« ΠροηγούμενηΣυνέχεια »
4. In the equation zo — 2cx=a? to find the two values of r.
Eramples for Practice. 5. Let x-2x=15, to find the two values of x... Ans. x=5 or-3. 6. Given r?- 16x=-15...
Ans, x=15 or 1. 7. Let x + 6r=135.....
Ans.x=9 or -15. 8. Given r–2x= -2
..Ans. x=l+v-1. 9. Let x? +12x=-50..
Ans. r=-6+v - 14. 10. Given ro - Bax=31°.....
Ans. x=40+ ✓ 16a? +31%. (73). Every equation, in which the unknown quantity is found in only two terms, and the index of the highest power is double the index of the lowest, may be reduced to the form of a quadratic equation, and solved according to the foregoing rules.
(74). For Example. Let x2 + Px=Q be any general equation.
Solu. Let xa=y, substitute now these values for 22n and 2", and then run=y; the given equation becomes yo + Py=Q.
+vP2 +40-P Then yo + Pyt
; consea 4 4
2 quently, since r"=y, x must be equal to py;
+P+4Q-P Therefore, I=
Examples. 1. Let 3* - 6r=27, to find the values of x in this equation.
Solu. Let x'=y, } whence the equation becomes y: -6y=27; then I'=yo;
x Therefore y-3= + V36=+6, and y=+6+3=9 or --3.
But, because x'=y, r=+vy=+g, or +V-3= +3, or +3.
2. Find the values of x in the equation 26—2x=48.
Solu. Let r=y, } therefore, by substituting ye- 2y=48. then ro=yo;
But y-1=+ 49=+7; hence, y=+7+1=8 or-6; consequently, r=y=38, or -6=2, or 3-6.
Examples for Practice 3. Find the values of x in the equation x* +4x+=12
Ars. r=+vz, or +-6 4. Find the values of x in the equation 20 – 8x9=513.
Ans. x=3 or 31 - 19. 5. Find the values of x in the equation 12 --4x"=10.
As. x=2+ V14.
Solu. First, then, *.*-11==1=42°–33=r, multiplying by 3
dividing by 4. 4 4
1 33 1 528 Coniplete the square, x?. * +
64 4 64 64
2 2. Given the equation
+ =5, to find the values of r x+1
7 2 Solu. First, multiply by r+1, and the equation +
becomes 7x + 2x +2=5x + 5x.
4 4 2 4 14
2 Extract, now, the square root, and x
5 +74+2 Therefore, I =
=1.14, or —.34.
3. Let 5.20 -90.x4-270=945, to find in this equation the values
Solu. By transposition,
570-90r-270=945=570-90x'=945+270= 1215. Then divide by 5, and we get 208-18% = 243. Complete the square, and it becomes
-18.2+81 = 243 +81=324. Extract now the square root, and 24-9=+324=+18.
Therefore = +18+9=27, or -9;
And r= 27, or 9=3, or $9. Nole. Without the formality of substituting yo'for 36, and y for 30s, we proceeded at once to the solntion of the equation, 26 - 18x3 = 243, as exemplified in the second step of the operation. 4. Find, in the equation 3ax—21=4x—56, the values of x. Solu. First, by transposition, we get 3ax®-4x=21-50.
26-50 Then divide by 3a, and it is .
30 Complete the square, and 4 4 26-50 4 6ab-15ac+4
2 Voab - 15ac+4 Extract the root, and we have I—
За +vab - 15ac+4
+2 Therefore ==
Examples for Practice 3. Let i? — 7x+1=9, to find the values of x.
Ans. x=8, or 6. Given 4ro-73=492, required the values of r.
7. Let 3-6r+19=13, to find x's values.
Ans. I=4.732, or 1.268. 8. Given 51% +4.5=25, required the values of r.
Ans. x=1.871, or 2.671.
9. Let us - 1=1+11, to find the values of x.
en + =3, to find the values of 3
Ans. I=2, or
Given 30—34= =, required the values of x
+1 13 14. Given
to find the values of x. It)
Ans. x=2, or – 3. 15. Given 2x*-77%=9, to find the values of x.
Ans. r = +3, or +v
16. Let 2ro_r=496, required the values of x.
· Ans. x=+4, or + V =31
17. Given mora to find the values of x.
Ans, x= 3
1+140 + 1
2 18. Let 2x +15=3x required the values of r.
4 19. Given 35-**=10, to find the values of r.
Ans. x=6+V=4. 20. Let 4ar-20x=c, required the values of r.
6+1 12+ 4ac Ans. II.
21. Given += 2, to find the values of r. +
Ans. x=1+VI-a. $75) PROBLEMS, or questions, involving quadratic equations, are solved by the same process as that for the solution or those which contain simple equations. But as every quadratic
equation contains two values of the unknown quantity ; ques. tions of this class may admit of a double solution. And as the particular nature of the question must determine whether one or both of these values apply to it, no general rule can be laid down ; yet may the processes of solution of Problems involving quadratic equations be sufliciently understood by observing the conclusions deduced from the following examples.
Exam. I. To find that number to which if you add 12, and multiply the sum by the number required, the product shall be 589.
Solu. Let r = the number sought ;
Then (r+12) x =589, or x2 + 12x 589 by the question. Complete the square and we have x! + 12x+36 = 625; hence, Extract the root, we get +6 =V625 = 25
Therefore, r = 25—6= 19, the number sought.
Exam. II. To divide the number 56 into two such parts, that their product shall be 640.
Solu. Let I = one part; then 56 x = the other part ; And I x (56 - x) the product of the two parts.
Hence, x x (56 — x) 640, or 56x x2 = 640 by the question, or r? 56.0 - 640; hence By completing the square, x2 56r + 784 = 784 - 640 = 144 ; Therefore, x - 28
+ 12 ; and, consequently, r = 28 + 12 = 40 or 16.
Note. The two values of the unknown quantity are, in this instance, th two parts into which the given number was required to be divided.
Exam. III. Let the difference of two numbers be 7, and half their product plus 30 be equal to the square of the less number; it is required to find those two numbers: Solu. Let x = the less, then x + 7 = the greater number, r
+ 30 = half their product plus 30. Hence, 1(t + 7) + 30 = xe (square of the less) or
I+ z by the question.
Multiply now by 2, and x? + 7x + 60 = 2x,
289 By completing the square - 7r + 192