21. In how many ways can a man pay a bill of $53 using $2 bills and $5 bills? 22. Find the smallest number that when divided by 4 or divided by 7, leaves a remainder of 3. 183. Graphical Solution of Indeterminate Equations. Any indeterminate equation of the first degree may be represented graphically by a straight line. The positive integral values of x and y are then determined by noting at what points the line crosses the intersections of the ruled lines of the graph paper, in the first quadrant, since the values of x and y are both positive in the first quadrant only. = Thus, the equation of Problem 2, yields the graph shown in Fig. 1. When x = 2, y 3, the line cuts the corner of one of the squares. Since this is the only point in the first quadrant, the equation 5x+8y = 34 has only one set of positive integral values of x and y. Note. The success of this solution depends entirely upon the accuracy of the drawing. The line should be accurately determined by locating two points some distance apart. The equations of Exercise 134 furnish ample material for the application of this method. REVIEW EXERCISE XXV 1. In how many ways may $2.75 be paid with quarters and half dollars? 2. Find the smallest number that, when divided by 5 or divided by 7, gives 4 for a remainder. 3. Solve in positive integers, 4x + 7y = 94. 4. A farmer spends $752 in buying horses and cows; if each horse cost $37 and each cow $23, how many of each kind does he buy? 5. Find two fractions whose denominators are 9 and 5 respectively, and whose sum shall be equal to y + 4z = 27 2x + 3y 5z = -8. 7. A party of 20 people, consisting of men, women, and children, pay a hotel bill of $67. Each man pays $5, each woman $4 and each child $1.50. How many of each were there? = 107. 8. Solve in positive integers, 8x+5y 9. A woman expended 93 cents for 14 yards of cloth, some at 5, some at 7, and the rest at 10 cents a yard. How many whole yards of each did she buy? 10. What is the least number that will contain 25 with a remainder of 1, and 33 with a remainder of 2? CHAPTER XXVI DISCUSSION OF EQUATIONS 184. Character of the Roots of a Quadratic Equation. Every quadratic equation may be reduced to the form ax2 + bx + c = 0 in which a is positive, while b and c may be positive or negative. If the roots of this general form are denoted by r1 and r2 then, as previously shown, These values of the two roots show that the character of these roots, as real or imaginary, rational or irrational, equal or unequal, depends upon the value of b2-4ac. 1. To determine whether the roots are real or imaginary. If the quantity under the radical sign is positive, the roots are real, if negative, the roots are imaginary. Hence, when b2-4ac is positive, the roots are real, when b2-4ac is negative, the roots are imaginary. 2. To determine whether the roots are rational or irrational. If the roots are imaginary, they are also irrational. If the roots are real, they may be rational or irrational, according to whether the quantity under the radical sign is a perfect square or not. Hence, when b2-4ac is a perfect square or equal to zero, the roots are rational; otherwise, they are irrational. 3. To determine whether the roots are equal or unequal. It is evident that the two roots r1 and r1⁄2 differ from each other in that the radical is added to -b in the one and subtracted from -b in the other. If the radical vanishes, the difference between the two roots vanishes and the roots are equal. quadratic equation ax2 + bx + c = 0. 4. To determine the signs of the roots. When a is positive, b and c either positive or negative, the signs of the roots r1 and r2 may be determined from the signs of b and c. If c is positive, √b2 — 4ac < b, hence both roots have the sign of -b. If c is negative, b2-4ac > b, and the roots will have √b2 4acb, opposite signs, since ri will be positive and r2 negative. If b is negative, -b is positive and therefore r1 will be numerically greater than r2. If b is positive, -b is negative and therefore r2 will be numerically greater than r1. In general, if c is negative, the roots have opposite signs and the numerically greater root has the sign opposite to that of b. Determine the character of the roots of 2x2 + 3x Reduce the equation to the general form, 2x2 3x5 = 0 = 5. As b2 = -- 4ac is a perfect square, the roots are real and rational. 4ac is not equal to zero, the roots are unequal. As c is negative, the roots have opposite signs, and as b is positive, the negative root is the numerically greater. EXERCISE 135 Without solving, determine the character of the roots of: 13. For what values of m wil x2 + 2mx When b2 0, the roots are equal. 0 +16 = 4 m = +2 14. For what values of a will the equation 2x2 + 3ax + 2 = 0, have equal roots? 15. For what values of m will the equation 4x2 + (6 2a)x-1 have equal roots? 16. For what values of m will the equation (m + 1)x2 + (m − : 1)x + m + 1 = 0 have imaginary roots? RELATION OF ROOTS AND COEFFICIENTS 185. Any quadratic equation in the form ax2 + bx + c = 0, may be reduced to the form, x2 + px + q = 0, by dividing by the coefficient of x2. By solving x2 + px + q = 0, we find Hence, the sum of the roots of a quadratic equation in the form x2 + px + q = 0, is equal to the coefficient of x with its sign changed, and the product of the roots is equal to the absolute term. As the above principle corresponds exactly to the rule for writing the product of two binomials, the equation x2 + px + q = 0, may be written (xr1) (x — r2) = 0. Therefore, when the roots are known, the equation may be |