2 Equations in which the Unknown Quautity is containea in a Surd Expression. We have now a pure cubic equation, in which x has necessarily three values or roots, (255). Extracting the cube root of each side of the last equation, we find x=4, which is one value of x. To find the other two values of x, we must reduce the cubic equation to a quadratic by division, (253). Dividing each side of the equation x3-64-0 by x-4, we find x2+4x+16=0, or x2+4x=−16; hence x=-2±√−12. We have thus found x=4, or -2+√−12, or -2-√12; the first value being real, the other two imaginary. Each of these imaginary values of x, as well as the real value 4, will satisfy the equation 23-64. Thus a being equal to -2+-12, we have, (247), (249), x2=(−2+√−12)2—4—4 √ −12—12=—8—4√√—12. x3 —(—8—4√—12)(−2+√−12)=16+48=64, and In like manner the other imaginary value of x may be verified. 35. Find the value of x in the equation x+√x=20. From this equation we shall find x=25 or 16. The value 25 will not satisfy the original equation if the square root of a be restricted to its positive value; but this root is ±√x, and 25 satisfies the equation for the negative root. Two values of x may be found in each of the next three Exercises; but only that value is given in the Ans. which corresponds to positive roots in the given Equation. (x-2)2= By rationalizing the denominator, (243...2), we shall find Extracting the square root of each side of this equation, we have (x+√(x2-9))2 9 42. Find the value of x in the equation 3/x3+37×(x3+37)33–64. Ans. x=3, or-±√3. Equations of a Quadratic Form with reference to a Power or Root of the Unknown Quantity. (263.) Any Equation containing the unknown quantity in bat two terms with its exponent in one double its exponent in the other -is a Quadratic with reference to the lower power of x; and the value of such power may be found accordingly. To find the value of x in the equation x+4x=21. The higher fractional power is the square of the lower x; and the equation is therefore quadratic with reference to a Completing the square, we have x+4x+4=21+4=25. Extracting the square root of each side, x+2=±5; from which x-3, or -7. By raising each of these values to the 4th power, we find x=81, or 2401. The first of these two values of x is easily verified. In verifying the value 2401 it must be observed that its 4th root is -7, and that 4x4 is therefore —28. When x is in a fractional power, in the following Exercises, only that value will be given in the Ans. which satisfies the given form of the Equation.-Imaginary values of x are also omitted. 48. Find the value of x in the equation (x+12)+(x+12)=6 In this equation we must regard the binomial (x+12) as the unknown quantity; and to simplify the operation we may represent this binomial by y. 49. Find the value of x in the equation (2x+6)‡—6——(2x+6)a. 50. Find the value of x in the equation х x2+11+√x2+11=42. 51. Find the value of x in the equation Ans. x=4. Ans. x=5. Ans. x=5 This equation may be reduced to the form of a quadratic, thus.— The first two terms of the square root of the first side will be found to Now the square of the root found, and the remainder, are together equivalent to the first side of the equation; hence we have (2x2 — x)2 — 1(2x2 —x)=33. Ans. x=2, or —14, A Biquadratic equation may be reduced to the form of a Quadratic, as above, whenever the remainder-after having found the first two terms of the square root of the first side- -can be resolved into two factors, one of which is the same as the part of the root thus found Transposing the 12, and multiplying by x, we have x4-8x3+19x2-12x=0. We have now a Biquadratic equation which may be reduced to the form of a Quadratic by the method just exemplified. We find the first two terms of the square root of the first side of the Equation, by the common Rule. The remainder, 3x2-12x, may be resolved into 3(x2-4x), in which the binomial factor is the same as the part of the square root above found. Then (x2-4x)2+3 (x2-4x)=0. Ans, x=4, 3, or 1. 54. Find the value of x in the equation x4-2x3+x=5112. 55. Find the value of x in the equation Ans. x=9 Ans. x 2, -3, 1, or —2 56. Find the value of x in the equation 57. Find the value of x in the equation = Ans. 1, 2, 3, or 4. x4-12x3+44x2-48x-9009. 58. Find the value of x in the equation x2-8ax3+8a2x2+32a3x=s. Ans. x=13 Ans. x=2a±1 8a2±√s+16a*. |