Observe that we may duplicate the expression x2 + 5x + 3, which appears under the radical sign, by adding 3 to the expression x2 + 5x which appears outside. Accordingly, adding 3 and also subtracting 3 from the first member of (1), we obtain the equivalent equation = x2 + 5 x + 3 − 2√x2 + 5 x + 3 15 0. Factoring with reference to √x2 + 5 x + 3, we obtain (2) (3) The roots of equation (3) include all of the roots of the following equations: V+5 +3–5=0, (4) and x2+5x + 3 + 3 = 0. 5 ± √113 The roots of equation (4) are found to be x = values will be found by substitution to satisfy the given equation. From equation (5) we obtain (5) These Since the right member, 9, of equation (7) was obtained by squaring a negative number in the second member of equation (6), it follows that when the values 6 and + 1 are substituted for x in the expression x2 + 5x +3 appearing under the radical sign in equation (6), the expression will represent the square of a negative number. Accordingly, when finding the value of √x2+5x+3, after having substituted — 6 and + 1 for x, it is necessary to consider that the result is a negative number. With this understanding, it may be seen that the values 6 and +1 satisfy the given equation. For, substituting - 6, we have, 36 – 30 — 2(−3) -12=0. Hence, 0=0. Substituting 1, we have, 1 + 5 – 2(− 3) — 12 = 0. Hence, 0 = 0. Equation (1) may be written in the equivalent form Equation (3) is quadratic with respect to the expression √x2 - x + £. Hence, factoring with reference to (2√/x2−x + 4 + 3) ( √/ x2 − x + 4 − 2) = 0. x2 x+4, we have, (4) Placing these factors equal to zero, we obtain the equations It should be observed that when these values are substituted for x, √x2- x + 4 must be a negative number, — §. Accordingly, with this understanding, these imaginary values will be found to satisfy the given equation. Equation (6) is found to have the solutions x = 0 and x = 1, both of which satisfy the given equation. EXERCISE XXIV. 1 Solve the following irrational equations, verifying integral or fractional results and rejecting "extra roots": (b) 2-√3x-2-√8 x 0. 13. (a) √x + 2 + √x − 13 + √x-5=0. x (b) √x + 2−√x-13-√x-5=0. (c) √x + 2 - √x - 13+ √x-5=0. (d) √x + 2 + √x - 13 - √x-5=0. 14. V+12+ √x 12 - √x + 23 = 0. 15. √5x + √2 + x = √14. 11. At least one solution of certain equations which have the p special form a may be obtained by the following process: 12. It should be observed that more solutions exist than are commonly obtained by the process above. Ex. 1. Find one or more solutions of x = 3. In this case the solution obtained is the only one which exists for the given equation. Ex. 2. Find one or more solutions of yž = 8. The solution obtained, y=4, is in this case one of the three possible solutions of the given equation. The complete solution of the equation may be obtained as follows: Solving the equations obtained by placing these factors separately equal to zero, we have Accordingly, the three solutions of the given equation are the real number 4 and the conjugate complex numbers - 2 + 2-3 and -2-2-3. These solutions will be found, by substitution, to satisfy the given equation. MENTAL EXERCISE XXIV. 2 Obtain one or more solutions of each of the following equations, regarding x, y, z, and w as unknowns and all other letters as representing known numbers: |