The remainder, -9, shows that the expression lacks 9 of being a 12. (x21)2 + 24 = 11(x2 – 1). 13. x+2x-3= 6. 3x + 4 + 2x2 - 3x + 6 = 14. 14. x2 Jerome Cardan was one of the greatest mathematicians of his time. His name is usually applied to the general solution of the cubic equation, although this was really the work of Tartaglia in 1541. Çardan was astrologer, physician, philosopher and mathematician. HIGHER EQUATIONS SOLVED BY FACTORING 143. Equations of a higher degree than the second can often be solved by factoring and placing each factor equal to zero, as previously explained. Note that any method of factoring may be used, also that the degree of the equation indicates the number of roots. 144. Since a quadratic equation has two roots, problems involving quadratics seem to have two solutions. If both roots are positive integers, there will usually be two solutions of the problem. If the roots are fractional or negative, they may or may not give solutions, depending upon the conditions stated in the problem. If the roots are imaginary, the problem is impossible. 1. The difference of two numbers is 4 and their product is 12. Find the numbers. When x2, x + 4 When x=- −6, x + = 4 : = x 2 = 0 x = 2 6, hence the two numbers are 2 and 6. -2, hence the two numbers may also be Evidently both solutions satisfy the conditions stated in the problem. 2. Find two consecutive numbers, the sum of whose squares is 145. When x = 8, x+1 9 and the numbers are 8 and 9. The negative value of x cannot be used here as consecutive numbers are understood to be integers which follow each other in the common scale 1, 2, 3, 4. 3. A club ordered a dinner that cost $150.00. Ten of the members failed to attend the dinner and each of the other members had to pay 75 cents more. How many members were there in the club? Let x = the number of members in the club, Then 150 x = the number of dollars each would have paid. As 10 members were absent, 10 150 = And = -- x 10 the number of members who attended. the number of dollars each paid. But as each had to pay 75¢ more. 4. A clothing merchant sold a suit for $39 and gained as many percent as the suit cost him in dollars. What |