Note. All equations, which have the index of the unknown quantity, in one of their terms, just double that of the other, are resolved like quadratics, by first finding the value of the square root of the first term, according to the method used in the above rule, and then taking such a root, or power of the result, as is denoted by the reduced index of the unknown quantity. Thus, if there be taken any general equation of this kind, as x2m+axm=b, we shall have, by taking the square root of x2m, and observing the latter part of the rule, of the second term of the equation be added to each of its sides, so as to render it of the form that side which contains the unknown quantity will then be a complete square; and, consequently, by extracting the root of each side, we shall have ≈±za=±√‡a2 ±b, or x=‡‡a±√‡a2±b, which is the same as the rule, taking a and b in + or - as they may happen to be, It may here, also, be observed, that the ambiguous sign ±, which denotes both and, is prefixed to the radical part of the value of x in every expression of this kind, because the square root of any positive quantity, as a2, is either a or-a; for(+a) x (+a), or ( a)×( − a) are each +a2 but the square root of a negative quantity, as - -a2, is imaginary, or unassignable, there being no quantity, either positive or negative, that, when multiplied by itself, will give a negative product. To this we may also further add, that from the constant occurrence of the double sign before the radical part of the above expression, it necessarily follows, that every quadratic equation mus have two roots; which are either both real, or both imaginary, according to the nature of the question. And if the equation, which is to be resolved, be of the following form m xmax=b, we shall necessarily have, according to the same prin 1. Given x2+4x=140, to find the value of x. Here x+4x=140, by the question, Whence x-2/4+140, by the rule, Or, which is the same thing, x=−2±√/144, Wherefore x= →2+12=10, or —2—12——14, Where one of the values of x is positive and the other negative. 2. Given x2. 12x+30=3, to find the value of x. Whence x=-2±4+45, by the rule, Or which is the same thing, x—— −2±√49, Therefore x=— −2+7=5, or =— -2-7=-9, Where one of the values of x is positive and the other negative. 4. Given 3x2 -3x+6=51, to find the value of x. 2 Here 3x-3x=54-6-by transposition. And x2 X=- by dividing by 3, 1 Whence x= = ±√(3), by the rule, Or, by subtracting from 14 1 1 x= 36' 2 9 1 Whence we have x=√(+444), by the rule, 3 1 Or, by adding and 441 together, x= 9 1 3 Therefore x= +63=7, or =1—63=−6}, 400 Where one value of x is positive, and the other negative. 6. Given ax+bx=c, to find the value of x. b C Here x2+ x=-by dividing each side by a. α α 7. Given ax2 bx+c=d, to find the value of x. Here ax2-bx=d-c, by transposition, 8. Given x4+ax2=b, to find the value of x. Here x+ax2b, by the question, α · 1 Or x2 = • = ± √(22 + b) = = = = = √(a2+46), by the rule, Whence a=±√(-±√46+aa) by extraction of 10. Given 2x3+3x3 =2, to find the value of x Here 2x3+3x3 =2, by the question, Therefore x=({})3='1', or (−2)3— — 8. 11. Given x-12x3+44x2-48x=9009(a), to find the value of x. This equation may be expressed as follows, Whence x2-6x=-4±√16+a, by the common rule, Or, by extraction of roots, x=13, the Ans. EXAMPLES FOR PRACTICE. (d) 1. Given x2 - 8x+10=19, to find the value of x. Ans. x=9 2. Given x2. -x-40=170, to find the value of x. Ans. x=15 3. Given 3x2+2x-9=76, to find the value of x. 1 1 Ans. x=5 4. Given 3 x+7=8, to find the value of x. Ans. x=11⁄21⁄2 (d) 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, are here suppressed. |