4. Given (x+2x)—23(x2+2x)=-120, to find the values of x. This equation is in the quadratic form, for it contains the first and second powers of the compound quantity, x+2x. For convenience, let us assume y = x2+2x. Then by substitution in the given equation, we have and it will be found by trial that any one of these four values will satisfy the given equation. Equations of the third and fourth degrees may often be solved like quadratics, even if they do not, at first, present themselves in the quadratic form, like the last equation. If any equation is susceptible of such a solution, it will be found to contain the first and second powers of some compound quantity, with known coefficients. To determine whether this be the fact in any particular case, we may proceed as follows: Transpose all the terms to the first member; and if the highest power of the unknown quantity is not even, multiply the equation through by the unknown letter, to render it even. Then extract the square root to two or more terms, as the case may require; and if at any time a remainder occurs, which, with or without the absolute term, is a multiple, or an aliquot part of the root already obtained, a reduction to the quadratic form may be effected. Otherwise it will be impossible. 5 Given x-4x-14x+36x+45= 0, to find x. Hence the given equation may be written thus: (x3 —2x)3—18(x3—2x)+45 = 0. Without substituting any letter for the compound quantity, x2—2x, the remaining part of the operation will be as follows: (×3—2x)3—18(x2—2x) = −45, (x2-2x)-18(x2-2x)+81 = 36, 6. Given x+4ax2+2a3x-4a3 = 0, to find x. As the highest exponent of x is odd, we multiply the equation by x, and obtain Taking the square root of two terms, and factoring the remainder, (x2+2ax)3—2a2(x2+2ax) = 0. Hence we have two equations, solved as follows: Hence, x = a(1-3), −a(1+√/3), or -2a, Ans. We The two extremes in the first member are perfect squares. will therefore seek for a middle term which will render the square complete. This will be twice the product of the square roots of the extremes; or We therefore add 1 to both members, and solve as follows: 8. Given x+4x= 21, to find the values of x. OPERATION. x+4x=21, √x+2=±5, It should be observed here that when the equation contains a radical, as in the last example, it cannot be satisfied by the roots obtained, without a trial of signs. The roots found in the last solution are 9 and 49; but we have Now we may verify the given equation, if we take √x =+3 or -7; but not otherwise. Thus, putting x = 9 and x = 3 in the given equation, we have 9+12=21; also with x = 49 and √x=-7, we have In general, it will be found that a radical equation can be satisfied by each of the roots of solution, under at least one of the possible combinations of signs. We have here a radical equation which is not in the quadratio form. In such cases, it is generally better to clear the equation of radical signs, either entirely or partially. Thus, 11. (x+a)*+2b(x+a)* — 36′. Ans. x = b'—a or 816*—a. 12. +V/5 +10 = 8. = 13. 9x+4+219x+4=15. 14. V/10+x-10+x=2. = 15. (x—5)3—3(x—5) — 40. -5)3. 20. x+2x-7x-8x+12= 0. Ans. x = 1, 2, -2, or 3. 21. x-8x+19x-12= 0. Ans. x 1, 3, or 4. = 22. x-10x+35x-50x+24=0. Ans. x=1, 2, 3, or 4. 23. x*-8ax+8a2x2+32a3x-9a1 = 0. |