292. Either of the two preceding rules is sufficient for the solution of any quadratic equation whatever. 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 example. 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 t. Then Now in any binomial square, the middle term is twice the product of the square roots of the extremes. Hence, which case. or a2a2+bx+(2a)2 = c + (2a)", 4a (3) may be used as the formula for completing the square, ir this Or we may proceed according to the following RULE. Divide the coefficient of x by twice the square root of the coefficient of x3, and add the square of this result to both members 1. Given 25x-20x=-3, to find the values of x. In this example the number to be added to complete the 294. When the equation is in the form of x2+2ax = (2a+m)m. (1) In this case we may avoid tedious numerical operations, by the use of the auxiliary quantity, 2a. 1. Given x2-5x= 6, to find the values of x. Put 2a 5; then 2a+1=6; = Assume 2a = 19; then 4(2a+4)=8a-+16=92; and the solution will be as follows: x+2ax=8a+16, x4 or -23, Ans. Let it be observed that we always put the coefficient of x equal to 2a. Then the method will apply, provided the second member is 2a+1, or 4a+4, or 6a+9, or 8a+ 16; or, in general, 2am+m2; that is, any multiple of 2a plus the square of the multiplier. 5. x-75x = 76. Ans. x=76 or -1. 6. x+72x385. 7. x2-325x = 3350. Ans. x 5 or -77. Ans. x= 335 or -10. 295. When large numerals may be avoided in the operation, by the use of an auxiliary quantity. As all the examples of this kind can not be included in any general classification, we give the following illustrations : 1. Given x2+9984x = 160000, to find the values of x. Put 2a = 10000; then 2a -16 = 9984, and 32a = 160000. Whence, x2+(2a-16)x = 32a, x2+(2a—16)x+(a−8)2 = a2+16a+64, x+(a-8)=(a+8), 2. Given x2+45x = 9000, to find x. x = 16 or 10000, Ans. = 9000. Put a 45; then 200a x2+ax=200a, 4x2+4ax+a2 = a2+800a, 2x+a = ±Va(a+800) = √/45×845. Multiply one of the factors under the radical by 5, and divide the other by 5; then we have 4(a+1)2x2—4a2(a+1)x+a* = a*+4a3+4a”, 2(a+1)x-a' = ±(a2+2a), 2(a+1)x2a'+2a or -2a, EQUATIONS IN THE QUADRATIC FORM. 296. There are many equations which, though not really quadratic, or of the second degree, may be solved by methods similar to those employed in quadratics. All such equations are reducible to the following form: x2+2ax" = b; in which x represents a simple or a compound quantity, and n is positive or negative, integral or fractional. It is always necessary that the greater exponent should be twice the less. 1. Given x-16x3-28, to find the values of x. 2. Given x-6x=-5, to find the values of x. 3. Given x2+10x-1=24, to find the values of x. |