by aid of these general formula; or the solution, in each individual case, may be discovered by applying to it the process whence those formulæ have arisen; and this is the plan usually recommended in books of instruction; thus, if the equation x2+ 12x+11=0 were proposed, we should proceed as follows:By transposing, a2+12x=-11; Completing the square, that is, adding 6o to each side, x2+12x+36=25.. extracting the root, x+6=±5; consequently, x=—6+5, or x=- -6-5: that is, x=-1, or —11; and these are the two roots of the proposed equation. If be put for x in the left hand member, we have (-1)2—12+11, which we see makes 0, and therefore satisfies the equation; and if 11 be put for x, we have (-11)2-12×11+11, which equally makes O, and which therefore equally satisfies the equation. As another illustration, let the equation x2-3x=10 be proposed. 2 Completing the square, a2—3a+ (3)2=10+ (3)° L 49 = 2. 4 ; that is 5, or -2. Substituting the first for x, the left-hand member of the proposed equation becomes 52-3×5, which is 10; and by substituting -2, the same expression becomes (-2)2+3× 2, which is also 10. It will of course be observed that, in extracting the root of the unknown side, after it has been converted into a complete square, no formal process is gone through: the extraction is effected by simply writing down x, and, against it, half the coefficient of x, with its proper sign. The following, therefore, is the rule for solving a quadratic equation: 79. RULE. Place the unknown terms on one side of the equation and the known terms on the other. Divide each side by the coefficient of the unknown square if it have a coefficient other than positive unity. Add the square of half the coefficient of the simple unknown to each side of the equation, and the unknown side will then be a complete square. Extract the square root of each side, affixing the double sign to the root of the known side, and the two values of the unknown quantity will be immediately deducible from the result. (1) Given x2-6x+7=47 to find the values of x. By transposing, x2-6x=40; Completing the square, x2-6x+9=49; .. extracting the root, x-3=√49±7.x=3±7= 10 or 4. Verification. Putting 10 for a in the proposed equation, the first member becomes 10-6.10+7. Putting -4 for x, it becomes (-4)+6.4+7: the result in each case is 47, which shows that x=10, and 4, both satisfy the equation. (2) Given 2x2+3x+44-46, to find the values of x. By transposing, 2x2+3x=2; Dividing by 2, x2+3x=1; Completing the square, x2+3x+(2)2=1+1=16; Verification. Putting for x, the first member of the equation becomes 2(--)2+-3()+44; and putting -2 for x, it becomes 2(-2)-3.2+44; and the result in each case is 46, as it ought to be. 36-x (3) Given 4x -46, to find the values of x. and these values may be verified as in the preceding instances. Transposing, +4x=26; Clearing fractions, and dividing by 2, 3(2x-11)+2x(x−3) =13(x-3); Removing brackets, and transposing, 6x+2x-6x-13x= 33-39; that is, 2x2-13x——6 .. x2- x=- -3; 13 2 y(y-1)=(y-1) .. dividing by y-1, y(y+1)= or y2+y = Completing the square, y2+y+=+1=1; ..x=2y=-13; that is, x=-1+√3 or 1-3. 2 (6) Given x-1=2+ to find the value of x. Putting y for, the equation becomes y—1—2(y+1). = y .. yy2.. completing, Extracting the root, y± &• y={+}; that is, y=2 or 1.. x=y2=4 or 1. It is necessary to notice here that these results, 4 and 1, are got by squaring 2 and -1; so that the square root of the 4 must be regarded as exclusively plus 2, and the square root of the 1 as exclusively minus 1: attending to this, the verification will be as follows (7) Given x=56x 256x ៖ +x to find the values of x. Multiplying by a3, x3=56+x3 ......(A); x Or, putting xy, y2=56+y .. y2—y=56; 225 Completing and extracting, y2-y+1= y== ± ..y=1(1±15)=8 or —7 .. x3=y2=64 or 49; .. x 4 or /49. NOTE.-Every equation, containing only two unknown terms, and where the index of the unknown quantity in one is double that of the other, as in the equation (A) above, may be reduced to a quadratic by substituting, as here, a new symbol for the lower of the two powers. It was with a view to this reduction that y was put for 2, and consequently y2 for r3, in this solution. The equation might indeed be treated as a quadratic in such circumstances, without substitution; but, in general, the solution would have a more complicated appearance. (8) Given x-9x2+20=0, to find the values of x. Putting y for x-2, and transposing, y2-9y=—20, Completing the square, ya—9y+ (2) * 2 81 Dividing by numerator of the second fraction and multiplying by denominator of the first, we have = .. clearing, a3—a2x—ax2; Dividing by a and transposing, x2+ax=a”; Completing the square, (10) Given 9x+ √ (36 x3 +16 x2)=15x2-4, to find the values of x. Taking the factor (2x)2 from under the radical, and transposing the 4, we have 9x+4+2x√(9x+4)=15x2.... (1); that is, putting y for √(9x+4), y2+2xy=15x2; Completing the square, y2+2xy + x2=16x"; Extracting the root, y+x=4x..y-3x or -5x. that is, √(9x+4)=3x or —5x .. 9x+4=9x2 or 25x2..(A); We have therefore to solve the two quadratic equations, 9x2-9x=4, and 25x2—9x=4. From the first, x2-x-.. completing, x2-x+4=÷+1=38: Extracting the root, a&.. x=1±1 =} or -. Hence, the proposed equation is satisfied by either of the following four values of x, namely:— It must be carefully observed, however, in substituting these values in the proposed equation, for the purpose of verification, that the conditions (A) by which they are governed be not violated: that is to say, that in using either of the first pair of values of x we must take care that the condition (9x+4)=3x be complied with, and that in using either of the second pair, that the condition (9x+4)=-5x be complied with, as these conditions overrule the sign of the radical. Suppose, for instance, we substitute the value x- ; then if we disregard the restrictions here adverted to, the first member of the proposed equation (1) would be 3+4— √1, that is, —3—4—3=3, while the second member would be 15(—})o 5;so that the value x would seem not to satisfy the equation: |