13. The roots of x2+px+9=o are p± √p2 -q. Hence, (i) If >q, we shall have p2-q positive, and therefore √p2 -q a possible quantity; and since in one root it is taken with +, and in the other with, the two roots will be real and different in value. (ii.) If 29, we shall have 2-q=o, and therefore the two roots will be real and equal in value. (iii.) If p2<q, we shall have p2 -q negative, and √p2 - q impossible, and so the two roots will be impossible. Hence, if any equation be expressed in the form x2+px+9=0, its roots will be real and different, real and equal, or impossible, according as p2>,-, or <49. PROBLEMS LEADING TO QUADRATICS OF ONE 6. Let x=number of party, then 120 120 = number of shillings each had to pay, x-4= number after four left. Then +2=what each had to pay = 240+50 x = 2x x2=11900 - 100X, x2+ 100x = 11900, x2 + 100x+(50)2 = 11900 + 2500 = 14400. 9. x2+(x+5)2=1525. x2 + x2+10x+25=1525, 2x2+10x=1500, x+5x + ( ) = % +750=3073, x + 2 = 55, x= 50 = 25. If x= number of gallons in one cask, and (x+5) number in the other, and the price of each kind of wine per gallon is in shillings the number of gallons in the cask, the total cost must be x2+(x+5), which by question is £76, 55. = 1525 shillings; therefore the above equation gives the number of gallons, 25, in one cask, and 30 must be the number in the other. MISCELLANEOUS EQUATIONS. 1. (1) (i.) 2x-5y=2, (ii.) 8y-3x= I. Multiplying (i.) by 3, and (ii.) by 2, we have 2. (1) 2x-y-10 (i.). 5x+2y=63 (ii.). Multiplying (i.) by 2, and adding result to (ii.), we get 9x=83, ...y-8. |