The two extremes in the first member are perfect squares. We will therefore seek for a middle term which will render the square complete. This will be twice the products of the square roots of the extremes; or We therefore add 1 to both members, and solve as follows: 8. Given x + 4√x=21 to find the values of x. OPERATION. x + 4√x = 21, x + 4√x + 4 = 25, √x + 2 = ±5, √x=3 or -7, x = 9 or 49, Ans. 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 492821; and the equation is satisfied in both cases. But putting x = 9 and √x=-3, we have 9-12=21; also with x 49 and √x = +7, we have both of which are false. 49 + 28 = 21; 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 quadratic form. In such cases, it is generally better to clear the equation of radical signs, either entirely or partially. Thus, |