129. 1. In what time will $ 1 double itself at 4 %, interest compounded annually? At the end of the first year the interest is $.04, the amount is $1.04. At the end of the second year the interest is $ (1.04) (.04), the amount is $1.04 (1.04) (.04) = $ (1.04)2. At the end of the third year the interest is $ (1.04)2 (.04), the amount is $ (1.04)3. At the end of x years the amount will be $ (1.04)*. We have the equation, (1.04)* = 2. x = 17.7 nearly. 2. In what time will $1 treble itself at 4 %, interest compounded annually? 3. A sum of $1000 bears 5% interest compounded annually. What is the amount after 20 years? 4. Compute the compound interest on $879 at 4 % for 13 years. HISTORICAL NOTE 130. Logarithms constitute one of the most useful inventions in mathematics. A great French astronomer once said that logarithms "by shortening the labors, doubled the life of the astronomer." Logarithms were invented by John Napier, Baron of Merchiston, in Scotland, who in 1614 published a table of logarithms and described their use. An independent inventor of logarithms was the Swiss watchmaker and astronomer, Joost Bürgi, who published a table in 1620, at a time when Napier's logarithms were already known and admired throughout Europe. Napier's and Bürgi's logarithms differed somewhat from each other and from the logarithms described in this book. Our system of logarithms to the base 10 was designed by John Napier and his friend Henry Briggs, conjointly, before the year 1617. At one time Napier lived in a beautiful castle on the banks of the Endrick. Yet even in such surroundings he was not free from annoyances which hindered intellectual work. On the opposite side of the river was a lint mill, and its clack greatly disturbed Napier. He sometimes desired the miller to stop the mill so that the train of his ideas might not be interrupted. Much of Napier's time was taken up in the management of his estate. Joost Bürgi was not rich, like Napier; the necessity of earning a livelihood greatly hampered his scientific work. But both Bürgi and Napier were men who loved mathematics and were able to achieve great results in spite of many hindrances. CHAPTER V QUADRATIC EQUATIONS AND THEIR PROPERTIES REVIEW 131. In solving a quadratic equation of the form ax2 + bx +c=0 by "completing the square," the first step in our explanation was to divide both sides of the equation by the coefficient of x2. There are other methods of procedure which possess certain advantages; fractions having large denominators may be avoided, or, fractions may be avoided altogether. These methods will be explained in § 191. In the following exercises solve by any method. To find the square roots of numbers, use the tables in § 197. EQUATIONS QUADRATIC IN FORM 132. Equations like x2+ax+b=0 are said to be quadratic in form, because the unknown x occurs only with the exponents n and 2 n, where one exponent is twice the other. Equations of this form can be solved by the methods used in solving ordinary quadratic equations. From the tables in § 197 we obtain x = 2.236, x = ± 1.414, approxi mately. As there are in general three cube roots to a number, and there are in this case two different numbers under the radical sign, it is seen that there are six roots of the given equation. Of these, four are imaginary. As we wrote the answer, only the two principal roots are indicated. RELATIONS BETWEEN ROOTS AND COEFFICIENTS 134. From the solution of ax2 + bx + c = 0, we obtain If we designate the two roots by x1 and x2, we have If we divide both sides of ax2 + bx + c = 0 by a, we obtain in which the coefficient of x2 is unity. We see that, in the last equation, (1) The sum of the roots is the coefficient of x with the sign changed; and (2) The product of the roots is the absolute term. To illustrate: Solve x2 + 5 x + 6 = 0. The roots are 2 and - 3. Their sum is 5, the coefficient of x with its sign changed; their product is +6, the absolute term. From this relation between the coefficients and the roots, we may form an equation, if the roots are known. Form the equation whose roots are 1 and 6. We assume that, in the required equation, the coefficient of x2 is 1. 5, the coefficient of x with its sign changed; x1X2: Then, x1 + x2 = the absolute term. Hence, the equation is x2 - 5 x − 6 = 0. Whence, x2-(-16)x+(−1·6) = 0. == - 6, |