In the following examples the roots will be limited to principal values: Ex. 1. Solve the equation x + √(25 − x2) = 7. Before squaring, it is better to have the radical by itself in one member. Transferring x, Squaring, √(25x2)=7-x. 25x249 – 14 x + x2. The roots of this equation are 3, 4. (1) (2) Both roots of (2) satisfy the given equation, since 3+ √(25 — 9) = 7, and 4 + √(25 – 16)=7. Therefore no root was introduced by squaring both members of the given equation. This is also evident from the following considerations: Any root of the additional equation, √(25 — x2) = −(7 — x), or −√(25 − x2) = 7 — x, (3) obtained by changing the sign of one of the members of the given equation when prepared for squaring, must be a root of the rational integral equation (2). But both roots of this equation, 3 and 4, make the first member of (3) negative, and the second member positive. That is, equation (3) is an impossible equation. Ex. 2. Solve the equation x√(25 — x2) = 1. The roots of this equation are 4 and 3. The number 4 is a root of the given equation, since 4 - √(25 — 16) = 1; but the number - 3 is not a root of the given equation, since Therefore, the root 3 is a root of the additional equation √(25 x2)= − (1 − x), or √(25 − x2) = 1 − x, introduced by squaring. That -3 is not a root of the given equation is also evident from the form of the equation. For any real value of x which makes x - √(25 - x2) equal to 1 must be greater than 1, and therefore cannot be equal to 3. - The preceding examples illustrate the following method of solving irrational equations: Transform the given equation so that one radical stands by itself in one member of the equation. Equate equal powers of the two members when so transformed. Repeat this process until a rational equation is obtained. ALGEBRA. [CH. XXIII 4. In the preceding article the indicated roots in the equations were limited to principal values. At the same time an irrational equation, if written arbitrarily, may be inconsistent with the laws governing the relations between numbers. In such a case the equation is impossible, that is, it cannot be satisfied by either real or imaginary values of the unknown numbers. E.g., √(x+6) + √(x + 1) = 1 is an impossible equation. For it cannot be satisfied by any complex value of x, since by Ch. XX., Arts. 31 and 22, √(x + 6) + √(x + 1) must be complex if x be complex, and hence cannot be equal to 1. It cannot be satisfied by any real positive value of x, since, in that case, either √(x + 1) or √(x + 6) is greater than 1. It cannot be satisfied by any real negative value of x, since, if z be negative and its absolute value be less than 1, √(x+6) will be greater than 1, and if x be negative and its absolute value be greater than 1, √(x + 1) will be imaginary. 5. But if the restriction to principal roots be removed, any irrational equation contains in itself the statements of two or more equations. E.g., if both positive and negative square roots be admitted, the equa√(x+6) + √(x + 1) = 1 tion is equivalent to the four equations √(x+6)+√(x+1)=1 (1), −√(x+6)+√(x+1)=1 (3), √(x+6)=√(x+1)=1 (2), −√(x+6)=√(x+1)= 1 (4), in which the roots are limited to principal values. The same rational integral equation will evidently be derived by rationalizing any one of these equations. Therefore the roots of this rational equation must comprise the roots of these four irrational equations. Consequently, in solving an irrational equation, we must expect to obtain not only its roots, but also the roots of the other three equations obtained by changing the signs of the radicals in all possible ways. Some of these equations can be rejected at once as impossible. The roots of the other irrational equations will be the roots of the rational equation. Thus, of the above equations, (1), (3), and (4) can be rejected at once as impossible. The rational equation derived from any one of the four equations is x+1=4; whence x = 3. The number 3 is a root of the one equation not rejected, since √(3+6) − √(3 + 1) = 1. The same conclusions could have been reached by substituting the roots of the integral equation successively in the irrational equations, rejecting those which are not satisfied by any root. we may take √(3 x2 − 2 x + 4) as the unknown number, replacing it temporarily by y. We then have y − y2 + 4 = − 16. The roots of this equation are 5, and — 4. (3x2 - 2 x + 4) to each of these roots, we have √(3x2 − 2 x + 4) = 4, whence x = } (1±√37). The numbers 3, satisfy the given equation, and are therefore roots of that equation. The numbers √(1±√37) do not satisfy the given equation. But if the value of the radical be not restricted to the principal root, the given equation comprises the two equations √(3 x2 − 2 x + 4) − 3 x2 + 2 x = — 16, −√(3x2 - 2 x + 4) - 3 x2 + 2 x = 16. Then (1±√37) are roots of (2). Ex. 2. Solve the equation (3 x2 + 13) + √(3x2 + 13) = 6. (1) (2) Assuming (3x2 + 13) as the unknown number, and representing it by y, we have y + y2 = 6. Equating (3x2 + 13) to each of these roots, we have √(3x2 + 13) = 2, whence x = ±1, √(3 x2 + 13) = − 3, whence x =±√2 =±}√51. The numbers ±1 are roots of the given equation, since 16+ √16=6. The numbers±√51 are evidently not roots of the given equation, but are found to be roots of the equation − √(3 x2 + 13) + √(3 x2 + 13) = 6. The preceding examples illustrate the following principle: If a radical equation contain one radical, and an expression which is equal to the radicand or which can be made to differ from the radicand (or a multiple of the radicand) by a constant term, it can be solved as a quadratic equation. The same is true if the equation contain two radicals, one the square of the other, and in addition only constant terms. In both cases, the radicand must, in general, be a linear or a quadratic expression. 7. Irrational equations containing cube and higher roots in general lead to rational, integral equations of a higher degree than the second, and therefore cannot be solved by means of quadratic equations. But in some cases their solutions can be effected by special devices. Ex. Solve the equation (8x + 4) − √ (8 x − 4) = 2. Cubing, 8x+4-3[(8x+4)]2 (8 x −4) + 3 √(8 x + 4) [ √(8 x − 4)]2 - 8x+4=8. Transferring and uniting terms, and dividing by - 3, (1) [√(8x+4)]2(8x-4) (8x + 4) [ √(8 x − 4)]2 = 0. (2) Factoring, (8x + 4) (8 x − 4) [ † (8 x + 4) – † (8x-4)] = 0. This equation is equivalent to the three equations (3) The numbers (8x+4) − (8 x − 4) = 0, 8x+4=8x 4. Equation (6) is not satisfied by any finite value of x. and are found to satisfy the given equation. (6) EXERCISES. If a Solve each of the following equations, and check the results. result does not satisfy an equation as written, determine what signs the radical terms must have in order that the result may satisfy the equation. 23. 3x-2√x − 1 = 0. 24. √(x+2)=√(x2+2x)= 0. 31. √(x-2)+ 2 √(x+3) − 2 √(3 x − 2) = 0: 40. x2-x+2√(x2−x−11)=14. 41. x2+24=2x+6√(2x2-4x+16) 42. √(2x2 - 3 x + 5) + 2 x2 − 3 x = 1. 43. 2x(4x2- 27 x)=-5x2+27x+9. 44. √(3x2+7x1)- √(3x2-4x+5)=8. 45. √(2x2-7x+7) + √(2x2+9x-1)= 6. |