II. SOLUTION OF PURE EQUATIONS OF THE SECOND, AND OTHER higher degREES, BY EVOLUTION. RULE. 396. Transpose the terms of the equation in such a manner, that the given power of the unknown quantity may be on one side of the equation, and the known quantities on the other; then extract the root, denoted by the exponent of the power, on each side of the equation, and the value of the unknown quantity will be determined. In the same way any adfected equation, having that side which contains the unknown quantity, a complete power, may be reduced to a simple equation, from which the value of the unknown quantity will be ascertained, by the rules in Chap. III. Ex. 1. Given x2-17-130-2x2, to find the values of x. By transposition, 3x2=147; .. by division, x2 =49, and by evolution, x= ±7. 397. It has been already observed, (Art. 295), that a may be either + or —, where n is any whole number whatever; and, consequently, all pure equations of the second degree admit of two solutions. Thus, +7x+7, and—7X-7, are both equal to 49 ; and both, when substituted for x in the original equation, answer the condition required. Ex. 2. Given x2+ab=5x2, to find the values of x. By transposition, 4x =ab; ..2x=ab, and x=√ab. Ex. 3. Given x2-6x+9=a2, to find the values of x. By evolution, x−3=±ï; .. x=3±α. Ex. 4. Given 4x2-4ax+u2x2+12x+36, to And the value of x. By extracting the square root on both sides, we have 2x-α=x+6; ... by transposition, x=ɑ+6. Ex. 5. Given 2+y213,) to find the values of and x-y=5, § x and y. By addition, 2x2 = 18; ... x=±√9=±3. By subtraction, 2y8;.. y=±√/4= ±2. Ex. 6. Given 81x=256, to find the values of x. By extracting the square root, 9x2=±16; By extracting again, 3x=/16±4, or 4√-1; •• x= ±4, or x=±4√ −1. Ex. 7. Given x-3x4+3x-1=27, to find the values of x. By evolution, x2-1=3; .'. x=4, and x=±2. Ex. S. Given 36x2a2, to find the values of x. Ans. x=. Ex. 9. Given x3=27, to find the value of x. Ans. x=3. Ex. 10. Given x2+6x+9=25, to find the values Ans. x=2, or--8. of x. Ex. 11. Given 3x-9=21+3, to find the values of x. Ans. x=√11. Ans. xab· x3-x2+x-a3. to find the Ans. xa+}. Ex. 13. Given x2+3x+1=a2b2, to find the values of x. Ex. 14. Given x2+bx+1b2=a3, to find the values of x. Ans. x= =±a—1b, Ex. 15. Given x-2x2+1=9, to find the values of x. Ex. 16. Given x 4 of x. གྱྱི Aus. x2, or ±√−2. ·4x2+4x=4, to find the values Ans. x2, or ±0. Ex. 17. Given 5x2-27-3x2+215, to find the va lues of x. Ans. x= =±11. Ex. 18. Given 5x2-1=244, to find the values Ans. x=7. of x. Ex. 19. Given 9x2+9=3x2+63, to find the va lues of x. Ex. 20. Given 2ax2+b-4 find the values of x. i Ans. x3. cx2-5+d-ax2, to d-b Ans. x= 4 Ex. 21. Given x+y=a and x-y=b, to find the values of x and y. Ans. x=±√(√(2a+2b)) and y=±√(±√ (2a-2b)). $ III. EXAMPLES IN WHICH THE PRECEDING RULES ARE APPLIED IN THE SOLUTION OF PURE EQUATIONS. 398. When the terms of an equation involve powers of the unknown quantity placed under radical signs. Let the equation be cleared of radical signs, as in Sect. I; then, the value of the unknown quantity will be determined by extracting the root, as in Sect. II. And by a similar process, any equation containing the powers of a function of the unknown quantity, or containing the powers of two unknown quantities, may frequently be reduced to lower dimensions. Ex. 1. Given /x2=3⁄4/(a+b), to find the value of x. Here, the given quantity may be exhibited under the form (x2—8)* = (x —3); then, by squaring both sides, (x-9) (x-3) or (x2 — 9)2=x-3; by squaring again, x2-9=x2-6x+9; ... by transposition, 6x=18; and x=3. Ex. 3. Given x2-y2=9, and x-y=1; to find the value of x and y. Dividing the corresponding terms of the first equation by those of the second, we have x+y=9; adding this equation to the second, 2x=10; ..x=5, and y=9-x; ..y=4. Ex. 4. Given√x+y=5, to find the values of -y=1,5 x and y. and Adding the two equations, 2x=6, •'• √/x=3, and by involution, x=9. Subtracting the two equations, 2/y=4, and y=2; .. By involution, y=4. Ex. 5. Given x2+xy=12.) to find the values of and y+xy=24,5 x and y. By addition, x2+2xy+y2=36; ... extracting the square root, x+y=±6. Now x2+xy=x.(x+y)=±6x ; ..+6x=12, and x=2; •*•y=±672=±4. 2a2 Ex. 6. Given x+√(a2+x2)=. to find √(a2+x3) the values of x. Multiplying by √(a2+x2), we have x/(a2+x2) +a+x=2a2; by transposition, x√(a2+x2)=a2 —x3, and squaring both sides, a2x2+x=a*—2a2x2+x1 ; α .*. 3a2x2=a1, and x=± √3 From the 1st equation subtracting twice the 2d. 1 x2_2x¢+y2=(x−y)2=x—ÿ' ••• (x—y)3=1, and x—y=1; ••. x2 + y2 = 13; and 2xy=12; .. by addition, x2+2xy+y2=25, by extracting the square root, x+y=±5; but x-y=1; .. by addition, 2x=6, or -4: and x=3, or -2; by subtraction, 2y=4, or -6; and y=2, or -3. ៖ Ex. 8. Given x+y = 13, to find the values of and x+y=4, x and y. 3 ៖ Squaring the second equation, x3+2x3y3+y =25 23 ... by subtraction, 2x*y*=12. Subtracting this from the 1st equation, x -- -2x3y3+y3=1 = .. extracting the square root, x-y3±1; 8 10 and y=2, or 3; .'. y=8, or 27. Ex. 9. Given x+x+y+y=273, } to find the and x1+x2y2+y1=21, values of x and y. Dividing the first equation by the second, x-x2 y2+y1=13; subtracting this from the second equation, 2x2 y2=8; •'. x2y2=4; by adding this equation to the second, x+2x2 y2 + y+=25; ...x2+y2±15. Subtracting the equation x2y2-4, from x-x2 y2 +y+=13, x1—2x2y2+y^—9 ; .'. x2 —y2=±3, ... by addition, 2x2+8, and x=12, or ±2-1; and by subtraction, 2y2±2, and y=±1, or ±√ 1. Ex. 10. Given ✔(a+x)+√(a−x) value of x. √(a+x)+√(a−x)—b, to find the √)α+x)−√(a—x) . Multiply the numerator and denominator by √(a+x)+√(a−x), [√(a+x)+√(a−x) ]2 2x or 2a+2(a2—x2)=2bx; •*• √ (a2 —x2)—bx—ɑ, and squaring both sides, a2x2-b2x2-2abx+a2, |
