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16 12. =l+
Ans. x=3 or 1. (2x-4) (22—4) 13. x+11+V+*+11=42. Ans. x=+5 or +V 38. 14. x'—2x+61 +23+5=11. Ans. x = 1 or 1+2V 15.
172 15. *+
= 343+16. Ans. x=2,-2,-8, or -1. 2
(EY 23. 28+xt = 756.
(+12_911 22. =x-2.
= 5 or 3. 24V 249
Ans. x=243 or -8°33614. 24. 62-13* = 6.
Ans. x= Vor V.
-2 25. 2+2x+2x = c(1-x). Ans. x = 1 or
(c+2) 26. (x+a)—(ma)' = 352a'.
Ans. x = tay 5 or EaV7. 1 b+c
1 27. axt
SIMULTANEOUS EQUATIONS CONTAINING QUADRATICS.
297. Having treated of quadraties containing one unknown quantity, we will now consider certain cases of simultaneous equations where one or more is of a degree higher than the first.
298. In general, the solution of two quadratic equations, involving two unknown quantities, depends upon the solution of a single equation of the fourth degree, containing one unknown quantity.
To show this, let us represent the two equations under a general form, as follows: 2* +axy+by+cx+dy+c= 0,
(1) x* ta'xy+by+c'x+d'y+c = 0,
(2) in which the coefficients a, b, c, etc., and a', ', c', etc., may have any value, positive or negative, integral or fractional.
Arranging the terms in these equations with reference to x, and factoring, we have x'+ay+c)x+by+dy+c = 0,
(3) **+(a'y+c')x+by+d'y+d' = 0. (4) Subtracting (4) from (3), we have
[(a—a')ytec']x+(6—6%)yo+(d^d')y+(6—) = 0. By transposition, we have
[(a~a')yte-c'] x=(b’—b)y*+(d'—d)y+(c'—c); whence we obtain
(a<a' )+(c−c) This value of x substituted in either equation, (3) or (4), will give a final equation involving only, y. Without actually making this substitution, which would lead to a complicated expression, it is obvious that the resulting equation would be of the fourth degree. For the value of x is in the form of
ryts in which y is involved to the second power. Therefore the term containing xin (3), or (4), must involve y to the fourth power.
Hence, two equations, essentially quadratic, and containing two unknowư quantities, can not, in general, be solved by the rules for quadratics.
299. There are certain cases where simultaneous equations, involving one or more of the unknown quantities to a higher degree than the first, may be solved by means of a final equation in the quadratic form. Most of the examples of this kind are embraced in these three cases :
1st. Where one of the equations is simple, and the other quadratic.
2d. Where both of the equations are quadratic, but homogeneous.
3d. Where one or both of the equations are symmetrical, involving the different letters in a similar manner with respect to coefficients and exponents. The following are illustrations :
1st. SIMPLE AND QUADRATIC EQUATIONS. The solution is effected in this case by the ordinary methods of elimination.
5x* 1. Given
to find w and y. 3x 2y = 6
6+2y From equation (2),
; from (1), 5(6+2y), 6y(6+2y)
16ye-12y = 108;
4y = 12 or whence,
y = 3 or 4: and
x = 4 or
2d. HOMOGENEOUS EQUATIONS. In the case of homogeneous equations, an auxiliary quantity is employed in the elimination. 2. Given (23*.
2x*—xy = 6
Put x=vy; then the given equations become
6 2voy—vy* = 6, or y*
8 2yo+3vyo = 8, or yo = 2730 6
v = 2 or -5. Taking v = 2, equation (2) gives
y=+l; whence x = +2. Taking v=-, the same equation gives
77 It may be observed here, that in this example, as in all other examples of simultaneous equations, the different sets of values which are capable of satisfying the equations, will be found by taking the signs in their order ;—that is, the upper signs should be taken throughout, and the lower signs throughout. Thus,
when y : +1, w = +2;