13. Find a mean proportional between 7 and 9, carried to three decimals. 14. Find a mean proportional between 75 and 425, accurate to three decimals. 15. Find the sum of the series 1, 1, 1, &c., carried to infinity. 16. Find the sum of the series, 4, 1, §, &c., continued to infinity. 17. Find the sum of 7, 14, 28, &c., continued to infinity. 18. What is the sum of 81, 9, 1, §, &c., continued to infinity? 19. Insert three mean proportionals between 2 and 162. 20. Insert two mean proportionals between 5 and 1080. 21. Insert a mean proportional between every two adjacent terms of the progression, 3, 75, 1875, 46875. 22. Insert two mean proportionals between every two adjacent terms of the series, 1, 8, 64, 512. (16+x2)=x+2; squaring both members, 16+x2x2+4x+4; transposing, x2 — x2 — 4 x —4—16; reducing, changing the signs and dividing, x=3. Ans. Clearing the equation of fractions and reducing, x+34√x +168 = x +42√/x + 152; transposing, x+34√/ x − x − 42√π = 152 — 168; reducing, x x −8√/ĩ =−16; changing the signs and dividing by 8, 2 m m 4. Given x2+5ax+b2=a+x; to find x, Raising both members to the mth power, ducing, √x2+5 ax + b2 = a+x; squaring, x2+5ax+b2= a2+2 a x + x2; transposing and re 5. Given (x+6)* = (x—6)2; to find x, Raising both members to the 4th power, x+6=(x-6)2, or x+6= x2-12x+36; inverting the members, · 12 x + 36 = x+6; transposing and reducing, x2- 13x=- 30. Now, by substituting in formula 4th, Art. 94, x=3±√30+169 = 33±. Hence, We see, from the preceding examples, that an equation containing radical quantities, may generally be freed from them, by raising both members to the power denoted by the degree of the radical, the operation being repeated, if necessary. When some of the terms do not contain radicals, it is usually best, in the first place, to make them constitute one member, and the remaining terms the other. Many of the problems in this and the following section will give several answers each, if the double sign be prefixed to The mode pursued in the preceding questions, frequently leads to equations of so elevated a degree, that their solution would be too difficult for an elementary work. Other expedients, therefore, are often to be preferred. Whenever an equation can be reduced to the form of x2m± pxmq, that is, to an equation, in which the unknown quantity is found in two terms only, and has an exponent in one of them double its exponent in the other, it may be solved after the manner of affected equations of the second degree. 12. Given x4+6 x2 = 135; to find x. First consider 2 as the unknown quantity, and make the first member a perfect square, 24+6x+9=144; extracting the square root, x2 = 9 or 15; taking the square root of this, x = ± 3, or x =±√−15. Hence, x=3, x=-3, x = √/— 15, or x=-15. This question might have been solved by means of formula 1st, Art. 94. Thus, x2=−3±√/135+36= −3 ± 129 or — 15; then, x=3, or x = ±√15. 13. Given x+4=32; or what is the same, xx+4x+= 4 equations, -- 4 x =3, or =; taking the 4th power of both x=81, or x=6561. Ans. 15. Given x+10=5x+4; to find x. Transposing and reducing, x3 — 5 x = == 6; completing the square, x+5x+25=-6+25; or, reducing the second member, power, x3 — 5 x3 + 2 = 4; taking the square root, 2 — § = ± 1; transposing and reducing, z=3, or x = 2; raising both equations to the 6th x=729, or x=64. Ans. If we had substituted in formula 4th, Art. 94, the operation would have been shorter. 16. Given√/x5 +√x3 =6√/ĩ; to find z. Taking the roots of the perfect squares, and placing them before the radical sign, This may now be solved like any affected equation; and the following equations may be solved like the preceding. 32. Given x+5=√x+5+6; to find 2. By transposition, x+5−√x+5 = 6. Considering +5 as the unknown quantity, and completing the square, |