizing the denominator, thus making the fraction take the form (V3 – V2)/1, or V3-V2. All we now need to do is to look up in the tables the values of V3 and V2 and subtract the latter from the former. The entire work may be written in the form -=V3-V2=1.73205+-1.41421+=0.31784+. Ans. V3+V2 If we had not first rationized the denominator, we should have had to find the value of 1.73205++1.41421+' OF 3.14626+ which would thus compel us to divide 1 by 3.14626+. Note how much more difficult this would be than the above, where all we needed to do in the end was to subtract 1.41421 from 1.73205, which is very simple. This illustrates the general fact that in finding the value of a fraction, its denominator (if it contains surds) should first be rationalized. * EXERCISES Find (by rationalizing denominators and then using the tables) the approximate value of the following expressions. CHAPTER XVI PART I. PURE QUADRATICS 150. Quadratic Equation. An equation which contains the unknown letter to the second (but no higher) power is called a quadratic equation, or briefly, a quadratic. Thus, 3 x2–2 x=4 and x2+2 x+1=0 are quadratics, but 3 x-1=0 and 2 23—4 x2+x=6 are not. 151. Pure and Affected Quadratics. When the quadratic contains the second power only of the unknown letter, it is called a pure quadratic. Thus, wa=25, 6 x2—24=0 and ax?=bc are pure quadratics, but x2— x=25 is not. When the quadratic equation contains both the first and second powers of the unknown letter, it is called an affected quadratic, or a general quadratic. Thus, x2+2 x—15=0 is an affected quadratic, but 2 x2–27=0 is not. 152. Solution of Pure Quadratics. The following two examples suffice to show how the solution of any pure quadratic may be obtained. EXAMPLE 1. Solve the equation x? — 25=0. SOLUTION. 22-25=0. Transposing, we find x2=25. eu. Taking the square root of both members of the last equation gives x=+5. Ans.t The meaning of this answer is that x may have either the value +5 or -5. That is, the given equation has the two different roots 5 and –5. See § 137. CHECK. Substituting 5 for x in the given equation gives 52= 25, which is correct. Similarly, if we substitute -5 for a we have (-5)2=25, which also is true. EXAMPLE 2. Solve the equation 2 x2—30=0. SOLUTION. Transposing and dividing by 2 gives x2=15. Taking the square root of both members gives x=+V15. Ans. It should be carefully noted that these answers are different in character from those in Example 1 because we are here asked to extract the square root of 15, and this can be done only approximately. (See § 139.) Since the tables give v15=3.87298, our answers, correct to five decimal places, are x==3.87298. CHECK. 2(V15)2–30=2X15–30=30–30=0, as required. 2(-V15)2–30=2X15-30=30–30=0, as required. An examination of the examples above shows that we have the following rule. RULE FOR SOLVING A PURE QUADRATIC. Solve for xạ, then take the square root of the result. There will always be two roots, the one being the negative of the other. † Strictly speaking, when we extract the square root of both members of the equation x2=25 we get +x=+5 instead of x=+5. But to say that -x=+5 is the same as to say that x=+5, so it suffices to write simply x=+5 to cover all possible cases. |