of Simple Equations, the methods of resolving pure equations of the first degree, in all cases, except when the quantity is affected with radical signs or fractional exponents, in which case the following rule is to be observed. RULE. 395. If the equation contains a single radical quantity, transpose all the other terms to the contrary side; then involve each side into the power denominated by the index of the surd; from whence an equation will arise free from radical quantities, which may be resolved by the rules pointed out in Chap. III. If there are more than one radical sign over the quantity, the operation must be repeated; and if there are more than one surd quantity in the equation, let the most complex of those surds be brought by itself on one side, and then proceed as before. Ex. 1. Given Ex. 2. Given (4x+16)=12, to find the value of x. sides of the equation, 4x+16=144; by transposition, 4x=144-16; ••x=32. (2x+3)+4=7, to find the value of x. By transposition, /(2x+3)=7—4=3 ; cubing both sides, 2x+3=27; by transposition, 2x=27 −3; ‚.x=12. Ex. 3. Given (12+x)=2+√x, to find the value of x. By squaring, 12+x=4+4√x+x ; by transposition, 8=4/x, or x=2; by squaring, x=4. Ex. 4. Given (x+40)=10~√x, to find the value of x. By squaring, x+40=100~20✓✅/x+x ; by transposition, 20/x=60, or x=3; by squaring, x=9. (x−16)=8—√x, to find the value of x. Ex. 5. Given By squaring both sides of the equation, x-1664-16x+x;.. 16/x=64+16=80; by division, √x=5; .'. x=25. Ex. 6. Given√(x—a)=√√xa, to find the value of x. Squaring both sides of the equation, x—α=x-√√(ax)+1a; .. by transposition, √(ax)=‡a; 25α2 16 25a 16 Ex. 7. Given/5×√(x+2)=√5x+2, to find the value of x. By squaring, 5x+10=5x+4/5x+4; by transposition, 6=4/5x, .. √5x=}; by squaring again, 5x=; x=2· √x Ex. 8. Given = to find the value of x. √x x Multiplying both sides of the equation by √x, 1 Ex. 9. Given 1 x= 1-a to find the value of x. √x+4 Multiplying both sides by (x+4)×(√√/x+6), we have x+34/x+168=x+42√x+152; by transposition, 16-8/x, or 2=x; ... by squaring, x=4. Ex. 10. Given Vax-b 3/ax-2b to find the value of x. = Vax+b 3ax+56' Multiplying both sides by (ax+b)×(3✓ax+56), 3ax+2b/ax-562-3ax+bax-263, .. by transposition, bax=362 ... by squaring, ax=962, and x= Ex. 11. Given√(x+√x)−√(x−√x)= Multiply both sides of the equation by √(x+√x), x+✔ .. by transposition, x— √(x)=√(x*—x) ; and dividing by√x, √x −1=√(x−1) ; .. by squaring, x−√x+1=x−1 ; .'. √x=5, Ex. 12. Given √(x—24)=√✓✓x—2, to find the value of x. √(4a+x)=2√(b+x)−√x, to find the (b-a)? Ans. x= 2α-b Ex. 14. Given x+a+√(2ax+2)=b, to find the value of (b-a)2 Ans. 26 Ex. 17. Given x=√[a2+x√(b2+x2)]-a, to find the va lue of x. Ans. x 4α Ex. 19. Given Ex. 20. Given Ex. 21. Given x. x. /(10x+35) −1=4, to find the value of x. Ex. 22. Given √(x−32)=16— √x, to find the value of Ans. x=81. Ex. 23. Given √(4x+21}=2√x+1, to find the value of lue of x. Ans. x 25. [1+x√(x2+12)]=1+x, to find the va Ex. 24. Given Ans.x=2. 36 to find the va √(x-9) lue of x. Ans. x=25. Ex. 26. Given (a+x)=2m/(x2+5ax+b2) to find the va § 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. 2n 397. It has been already observed, (Art. 295), that 2/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 —7 ×—7, are both equal to 49; and both, when substituted for x in the original equation, answer the condition required. Ex. 2. Given x2+ub=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=±a ; ... x=3+a. Ex. 4. Given 4x2-4ax+a2=x2+12x+36, to find the va lue of x. By extracting the square root on both sides, we have 2x --=x+6; x .. by transposition, =a+6 Ex. 5. Given 2212-13, to find the values of x and y. By addition, 2x2=18; ... x=±√9=±6. By subtraction, 2y=8; .. y=±√4=±2. Ex. 6. Given 81x4=256, to find the values of x By extracting the square root, 9a2+16; By extracting again, 3x=±√±16=+4, or ±4√−1; ...x=±, or x=±‡√−1. Ex. 7. Given xo—3x2+3x2-1=27, to find the values of x. By evolution, 3-1=3; .. x2=4, and x=+2. Ex. 8. Given 36x2=a2, to find the values of x. Ex. 9. Given x3-27, to find the value of x. Ans. x=+ja. Ans. x=3. Ex. 10. Given x2+6x+9=25, to find the values of x. Ex. 11. Given 3x2-9=21+3, to find the values of x. Ans. x+11. Ex. 12. Given x3—x2+}x—27=a3, to find the value of x. Ans. xa+1. Ex. 13. Given x2+3x+1=a2b2, to find the values of x. = Ans. x+ab x= Ex. 14. Given x2+bx+b2=a2, to find the values of x. Ex. 18. Given 5x-1=244, to find the values of x. Ans. x=+7. Ex. 19. Given 9x2+9=3x2+63, to find the values of x. Ans. x +3. x= Ex. 20. Given 2ax+b-4-cx2-5+d-ax3, to find the values of x. Ans. x= d-b-1 3a-c. Ex. 21. Given x+y=a and x-y=b, to find the values of x and y. Ans. x=+√(±√(2a+2b)) and y=±√(±1√(2a—2b)). § III. EXAMPLES IN WHICH THE PRECEDING RULES ARE APPLIED IN THE SQLution 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 pow |