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y=3 or -7. Therefore
x=9 or 49. Although the square root of 9 is generally ambiguous, and may be either +3 or -3, still, in verifying the preceding values, vô can not be taken equal to -3, because 9 was obtained by multiplying +3 by itself. For a like reason, Voc can not be taken equal to +7. A similar remark is applicable to sev. eral of the following examples.
263. The same method of solution may often be extended to equations in which any algebraic expression occurs with two exponents, one of which is double the other.
Ex. 4. Solve the equation (2c2 + x)2 — 26(22 +w)=-120.
y=20 or 6. We have now the two equations,
x2 +x=20, and x2+x=6,
X=-3 or +2.
-5, +4, -3, +2, and any one of these four values will satisfy the given equa. tion.
Ex. 5. Solve the equation Væ+12+Vx+12=6.
*+12=16 or 81. Hence
x=4 or 69.
2.2 +1=9 or 16; therefore x=+2, -2, +v77, -V75.
264. Equations of the Fourth Degree.— An equation of the fourth degree may often be reduced to an equation containing the first and second powers of some compound quantity, with known coefficients, in the following manner: Transpose all the terms to the first member; then extract the square root to two terms, and see if the remainder (with or without the absolute term) is a multiple of the root already obtained.
Ex. 7. Solve the equation 24 - 1223 +44x2 — 48x=9009.
24 - 1223 +44.x2 -- 48x-9009=0(32-63
8(22—6x)-9009. Hence the given equation may be expressed as follows:
(ac2 - 6x)2 +8(x2–6x)=9009.
Ans. x=13 or -7, or 3+3V-10. Ex. 8. Solve the equation 04-226 +=132.
Ans. x=4 or -3, or 1+1V ---43. Ex. 9. Solve the equation ** +4.x2 =12.
Ans, x=IV2 or IV 6. Ex. 10. Solve the equation 26 – 8x=513.
Ans. x=3 or -19. Ex. 11. Solve the equation 25+7=756.
Ans. x=243 or – 285. Ex. 12. Solve the equation 426 – 7X=-31.
Ans. x=1V2. Ex. 13. Solve the equation 2x+3x3=2.
Ans. x= or -8. Ex. 14. Solve the equation fx-5Vx=227.
361 Ans. x=49 or
9 Ex. 15. Solve the equation V10+x-V10+x=2.
Ans. x=6 or -9.
Ex. 16. Solve the equation 206+ 20x3 — 10=59.
Ans. x=13 or v - 23. Ex. 17. Solve the equation 3.21 — 20" +3=11.
Ans. x=V2 or v Ex. 18. Solve the equation V1+c-22-2(1+-x2)=}.
Ans. x=1+1V41 or 1+1V11. Ex. 19. Solve the equation Vāc+Vx=20.
Ans. x=256 or 625. Ex. 20. Solve the equation 24 –4x3+702 — 6x=18.
Ans. x=3 or -1, or 1+V-5. Ex. 21. Solve the equation a2 +5x+4=5Vx2 +5x+28.
Ans. x=4 or -9, or +1V-51. Ex. 22. Solve the equation +3=2V22-22C+2+2.c.
Ans. x=1. Ex. 23. Solve the equation (x+Væ)*-(+ Vã)=20592.
Ans. x=9 or 16. Ex. 24. Solve the equation x+V25+x=157.
Ans. x=144 or 171. Ex. 25. Solve the equation Vx-1=x— 1.
Ans. x=1 or 2.
265. We have seen that every equation of the second degree has two roots; that is, there are two quantities which, when substituted for æ in the original equation, will render the two members identical. In like manner, we shall find that every equation of the third degree has three roots, an equation of the fourth degree has four roots, and, in general, an equation of the mth degree has m roots.
Before determining the degree of an equation, it should be freed from fractions, from negative exponents, and from the radical signs which affect its unknown quantities. Several of the preceding examples are thus found to furnish equations of the fourth degree, while others furnish equations of the second degree.
The above method of solving the equation 2n+px"=9 will not always give us all of the roots, and we must have recourse to different processes to discover the remaining roots. The subject will be more fully treated in Chapter XXI.
Problems producing Equations of the Second Degree.
2+8=20, the greater. Proof. 20-12=8, the first condition.
20 x 12=240, the second condition. Prob. 2. The Receiving Reservoir at Yorkville is a rectangle, 60 rods longer than it is broad, and its area is 5500 square rods. Required its length and breadth.
Prob. 3. What two numbers are those whose difference is 2a, and product b?
Ans. atva?+b, and -ava? +b. Prob. 4. It is required to divide the number 60 into two such parts that their product shall be 864.
Let x=one of the parts.
The parts are 36 and 24, Ans. Prob. 5. In a parcel which contains 52 coins of silver and copper, each silver coin is worth as many cents as there are copper coins, and each copper coin is worth as many cents as there are silver coins, and the whole are worth two dollars. How many are there of each ?
Prob. 6. What two numbers are those whose sum is 2a and product ?
Ans. at Val-b and a-Val-b.
Prob. 7. There is a number consisting of two digits whose sum is 10, and the sum of their squares is 58. Required the number.
Let x=the first digit.
And 2c2 +(10-2)=2x2 – 20x+100=58; that is,
x2 — 10x=-21,
x=5+2=7 or 3. Hence the number is 73 or 37.
The two values of u are the required digits whose sum 16 10. It will be observed that we put x to represent the first digit, whereas we find it may equal the second as well as the first. The reason is, that we have here imposed a condition which does not enter into the equation. If x represent either of the required digits, then 10—2 will represent the other, and hence the values of x found by solving the equation should give both digits. Beginners are very apt thus, in the statement of a problem, to impose conditions which do not appear in the equation.
The preceding example, and all others of the same class, may be solved without completing the square. Thus,
Let represent the half difference of the two digits.
Then, according to the principle on page 89, 5+x will represent the greater of the two digits, and 5-~ the less. The square of 5+x is 25+10x + x?,
5-*“ 25-10x+ 22. The sum is
50 +2x2, which, according to the problem,
dela = 4, and
5+2=7, the greater digit, and
5—X =3, the less digit. Prob. 8. Find two numbers such that the product of their sum and difference may be 5, and the product of the sum of their squares and the difference of their squares may be 65.