Subtraction. Ex. 1. 250-✓18=2√(25.2)—✔(9.2)=2.5√2—3√2= (10—3) √2=7√2. 2. √3—√3/7=√(}•§)+√(37•})=√}}−√}{=} √15 — 4. =5 a = 250 a3 x-16 a3 x=(125 a3. 2 x) (8 a3. 2 x) 2x- -2 a 2 x=3 a 2 x. 5. √45 s1 x—√20 s2 x3= √(9 s1. 5 x) — √(4 s2 x2. 5 x)= (3 s2 - 28 x) √5 x.. s Multiplication. Ex. 1. 18 x 54-5 (18.4)=5 (4.2.9)=5(8.9)= 5.29-10 9. 8. dx a b= d3×5a2 b2="/a2 b2 d3. 9. ✔a✔b— √3·× √a+ √b−√3= √(a3—b+ √3.) 5. (x3-xd-b+d √b)÷(x — √b)=x+ √b—d. 1 1 1 ac-ad α √(a3 x − a x2) ÷ 2b √ (a — x) = ac-ad X 27 √(+2)=27 √(76•2)=27·4 √2=188 √2. 3. (3+ √5)2={(3+ √5) (3+ √5)}=14+6 √5. 4. (ab)3=a3 — 3 a2 √b + 3 a b —bb. Evolution. Ex. 1. 10= √1000= √(100, 10)=√100 × 10 =10 ✔10. 2. /81 a1 y3 z= √(81 a1 y1·y2 z)=9 a3 y3 ✩/y3 z. 3. (a 4 a ✓b+b2) = a -2b, the operation being performed as in the arithmetical extraction of the square root. Note.-The square root of a binomial or residual a±b, or even of a trinomial or quadrinomial, may often be conveniently extracted thus :-Take d = √(a-b3); then Ja+d ✓ (a+b)= Jata± α d This is evident: for, if Jabe squared, it will give a + √(aa — d2) or a it ought and, in like manner, the square of √3/2+√2 = √2+ √1=1+ √2. 2. Find the square root of 6 2 25. + Here a=6, b=2 √5,d=√(3620)=√16=4, 3. Find the square root of 6+ √8−√12 -√24. Here a=6+8,b= √12+ √24, d= √(6+ √8)3 (√12+ √24)= √(44+12836-212.24) = √(44 — 36 +12 ✅8 · 128) = √8. But (Ex. 1), 2 = 3 + √8, and (3+2 √2) = √(3+ √8)=1+ √2. Therefore the root required is 1+ √⁄2 — √3.* SECTION VIII.-Simple Equations. An algebraic equation is an expression by which two quantities called members (whether simple or compound), are indicated to be equal to each other, by means of the sign of equality =placed between them. In equations consisting of known and unknown quantities, when the unknown quantity is expressed by a simple power, as x, x3, x3, &c. they are called simple equations, generally; * For the cube and higher roots of binomials, &c. the reader may consult the treatises on Algebra by Maclaurin, Emerson, Lacroix, Bonnycastle, J. R. Young, and Hine. and particularly, simple or pure quadratics, cubics, &c. according to the exponent of the unknown quantity. But when the unknown quantity appears in two or more different powers in the same equation, it is named an adfected equation. Thus 2a+15, is a simple quadratic equation: xa x = b, an adfected quadratic. It is the former class of equations that we shall first consider. The reduction of an equation consists in so managing its terms, that, at the end of the process, the unknown quantity may stand alone, and in its first power, on one side of the sign =, and known quantities, whether denoted by letters, or figures, on the other. Thus, what was previously unknown is now affirmed to be equal to the aggregate of the terms in the second number of the equation. "In general the unknown quantity is disengaged from the known ones, by performing upon both members the REVERSE OPERATIONS, "to those indicated by the equation, whatever they may be. Thus, If any known quantity be added to the unknown quantity, let it be subtracted from both members or sides of the equation. If any such quantity be subtracted, let it be added.t If the unknown quantity have a multiplier, let the equation be divided by it. If it be divided by any quantity, let that become the multiplier. If any power of the unknown quantity be given, take the corresponding root. If any root of it be known, find the corresponding power. If the unknown quantity be found in the terms of a proportion (Arith. Sect. 10), let the respective products of the means and extremes constitute an equation; and then apply the general principle, as above. This simple direction, comprehending the seven or eight particular rules for the reduction of equations given by most writers on algebra, from the time of Newton down to the present day, is due to Dr. Hutton. It is obviously founded upon the mathematical axiom, that equal operations performed upon equal things produce equal results. †These two operations constitute what is usually denominated transposition. Otherwise in appearance only, not in effect, By transposing the 3, and changing its sign, x + 5 =9+ 3. By transposing the 5, and changing its sign, x = 9 + 3 −− 57. = 20, to find x. = 2. Given 3 x + 5 3. Given2 + d = 3 b 2 c to find x. a First, transposing d, 2 = 3 b −2 c + d. a Then, multiplying by a, x = 3 a b −2 a c + a d. 4. Given (3 x + 4) + 2 = 6, to find x. First transposing the 2, (3 x + 4) = 6 − 2 = 4. = 64. Then, transposing the 4, 3 x = 64 5. Given 4 ax 5 b 3 d = 60 0 = 20. +4 c, to find x. - First, transposing 5 b and 3 d x, 4 ax-3 d x-5b+ 4 c. That is, by collecting the coefficients, (4 a 3 d) x = 56 + 4 c. 5b4c Therefore, by dividing by 4 a 3 d, x = 4a-3 d 6. Given + } x Multiplying by 120 = 4 x 5 x 6, we have That is, collecting the coefficients, 34 x = 360. 34 = 7. Given x: a:: 5b: 3 c, to find x. Mult. means and extremes, 2 c x = 5 a b, Dividing by 2c, x=5ab÷ 2c= 20 a b 9 c |