b b consequently, we have a + 2 = (1 + - or a2 + 20 a x a2 (x b12 2a4a2 b -30 a Hence, having reduced the equation to the form + consequently, extracting the square root (as in the case of a that the equation has two roots, or admits of two solutions, one of which results from taking + for ±, and the other from taking for 1. Hence, a complete quadratic equation may be solved by the following RULE. 1. By transposition, etc., reduce the equation to the form coefficient of the first power of x, which is called completing the square) to both members, and the first member will be 2. Extracting the square root, we get x+ which, being an equation of the first degree, is solved by b transposing 2a Thus, if we take the equation -8x= 9, and compare + 4a2 QUADRATIC EQUATIONS. x ” 9 and 1 for the roots of 2-8x=9; and since 9 and -1, when put for x, reduce it to 99 an identical equation, it follows that the roots have been correctly found. b a b a = 15 to x2 + Again, if we compare the equation 2 + 2 16 a and 15 and for the representatives of and of course x + 16 a 3 1 15 1 + = + =1, which gives x= and x = 4 16 16 4 for the roots of x2+ 2 (2.) The equations a2 = to the same general forms in x. For, dividing their terms by ax, and transposing, they bers are of like forms, and 0 is the right member of the first, b while is that of the second equation, which is the only way, it is clear that the values of x and must be the roots of the equation; and in like manner, it is clear that the values of x and tion. с must be the roots of the second equa ax To find the roots, we square the members of the equations, -9 a a = a2x2 = 2 = a2 ax a2x2 = a2 and ex tracting the square roots of the members of these equations If we take the half sum and half difference of the equa Hence we clearly perceive the reason why the square root b of x2 is usually written in the form «= ±√√2; viz., = a a because the sign + in is taken to express the arithmetical Again, if we take the half sum and half difference of the and it is clear that these roots are the same that we have found in (1). b equation + is reduced to that of the equations x + ; consequently, the solution of the QUADRATIC EQUATIONS. с and xy= a noticing that the values of x and y are the roots of the equation + x = b с Because (x + y)2 a — 4xy = (x − y)2 a2 = a 4c be we get x-y= + " α sequently, the solution of the equations x+y= с is reduced to that of the equations x + y = a2 and the values of x and y will be found to b a give the same values of the roots of the equation a2 + -x= as we have previously found. b Thus, if we compare the equation + 2 = 15 to a2 + c b с and will be represented by 2 and 15; conse a 15, and thence we shall have = √60+4=8; hence, by taking the -- half sum and half difference of the equations x + y = 2 the equation a2 + 2x = 15. If we represent the roots of the equation 2 + -= (x − r) (x — r') = 0; and the expression 2 + x a α с - is a resolved into the factors ar and x-r', agreeably to what Moreover, by taking the first derived equation of x2 + -x are real that b 2a α it is clear if the roots α must lie between them; which is in con formity what has previously been shown. b Hence, if does not lie between the roots of x2+x == 0 or x2 + α 2a b sible or imaginary. Thus, if we take the equation x2 + 2x + 3 = 0 for an exam ple; we have b 2a = 1, which being put for x in a2 + 2x +3, gives the positive result 2. Also, if we put = ±∞ an unlimitedly great number for æ, in a2+2x+3; the results are also positive. Hence, since x2 + 2x + 3 does 1 and x = it fol±00, not change its sign between x = lows that the roots of a2+2x+3= 0 are impossible or imaginary; indeed, the solution of the equation gives x=-1 ±√2; which are impossible roots. с (3.) The solution of the equation + _x = can be reduced to that of a pure quadratic. Remarks.-1. If we have the general equation " + Ax2-1 A + Px + Q = 0, and put x + = n equation, it will be changed into an equation of the form y" + B'y"-2 + C'y"-3+ etc. = 0; an equation which wants its A n |