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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 aRx+bx = c.
(1) Let the quantity to be added to complete the square in the first member, be represented by t. Then
a'a'+b3+= c+t". Now in any binomial
square, the middle term is twice the product of the square roots of the extremes. Hence,
2txax = bx;
And equation (2) becomes
(3) which 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 x', 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 square is
= (2)' = 4;
2X5 and the whole operation will be
25x*—20x = -3, 2529_203+4=1,
| or £, Ans.
294. When the equation is in the form of
x2 + 2ax = (2a+m)m. In this case we may avoid tedious numerical operations, by the use of the auxiliary quantity, 2a. 1. Given x*—5x = 6, to find the values of x.
Put 2a = 5; then 2a+1=6; and the equation becomes a' _-2ax 2a+1;
whence, 2-2axta' = a +2a+1,
x = 6 or -1, Ans. 2. Given 2*+19x = 92, to find x.
Assume 2a = 19; then 4(2a+4)=8a-+16=92; and the solution will be as follows:
2+2ax = 8a+16,
x=4 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+m'; that is, any multiple of 2a plus the square of the multiplier.
EXAMPLES FOR PRACTICE.
Solve the following equations : 1. x*—7x=8. 2. x+11x=26. 3. x*—17x = 60. 4. .c" +213=46.
Ans. x = 8 or -]. Ans. x=2 or
-13. Ans. X= 20 or -3. Ans. ~= 2 or —23.
5. X_75% 76.
Ans. X =
=76 or -1. 6. x+72x = 385.
Ans. x=5 or -77. 7. x?_325x 3350.
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 x*+9984x= 160000, to find the values of x.
Put 2a = 10000; then 2a -16 = 9984, and 32a = 160000. Whence,
x*+(2a—16)x = 32a,
x+(a—8) = +(a+8),
w = 16 or -10000, Ans. 2. Given ° +45% 9000, to find x.
Put a = 45; then 200a = 9000.
** +ax = 200a,
2x+a = +Va(a+800) = V 45X845. Multiply one of the factors under the radical by 5, and divide the other by 5; then we have
2x+a=V/225x169, 2.0+15.3 = +15. 13,
2x = 15 · 10 or — 15 · 16,
x = 75 or — 120, Ans. 3. Given 16x'-225x 225, to find x.
Put 15 = a; then 16 = a+1; whence,
(a+1)x*—a’x = a',
2(a+1)x—a’ = +(a'+2a),
2(a+1)x = 2a+2a or-2a,
x = 15 or -16, Ans
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 :
2can +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*— 16x = -28, to find the values of x.
OPERATION. 2* _ 16x* =
-28. Adding 8', or 64, x*-16x9 +64 = 36; extracting the square root,
28= +6; by transposition,
a = 14 or 2; whence,
x = =V14 or +12, Ans. 2. Given oc
-5, to find the values of x.