Examples of Substitution of Unknown Quantities. 10. Solve the equation (x2+5)2-4x2 = 160. Ans. x= 3, or 15. 11. What two numbers are they, whose product is 255, and the sum of whose squares is 514 ? 12. What two numbers are they, whose product is the sum of whose squares is b? its Ans. ±√ [}b+ √ († b2 — a2)], and ± √ [} b— ✓ († b2 — a2)]. 13. What number exceeds its square root by 20? Ans. 25. 14. What number is it, the excess of whose square above square root is equal to 56 divided by the number? 236. There are equations of higher degrees, which can be reduced to equations of the second degree by introducing other unknown quantities instead of those contained in them. Thus if the same algebraic expression is involved in different ways, it will often be found successful to consider this expression as the unknown quantity. (x2 — 23 y)3 + (x2 — 23 y)2 + (x2 — 23 y) (x — 2 y) = 18, (x2 — 23 y)2 + (x — 2 y) = 7. Examples of Substitution of Unknown Quantities. Solution. Consider (x2-23 y), and (x — 2 y), as the unknown quantities, making x=x2-23 y, y' = x-2y; and the equations become x' 3 + x' 2 + x' y' = 18, Hence, by the elimination of y', we have and, therefore, 2' 2 + 7 x' = 18, x' = 2, or =- 9; and the corresponding values of y' are x=(23v14001), y = (319±√14001). 2. Solve the equation x + (x + 6) 1 = 2 + 3 (x + 6)3. Examples of Substitution of Unknown Quantities. 238. Corollary. When there are two unknown quantities which enter symmetrically into the given equation, the solution is often simplified by substituting for them two other unknown quantities, one of which is their product and the other their sum. 239. EXAMples. 1. Find two numbers whose sum is 5, and the sum of whose fifth powers is 275. Solution. Let the numbers be x and y, represent their product by p, and we have x+y=5, x5 + y5 = 275. But we also have (x+y)=x5+5x+y+10x3 y2+10x2 y3+5x y1+y3 =x5+y5+5x y (x3 + y3) +10 x2 y2 (x+y); and 3 x3+ y3 = (x + y)3 — 3 x2 y — 3 x y2 Hence = (x + y)3 — 3 x y (x + y) 15 p. (x+y)5 = 275+5 p (125 — 15 p) + 10 p2 × 5=55 ; or, by reduction, and p2 - 25 p P = 19, or = 6; 114, x=2, or == 3, or = (5±√-51), y=3, or=2, or = 2. Solve the two equations (5√−51). Examples of Substitution of Unknown Quantities. Solution. These equations become, by development, x3 — x2 y — x y2+ y3=7, x3+x2y +x y2 + y3 = 175 ; 8. Find two numbers such, that their sum and product may together be 34, and the sum of their squares may exceed the sum of the numbers themselves by 42. Ans. 4 and 6; or (11✔―59,) and (-11—✔―59). 9. What two numbers are they, whose sum is 3, and the sum of whose fourth powers is 17? Ans. 2 and 1; or § (3+v'—55), and § (3 — √ — 55). 10. What two numbers are they, whose product is 3, and the sum of whose fourth powers is 82? 240. Corollary. In many cases, in which two unknown quantities enter into the given equations symmetrically except in regard to their signs, the solution is simplified by substituting for them two other unknown quantities, one of which is their difference, and the other is their sum or their product. |