In like manner other adfected equations are denominated according to the highest power of the unknown quantities. I. SOLUTION OF PURE EQUATIONS OF THE FIRST DEGREE BY INVOLUTION. 394. We have already delivered, under the denomination 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 (4x+16)=12, to find the value of x. Squaring both sides of the equation, 4x+16=144; by transposition, 4x=144-16; .'.x=32. Ex. 2. Given (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+√, to find the value of x. Ex. 4. Given Fue of x. By squaring, 12+x=4+4√x + x; by transposition, 8=4/x, or √x=2; .. by squaring, x=4. (x+40)=10-x, to find the va By squaring, x+40=100—20/x+x; by transposition, 20/x=60, or x=3; .. by squaring, x=9. Ex. 5. Given lue of x. (x−16)=8−√x, to find the va By squaring both sides of the equation, x-16=64-16x+x;.. 16x=64+16=807 by division, x=5; .. x=25. Ex. 6. Given √(x−a)=√x—a, to find the yalue of x. Squaring both sides of the equation, Ex. 7. Given the value of x. x- a=x−√(ax)+ia; .. by transposition, √(ax)=ža; 25a2 by squaring, ax = ; ..x= 16 25a 16 5×(x+2)=√✓5x+2, to find By squaring, 5x+10=5x+4/5x+4; by transposition, 6=4/5x, .*• √/5x=3; by squaring again, 5x=2, .. x = £• √, to find the value of x Ex. 8. Given x-ax = Ꮖ X- -ax=~=1, or (1—a)x=1; .'. x= 1 1 -α Ex. 9. Given +28_x+38, to find the va Multiplying both sides by (x+4)×(√x+6), we have x+34/x+168=x+42/x+152; by transposition, 16=8x, or 2=√x ; .. by squaring, x=4. Ex. 10. Given value of x. = Vax-b_3/ax-2b to find the Multiplying both sides by (ax+b)×(3/ax+5b), 3ax+2b/ax-5b2=3ax+bax-2b2, .. by transposition, b1ax=362 ; by division, ax=3b; 962 .. by squaring, ax=962, and x= α Ex. 11. Given √(x+√x)−√(x−√∞)= Multiply both sides of the equation by √(x+√∞), x+√(x)−√(x2 — x) = 3√/i0. 2 and dividing by √x, √x−1=√(x−−1); .. by squaring, x−√x+1=x-1; ··· √x=§, Ex. 12. Given √✓/(x—24)=√✓✓¤—-2, to find the va lue of x. Ex. 13. Given value of x. ✔(4a+x)=2✔✅✔✅(b+x)−√x, to find Ans. x=49. (b—a)2 Ans. x= 2a-b Ex. 14. Given x+a+(2ax+x)=b, to find the Ex. 17. Given x=√[a2 +x√(b2 +x2)]—a, to find the value of x. Ans. x= 4a Ex. 20. Given /9x-4)+6=8, to find the va lue of x. Ans. x=4. Ex. 21. Given (x+16)=2+√x, to find the value of x. Ans. x=9. Ex. 22. Given √(x-32)=16-x, to find the value of x. Ans. x=81. Ex. 23. Given ✔✅(4x+21)=2x+1, to find the value of x. Ex. 24. Given [1+x√(x2 +12)]=1+x, to find the value of x. Ans. x 25. Ans. x=2. 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 =19, and by evolution, x= ±7. 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 adınit 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=±~; .'. x=3±±α. Ex. 4. Given 4x2-4ax+u2x2+12x+36, to find the value of x. By extracting the square root on both sides, we have 2x-α=x+6; .. by transposition, x=a+ |