= 291. It frequently happens in reducing a quadratic to the form of x2 + 2ax b, that 2a, the coefficient of x, becomes fractional, thus rendering the solution a little complicated. In such cases it will be sufficient to reduce the first member to the simplest entire terms. The equation will then be in the form in which a and b are integral in form, and prime to each other, and c is entire or fractional. To render the first term of (1) a perfect square, multiply both members by a; thus, where the first member is a complete square. Now if b is even, 4 b2 will be entire; but if b is odd, will be fractional, a result which 4 we wish to avoid. To modify the rule to suit the latter case, suppose (3) to be multiplied by 4; thus, The first member is now a complete square, and its terms are entire. Moreover, we observe that (4) may be obtained directly from (1) by multiplying (1) by 4a, and adding 2 to both sides of the result. Hence, for the second method of completing the square in the first member, we have the following RULE.-I. Reduce the equation to the form of ax2 + bx = c, where a and b are prime to each other. II. If b is even, multiply the equation by the coefficient of x2, and add the square of one-half the coefficient of x to both members. III. If b is odd, multiply the equation by 4 times the coefficient of x2, and add the square of the coefficient of x to both members. The above rule may be considered as more general than the first; for if applied to equations in the form of x2+2ax=b, the operation will be the same as by the first rule, with the simple modification of avoiding fractions in the first member, when 2a is fractional. 1. Given 5x2-6x=8, to find the values of x. We will now apply this rule to an equation in the form of x2+2ax= b, where 2a is odd. Thus we may always operate in such a manner as to avoid fractions in the first member; and indeed in the second member, if we first reduce both members of the equation to entire terms. 49 292. Either of the two preceding rules is sufficient for the solution of any quadratic equation. There are certain cases, however, where the solution may be much simplified, either by a modification of one of the common rules, or by a special preparation of the equation. 293. When the coefficient of the highest power of the unknown quantity is a perfect square. In this case the equation will be in the form of Let the quantity to be added to complete the square in the first member, be represented by ť2. Then Now in any binomial square, the middle term is twice the product of the square roots of the extremes. Hence, which may be used as the formula for completing the square, in this case. Or we may proceed according to the following RULE.-Divide the coefficient of x by twice the square root of the coefficient of x2, and add the square of this result to both members. |