The first eight examples offer no difficulty beyond the exact performance of the operations indicated. 9. This may be shown in two ways: by extracting the square roots of each quantity and taking the sum of the results, and by squaring each side of the expression, and showing that the result when reduced is 32+68=100, an obvious identity. XXIX. The first five examples require no remark. 6. The expression may be put into the form X y-x-1 y-2x-12x-yv -1 (x − y√ −1)(y + x√ −1 · (x2 + y2)}. 10. 7±4/3, 26±15/3, 97±56/3, 362±209/3. 11. 5±2/6, 11√√/2±9√/3, 49±20√6, 109√2±89√3. 12. 77+30/6, 655✅/2+531√3, 11329 +4620✓✅/6, 98225✅/2+80187√3. XXXIV. {√2}={2})=2* or /2: {2}={2}-2 or /2. The square roots of the last four examples can be found as the example in the note, p. 15. The roots are respectively 3+ √3, √3+1, 3√3-2√2, 2+3√✓−1. √2 XXXV. 1. Here 3/147=3√(49×3)=3×7√3=21√3 -3√75=-3√(25 × 3) = −3×5√3=-15√3 2. 2√3. .. 3√147-375-3√/}=(21-15-1)√3=5√3. 3. 18+20/2-12/6. 4. 8√2+7√3+5√/5+2√30. 5. 1+√3+√5. See XXXIX. 8, p. 36. 6. 2. 7. 2/3. 8. 75/2. 9. /3+2/2. 10. 3. -- 11. Let (√5+2)*+ (√√/5 − 2)3=x, and put (√/5+2)3=a, (√√/5 − 2)3==b, then x=a+b, and x3 = (a+b)3=a3+b3+3ab(a+b), but as+b3=2√/5, and 3ab=3, .. x3=2/5+3x, and x3 - 3x=2√√/5. To find the value of x requires the solution of a cubic equation. 12. 17.69. 13. 90. XXXVI. Since (√a+b)(√a−√√b)=a−b, and (a\/x+b√y)(a√x−b√y)=a2x—b2y ; it is obvious that √/a+ √b may be made rational by multiplying it by √a−√b; ab by multiplying it by √a+√/b. and Similarly for the factors ax+b1y and a√√x-b√y. When there are three terms, √a+√√√/b+√c, since {(√a+√√/b)+√c}· {(√a+√√b)−√c}=(√a+√√/b)2 −c=(a+b−c) +2\/ab, and {(a+b−c)+2a√√/b}. {(a+b−c) − 2\/ab}=(a+b−c)2 — 4ab. And two factors will be required in general to render such a denominator a rational quantity. 1. 8-52_8-5\/2_3+2√2_24+√√2 − 20 ̧ 9-8 =4+√/2. 1. The expressions 24, 33, 4, 5, 63, are respectively equivalent to (230)ő, (320)a's, (415)a's, (512), (610); and by expanding the powers 230, 320, 415, 512, 610, their numerical values will be known and the given expressions can be arranged in the order of magnitude. 2 2. 2 √3 2√3 3 (3) 329 The extractions and divisions in each example are to be carried out to four places of decimals. 3. 4 is greater or less than /5. According as 17+2\/70 270 280 22+2/57, by squaring each side. but 280 is obviously less than 25+20√57 +228, .. 10+√7 is less than 19+ √3. (2) √2+√7 is greater than √3+ √5. 5. If every two of the three lines be greater than the third, a triangle can be formed with the three lines. 6. It is obvious that if the sides of the equilateral triangle, the square, the pentagon, be given in magnitude, the areas of these regular figures can be determined, and consequently the surfaces of each of the five regular solids. The surface of the tetrahedron is equal to the area of four equal and equilateral triangles. The surface of the hexahedron is the area of six equal squares. The surface of the octahedron is equal to the areas of eight equal and equilateral triangles. The surface of the dodecahedron is equal to the area of twelve equal equilateral and equiangular pentagons. The surface of the icosahedron is equal to the area of twenty equal and equilateral triangles. A sphere can be inscribed within, and another circumscribed about, each of the five regular solids, in the same manner as a circle can be inscribed within and another circumscribed about each of the plane regular figures, the equilateral triangle, the square, the pentagon. Suppose a sphere described about a regular tetrahedron, the surface of the sphere will pass through the four points of the trihedral angles of the tetrahedron. If lines be supposed to be drawn from these points to the centre of the circumscribing sphere, these lines will be each equal to the radius of the sphere. Hence the tetrahedron may be conceived to be divided into four equal triangular pyramids, whose common vertices are at the centre of the sphere, the three edges of each, radii of the sphere, and their four bases the equilateral triangles which make up the surface of the tetrahedron. Since the content of any pyramid is equal to one-third of the content of a prism on the same base and of the same altitude, the contents of the four equal pyramids can be found, and the content, or volume, of the tetrahedron determined. In a similar manner it may be shown that : The volume of the hexahedron is equal to the volumes of six equal pyramids whose bases are equal squares. The volume of the octahedron is equal to the volumes of eight equal pyramids whose bases are equal equilateral triangles. The volume of the dodecahedron is equal to the volumes of twelve equal pyramids whose bases are equal equilateral and equiangular pentagons. The volume of the icosahedron is equal to the volumes of twenty equal pyramids whose bases are equal equilateral triangles. c1+a+b+2a2c2 - 2bac2 - 2a2b2 = 4a2c2 - 4a2b2c2; See Section IV., XIII., Ex. 10, p. 32. 5. Since x(a2 -y2)*+y(a2 − x2)* = a2, x(a2 — y2)} = a2 − y(a2 — x2)1, by squaring each side, .'. a2 − (x2 + y2)=0 and x2 + y2=a2. 6. Since x+y=a3, x§+y1—ai=0, and x+y+a+2x1y‡ — 2a1x} − 2a1y=0 The truth of this also may be verified by numerical examples, as, for instance, when (16)+(25)*=(81)1. 8. If (a+b+c+d})}=x3+y1+zł, Then a+b+c+d1=x+y+z+2(xy)1+2(xz)1+2(yz)*, and x+y+za, 2(xy)*=b1, 2(xz)+=c*, 2(yz)*=d1. Hence 2(xy)1.2(xz)1.2(yz)*=bicid1, or 8xyz=(bcd)1, and x+y+z=α, But 4xy=b, 4xz=c, 4yz=d, .'. bc+bd+cd=16(x2yz+y2xz+z2xy)=16xyz(x+y+z). 9. Let {r+(r2+q3)*}*=a, {r−(r2+q3)+}}=b; then x3=(a+b)3=a3 +b3+3ab(a+b)=2r−3qx ; .. x3+3qx-2r=0. 10. Find y and z in terms of x, and substitute these values in a1+y1+z1=1, whence x is known, and y, z by substitution. Lastly, substitute these values of x, y, x in ax2+by2+cz2. |