And if the equation, which is to be resolved, be of the fol we shall necessarily have, according to the same principle, Wherefore x == 2 + 12 = 10, or - 2 12 14, Where one of the values of x is positive and the other ne gative. 2. Given a2 Here a2 12x + 30 3, to find the value of x. 12x33027, by transposition, Whence x = - 6 ±√ (36 27), by the rule, Or, which is the same thing, x = 6 ± √9, Therefore x = 6 +3 9, or = 6 3 3. Where it appears that x has two positive values. 3. Given 2x2 + 8x 20=70, to find the value of x. Here 2x2+8x 70+20= 90, by transposition. And x2 + 4x 45, by dividing by 2, Whence x = 2 ± √(4 +45), by the rule, 2+7= 5, or - 2±√ 49. 2-7= 9. Where one of the values of x is positive and the other 1 1 2 x 5. Given x2-x+20=42, to find the value of x. 4223 - 2012, 221, by transposition, 1 And a2--=443, by dividing by, or multiplying by 2, ±√ (+44), by the rule, 1 400 Or, by adding and 44 together, x = ㄓ/ 9 1 3 3 1 Therefore x = +632 = 7, or = 62 ——613, Where one value of x is positive and the other negative. Or, multg d-c and a by 4a, x= 8. Given a1 + αx2 = b, to find the value of x. 2a v( a b 1 / { 4a (d—c)+b2 }. α a2 2 Or, x2 ㄓˇ ±√(a2 +46), by Whence x =±√ {-±√ (46 + a2) } by extraction 2.x3 3x3== 2, to find the value of x. 10. Given 2x3 + 3x 3 Here 2x+3x=2, by the question, This equation may be expressed as follows: *(x2 — 6x)2 + 8 (x2 -- 6x) — a, Consequently, (x2-6x)2+8(x2 — 6x) X4 12x344x248 x = a; for since, in extracting the square root of any quantity, the square of the root thus found plus the remainder, is always equal to the proposed quantity. In a similar manner, the biquadratic equation x4+2ax3+5a2x2 +4a3x=d, may be exhibited under the form (x2+ax)2+4a2 (x2+ax)= d; which can be resolved by the rule, page 126, for resolving quadratic equations. Hence it follows, that if the remainder, after having found the first two terms of the square root, according to the rule page 50, can be resolved into two such factors, so that the factor containing the unknown Whence x2 6x = −4±√(16 + a), by the common rule, And, by a second operation, x=39-4±√(16+a)}, Therefore, by restoring the value of a, we have :3 XC 3 ± √(5±√9025), Or, by extraction of roots, x = 13, the Ans. EXAMPLES FOR PRACTICE.* 1 3. Given 3x2 + 2x — 9 = 76, to find the value of x. 4. Given 1/22-33 Ans. x 5. x + 73 = 8, to find the value of x. Ans. x 11. Ans. x= 3. 6. †Given x + √ (5x + 10) = 8, to find the value of x. quantity shall be equal to the terms of the root thus found; the proposed biquadratic may be always reduced to a quadratic form, as above. See Ryan's Algebra, page 396.-ED. *The unknown quantity in each of the following examples, as well as in those given above, has always two values, as appears from the common rule; but the negative and imaginary roots being, in general, but seldom used in practical questions of this kind, as here suppressed, + In some quadratic equations involving radical quantities of the form √(ax+b), both the values of x, found by the ordinary process, will not answer the proposed equation, except we take the radical quantity with the double sign. In resolving the above example, two values X, that is 18 and 3 are found: but it appears, that 18 does not answer the condition of the equation, except we take the radical quantity V(5x+10) with the sign of -54, Now, since these two values of x are found from the resolution of the equation 2 -21x=-54; it necessarily follows that each of them, when substituted for x, must satisfy that equation; which may be verified thus; in the first place, by substituting 18 for x, in the equation X2 21x 54. we have (18)2 — 11 × 18 54, or 324-378. that is, -54 54, or, by transposition, 0 0. Again, substituting 3 for x, we have (3)2-21×3 - -54, or 9-63 = 54, .. 54—54 = 0, or 0 = 0. And as the equation x2-21x -54, may be deduced from the equation +√(5x+19)=8—x, or ·√(5x+10)=8-x; it is evident that the radical quantity v (5x+10) must be taken with the double sign ±, in the primitive equation, in order that it would be satisfied by the values, 18 and 3, of x, found above; that is, 18 answers to the sign and 3 to the sign +. See Ryan's Elementary Treatise on Algebra, Theoretical and Practical, where this subject is clearly illustrated.-ED. 7. Given (10 + x) — ✓ (10 + x)=2, to find the value ✓(10+ of x. Ans. x= 6. 8. Given 2x4 x2 + 96 = 99, to find the value of x. 9. Given x + 20x3 1059, to find the value of x. 2 x 12. Given ∞ √(3 + 2x2) = +3a2, to find value of a. 1 2 2 the 1 Ans. x = √(1+ √2). 14. Given - ✔(1 − ∞3) = x2, to find the value of x. х 16. Given (1 + x − x2) — 2 (1 + x − x2) ✓ − Ans. x= 6 =x, to find the value. ✓ (x − 1 ) + √ (1 − 1 ) = x, X 19. Given ∞ — 2 x3 + x = a, to find the value of x. |