thus, are the two solutions. 7. Given x + 5 = √(x + 5) + 6, to find x. Instead of squaring each side, we may reduce the equation In this question 1 also appears as a solution, but on examination it will be found to correspond to the sign of the radical. 8. Given ax +√(a2x2 − bx) = cx + d, to find x. squaring, √(a2x2 — bx) = (c − a)x + d, b .. {(c − a)3 — a3} x2 + 2 { d(c − a) + ' } x = − đ, 10. Given a + x + Va− x = b, to find x. Squaring, Na+ x + 2 √ a2 − x2 + √ a − x = b2; transposing, √a+x+ √a-x= b2-2√α- x2; again squaring, a + x + 2 √ a2 − x2 + a − x = (b2 − 2 √√ a − x2)2 or Now, if 2 2 = b* — 4b2 √/a2 − x2 + 4 √√ æ2 − x2 ; α 2a + 2√axb-4b2 √a2x2 + 4 √√ a2 = x2. = a2 be considered as a simple quantity, √axis its square; and the equation will be of the usual form when written 2a-2-46 √α- x2 = 2a - b*; or, dividing by 2 and completing the square, extracting the root, Vax2- b2± √√/4a + 26* = 2 262± √4a + 26* 25 Na - x2= 2 Multiply each side of the equation by x, the result is But it must be observed that the equation itself is equivalent the that the sign of the square root in the given equation is +, former solutions only are held to be valid. 1 2 Here 4 is a common factor of every term, and as neither 4 nor x-√4, by which we have multiplied the given equation, involves it, the same quantity must be a factor. of the given equation. Now if a factor involving x enters into an equation, it is obvious that one way of satisfying the equation, that is, of solving it, must be to make this factor equal to zero. Hence 4 is a solution of the equation; which on inspection is found to be correct. x = If, now, we divide the equation by x 4, this solution no longer appears. By this division the equation becomes x √x − 4 = (x + 4) √x + 4 ; Squaring, x(x 4) = (x + 4)2 (x + 4), x3- 4x3 = x2 + 12x2+48x + 64, 16x2+48x+64 = 0, two impossible solutions, which satisfy the given equation. The given equation has therefore three solutions. Strictly speaking it is not a quadratic equation; but, as the treatment of equations of this class differs in no way from that which is applicable to ordinary quadratics, it is usual to include them in the same category. We shall frequently, in the examples which follow, omit impossible solutions, and set down only those that are possible. 15. x + √x : x − √x : : 3 ( √x + 5 ) : 4 √√x. |