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Example 3. Solve 2x-1-9x-1=-4.

9+V81-32 By formula,

x-1=

4
9+7
4

1
= 4 or

2* Raising to the fourth power,

1
x-1=256 or i

16

1 that is,

=256 or

16

i
1
.'. X =

256

ܪ

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or 16.

EXAMPLES XXVII. a. 1. 826 +65x3+8=0.

2. 312 — 3x-1=8. 3. 27xi -1=26.01.

4. 24 — 74x2= -1225. 5. -2–2x-1=8.

6. 9+x-4=10x-2.
7. 218+2.c-1=5.

8. 6x1=7x12x-1.
1
9. an +6=5an.

10. 3x2n — an-2=0.
11. 6x=5x-1-13.

12. 1+8x8 +9=0.

2

1

1

3

3

13.

82c2n - 8x2n=63. 219. Any equation which can be thrown into the form

ax2+bx+c+pvax? +bx+c=1 may be solved as follows. Putting y= Vax2+bx+c, we obtain

ya+py-q=0. Let a and ß be the roots of this equation, so that

Vax? +bx+c=a, Vax2+bx+c=ß; from these equations we shall obtain four values of x.

When no sign is prefixed to a radical it is usually understood that it is to be taken as positive; hence, if a and ß are both positive, all the four values of x satisfy the original equation. If however a or ß is negative, the roots found from the resulting quadratic will satisfy the equation

ax2 +bx+c-pVax2+bx+c=q, but not the original equation.

Note. The Greek letters a, B, 7, 9 (read Alpha, Beta, Gamma, and Phi), are of frequent occurrence in mathematical discussions.

Example 1. Solve 22–5x+2 væ2 – 5x+3=12.
Add 3 to each side ; then

22 – 5+3+2 V22 - 5x+3=15. Putting Vx2–5x+3=y, we obtain y2+2y-15=0; whence y=3 or -5.

Thus Væ2–5x+3=+3, or Væ2 — 5x+3=-5.

Squaring and solving the resulting quadratics, we obtain from the first w=6 or -1; and from the second x=

5€ /113

The first

2 pair of values satisfies the given equation, but the second pair satisfies the equation

x2 - 5x – 2 væ2 - 5x+3=12.

or

Example 2. Solve 3x2 — 7+3V3x2 – 16% +21=16x.
Transposing, 3x2 — 16x— 7+3V3x2 — 16x+21=0.
Add 28 to each side ; then

3x2 – 16x+21+3V3x2 – 16x+21=28. Proceeding as in Example 1, we have

y2+3y=28; whence y=4 or – 7. Thus V3x2 – 16x+21=4 or 73x2 – 16x+21= -7. Squaring and solving, we obtain

1

8+237
x=5,
3

3 The values 5 and į satisfy the original equation. The other values satisfy the equation

3x2—7-3 V3x2 – 16x+21=16x. 220. Occasionally equations of the fourth degree may be arranged in expressions that will be in quadratic form.

Example. Solve x4 – 8x3 + 10x2 + 24x+5=0.
This may be written 24 — 823 +16x2 - 6x2+24x=-5,

(x2 -- 4x) 2–6(x2 -- 4x)=-5;

6+ V36 – 20 by formula,

22 - 4x=

2 6+4

2

=5 or 1 ; whence

x=5, -1, or 2+ 15.

or

The student will notice that in such examples he should divide the term containing x3 by twice the square root of the first term and then square the result for the third term. In this case a third term of 16x2 is required, therefore we write the term 10x2 of the original equation in the form 16x2 - 6x2. 221. Equations like the following are of frequent occurrence.

22-6 5x Example. Solve

+

202-6
22 - 6
Write y for

5
y+-=6, or y2-6y +5=0;

Y whence

y=5, or 1.
x2 - 6

32 – 6
=5, or =1;

=6.

X

; thus

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that is,

Thus

X2 — 5x — 6=0, or x2 – 3—6=0.

x=6, -1; or x=3, -2.

42

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4. 500

+7

EXAMPLES XXVII. b.
Solve the following equations:

22-1 1. x2+x+1=

2.

+ 22+

22-1

=23.

XC

3 3. X+

X+
=5.

=223.

3
x+8
5 3x+14

3 2x
5.
+

6.

=516 X+12 x +4 3x+8

V2x

5 3x - 6 11- 2x 7. +

= 31. 10-4x

8. x2 + 2 vx2 +6x=24–6x. 6\2 9.

=5.

10. 27xi - 4=26x#.

5-X

=8.

20 11. 7V2-8-V21x+12=2 /3. 12. 22+ 3x

22+ 3x 13. x+2=V4+xV8-2. 14. 32c2 - 4x + 3x2 - 4x - 6=18. 15. 2x2 – 2x+2 v2x2 – 7x+6=5x – 6. 16. 22 +6Vx2 – 2x+5=11+2%. 17. 2 Vx2 - 6x +2+4x+1=x2–2x. 18. 14x2+2x+7=12x2+6x– 119.

[blocks in formation]

х

N

X

27. Vx-p+Vx-q=

р

9
+
2 - NT-p

2 28. V(x - 2) (x – 3)+5

Na? +6x+8.

- 3 29. Vx+4.0 - 4+2 +4.0 - 10 -6. 30. Vx-a-a-b=-a. 31. 204—833 — 12x2 +112x=128. 32. 24 + 2x3 – 3x2 - 4x - 96=0. 33. 24 — 10x3 +30x2–25x+4=0. 34. 204 – 14x3 + 613-— 84x+20=0.

CHAPTER XXVIII.

SIMULTANEOUS QUADRATIC EQUATIONS.

222. We shall now consider some of the most useful methods of solving simultaneous equations, one or more of which may be of a degree higher than the first; but no fixed rules can be laid down which are applicable to all cases.

X - Y= 9.5

EQUATIONS SOLVED BY FINDING THE VALUES OF

(x+y) AND (xy). Example 1. Solve 3+y=15...

(1), xy=36..

(2). From (1) by squaring, x2 + 2xy + y2=225; from (2)

4.ry=144; by subtraction,

2* — 2xy + yż=81; by taking the square root,

* - =9. Combining this with (1) we have to consider the two cases, *+y=15,1

2+y= 15,1

2 - Y=- 9.) from which we find

x=12,

x= 3,1

y=12. S Example 2. Solve 2 - Y=12 ..

(1), my=85

(2). From (1)

22 - 2xy + y2 =144; from (2)

4xy=340; by addition,

x2 + 2xy + y2=484; by taking the square root,

x+y= 22. Combining this with (1) we have the two cases,

++y=22, x+y=-22, )

x y=12. Whence x=17,1

5, y=-17.

y= 3.)

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Y= 5.

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