III. The second root is equal to half the coefficient of the second term, taken with a contrary sign, minus the square root of the second member increased by the square of half the coefficient of the second term. EXAMPLES. 1. Let it be required to solve the equation, These roots may be verified. Substituting 3 for x, in the given equation, we have, 2. Again, let it be required to solve the equation, Clearing of fractions, and reducing to the required form, Reducing and writing out the roots, we have, x= 24+√-432+576, and x" 24-V-432+576, or, x' = 36, and x" 12. In writing out the roots by the rule, it frequently happens that the quantity under the radical sign is made up of two fractions, or of an entire part and a fraction, as in examples 1 and 2. In such cases, the two parts must be reduced to the least common denominator that is a perfect square, and then added together. 144. Many equations of a higher degree than the second, may be reduced to the form of equations of the second degree, and then solved. One of the most important classes of such equations consists of what are called trinomial equations. Such equations contain three kinds of terms, viz.: terms involving the unknown quantity to any even degree, terms involving the unknown quantity to a degree half as great, and known terms. Such, for example, as 26 - 4x3 = 32, in the same way that equations of the second degree are reduced to the form of, After an equation is reduced to the form (1) we may regard x" as the unknown quantity, and then it may be solved by the rule given for the solution of equations of the second degree. Having found the values of x", we may find the values of x, by extracting the nth root of these. To illustrate, let it be required to solve the equation, 26. x64x3- 32. This is of the required form. Writing out the values of 3, by the rule, we have, Whence, by extracting the cube roots of these values, we have, This is of the required form. Writing out the values of x2, we Whence, by extracting the square roots of these roots, we have, = + √/56 + 1 = 8, and 2" = −√/56 + - squaring each value and extracting the cube root of the result, we have, |