226 CHAPTER VIII. ON PURE EQUATIONS. 325. Equations are considered as of two kinds, called simple or pure, and adfected; each of which are differently denominated according to the dimensions of the unknown quantity. 326. If the equation, when cleared of fractions and radical signs or fractional exponents, contain only the first power of the unknown quantity, it is called a simple equation. 327. If the unknown quantity rises to the second power or square, it is called a quadratic equation. 328. If the unknown quantity rises to the third power or cube, it is called a cubic equation, &c. 329. Pure equations, in general, are those wherein only one complete power of the unknown quantity is concerned. These are called pure equations of the first degree, pure quadratics, pure cubics, pure biquadratics, &c., according to the dimension of the unknown quantity. Thus, x=a+b is a pure equation of the first degree; x1=a1+a3b+ac2+d is a pure biquadratic; &c. 330. Adfected equations are those wherein different powers of the unknown quantity are concerned, or are found in the same equation. These are called adfected quadratics, adfected cubics, adfected biquadratics, &c., according to the highest dimension or power of the unknown quantity. Thus, ax=b, is an adfected quadratic; xax+bx=c, an adfected cubic; x2+ax3+bx2+cx=d, an adfected biquadratic. 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. 331. 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. 332. 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. Ex. 2. Given by transposition, 4x=144-16; ..x=32. (2x+3)+4=7, to find the value of 2. By transposition, 3/(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. By squaring, 12+x=4+4✓√x+x; by transposition, 8=4x, or √x=2; .. by squaring, z=4. Ex. 4. Given 4′(x+40)—10—, to find the value of x. By squaring, x+40=100-20√x+x; by transposition, 20/60, or x=3; .. by squaring, x=9. Ex. 5. Given ✓✓/(x—16)—8—✔x, to find the value of x. By squaring both sides of the equation, x-16-64-16 √ x+x; .•. 16 √x=64+16=80; by division, x=5; ...x=25. (x—a)=√x—a, to find the value of x. Ex. 6. Given Squaring both sides of the equation, x-u=x-(ax)+a; ... by transposition, (ax)=a; 25a2 by squaring, ax=' 16 25a 16 Ex. 7. Given ✓5×√(x+2)=√5x+2, to find the value of x. Ex. 8. Given By squaring, 5x+10=5x+4/5x+4; by transposition, 6=4√5x, ... √5x=3; by squaring again, 5x=;.. x=20. to find the value of x. x-ax √ x √x Multiplying both sides of the equation by x, √x+4 √x+6 a , to find the value of x. 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 ax-b 3/ax-2b √ax+b ̄3√ax+5b' = to find the value of x. Multiplying both sides by (ax+b)×(3√ax+5b), 3ax+2b/ax-5b2=3ax+b√ ax—2b2, .. by transposition, bax=3b2; by division, ax=3b; 962 .. by squaring, ax=962, and x= a Multiply both sides of the equation by √(x+√/x), x+ √(x)—√(x2-x)=· and dividing by√x, √x−1=√(x−1) ; .. by squaring, x-x+1=x-1; ·· √/r=§, √x+1=x−1; 25 and by squaring, z= 16 Ex. 12. Given √\/(x—24)=√/x-2, to find the value of x. Ans. x 49. Ex. 13. Given √(4a+x)=2√(b+x)—√√√x, to find the va lue of x. x. ·(b—a)3· Ans. x= 2a-b Ex. 14. Given x+a+√(2ax+2)=b, to find the value of Ex. 17. Given x=√/[a2+x√/(b2+x)]-a, to find the va lue of x. Ans. x= 4a 4 Ex. 18. Given √(2+)+√x=√(2+x) lue of x. x. x. Ex. 19. Given /(10x+35)—1=4, to find the value of x. Ans. x=9. Ex 20. Given 5/(9x—4)+6=8, to find the value of x. Ex. 21. Given √(x+16)=2+ √x, to find the Ex. 22. Given √(x-32)=16-√x, to find Ans. x=4. value of x. Ans. x= =9. the value of Ans. x=81. Ex. 23. Given √(4x+21)=2 √x+1, to find the value of An. sx=25. Ex. 24. Given [1+x√(x2+12)]=1+x, to find the va lue of x. Ex. 25. Given √x+√(x−9)= lue of x. Ex. 26. Given (a+x)=2(x2+5ax+b) to find the va § II. SOLUTION OF PURE EQUATIONS OF THE SECOND, AND OTHER HIGHER DEGREES, BY EVOLUTION. RULE. 333. 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 22—17—130—2x2, to find the values of x. By transposition, 3x2=147; .. by division, x=49, and by evolution, z=7. 334. It has been already observed, 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 +ab=5x2, to find the values of x. By transposition, 4x=ab; ..2x=ab, and rab. Ex. 3. Given 2-6x+9=a2, to find the values of x. By evolution, x-3=a; .. x=3a. Ex. 4. Given 4x2-4ax+a2x2+12x+36, to find the value of x. By extracting the square root on both sides, we have 2xa=x+6; .. by transposition, x=a+6. Ex. 5. Given 2+y=13, to find the values of x and y. Ex. 5. Given 22+ y2=13, and ---5, By addition, 2x2-18; .. x=√9±3. By subtraction, 2y=8; ... y=±√/4=±2. Ex. 6. Given 81x256, to find the values of x. |