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Again.

Bucl 2

Let ilaa-2baadc Care 2.

bbbb
I +33182-2ba+bb=de tobb
3 w?412-b=vdo+bb, &c. as before.

But in Case 3. you must change the Signs of all the Terms in the Equation,

Thus | 1/2 ba-aa=dc Case 3.

+12 aa- 2b2=dc
Then 31a2- 2ba+ob=bbdc, &c.

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For 2

And this Method of compleating the Square, holds true in those other Equations. Viz. 1 la a aa + 2 baa=dc Case 1.

bbbb, as before.
I+23 aga a + 2 baa+bb=dc+bb
3 wa 4 aa+b=vdobo
4-515 a 2 =V do toh:- b
5 w

216la=Vivdc tbbim ·b, and so on for the rest.

And 2

23

Or let 119 +2baaa=dc, as before, Cafe 1.

bb=bb
it2 a + 2baaa tobb=dc+bb

3

4 a aatba v dit bb
4-65 aca=v drt66: -6
5 w'l ola='v: v det06:, &c.

2

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COROLLAR Y.

Hence it is evident, that whatsoever Method is used in solving these (or indeed any other) Equations, the Result will fill be the same, it the Ilork be true; as you may observe from the Operations of this Staion: for both. these Methods here proposed, give the same Theoreins in their respective Cales for the Value of (a).

Thus

Thus, when a a + 2 ba=ds, then
Theorem 1. a=vdo+06: -6
And when a a 2 bardc, then
Theorem 2. a=b+vdo+bb
Again when 2 b 2.-aa=dc, then
Theorem 3. a=b-V

obdo .

The like Theorems may be easily raised for the rest.

If the known Co-ëfficients (of the second or loweft Term) be any single Quantity, as a atbardi, &c. then is į b it's Half, and 166 will be the Square of that Half; that is, t bixib=$b by and then the Work will stand

Thusi laatbardo iCo2ja a +bat, 15b=dct4bb

2 w3atib=vdo +100 3-ib|412=vdit:60:- ib, and so for the rest.

Note, Co placed in the Margin against the second Step, fignifies that the imperfect Square a atba in the first Step, is there compleated, viz. in the second Step.

Now by the Help of these Theorems, it will be easy to calculate or find the Value of the unknown Quantity (a) in Numbers.

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Suppose a a+2ba=z. Let b=16, and 2=4644.
Then a =Vz+00:- per Theorem 1.
But z +6b=4644 +256 = 4900, and V 4900 = 70.
Consequently a = 70 -- 16, viz. a = 54.

But every Adfected Equation; hath as many Roots (or rather Values of the unknown Quantity) either real or imaginary, as are the Dimensions (viz. the Index) of it's highest Power, and therefore the Quantity a, in this Equation, hath another Value either Affirmative or Negative ; which may be thus found.

The given Equation is aa +322= 4644, and it's Root a=54.

Let these two Equations be made equal or or equated to o, viz. to Nothing

Thus,

Thus, a a +32 2 - 4644 = 0, and a -54 = 0.

Then divide the given Equation by it's first Root, and the Quotient will fhew the second Value of a.

Thus, a-5450) a a+32 - 4644 = 0 (a +-86=0

a a — 544
+860 - 4644

4644
(0)

86 a

Hence the second Value of a is =86, or 86 Iwhich seems impoffible, viz. that an Affirmative Quantity should be equal to a Negative Quantity ; yet even by this second Value of a, and the same Co-efficient, the true (or first) Equation may be formed

Thus, Let Ija=- 86

aa=+7396, viz. -86% -86=+7396 I X 32 31 32 2 = -2752 2+3/4|aa +320 =4644, as at first.

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2.

E X AMPLE
Suppose | Ilaam=948,75, then per Theorem 2.
ICO2 aa-7a+=948,75+4= 961
2 w312

32 - (or 3,5) =V961 = 31 3+3,5 14 10 = 31+ 3,5 = 34,5

Again, for the second Value of a, let aa-70-948,75=0, and a-34,5=o. Then a-34,5=o)aa-72-948,75=0 (a +27,5=o. Consequently this second Value is a = -27,5 which will form the original Equation, aa-70=948,75 if it be ordered as the last was.

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Suppose 36 a-aa= 243, then per Theorem 3. a=18 -V ✓ 324 - 243, viz. half 36 squared is 324, &c. that is, a = 18-81; but ✓ 81=9, therefore a = 18—9= 9.

Now this third Form is called an ambiguous Equation, because it bath two Affirmative Values of the unknown Quantity (a), both which may be found without such Division as was used before. For in this Case, a = 18-4v 81, viz. a=18+9=27, or, a = 18-9=9, as before. And both these Values of a are equally true, as to forming the given Equation ; viz. 36a-aa = 243. For if a=9, then a a=81, and 36 a=324; but 324 -81=243, therefore a = 9.

before.

Again, if a = 27, then will a a=729, and 360=972: But 972-729=243, consequently it may be, a=27. Now either of these Values of a may be found by Division, as those were in the other two Cases, one of them being first found by the Theorem. Thus, let 36 a--a2- 243 = 0, and 9-a=o then 9-a=o) 360-02-243=0 (a - 2750

9aaa 27 a =0

243 27

243 (0)

(0) Hence, if a 27 =0, then a= 27, as before.

Notwithstanding all Quadratick Equations of this third Form have two Affirmative Roots (as in this) yet but one of those Roots will give a true Answer to the Question, and that is to be chosen according to the Nature and Limits of the Question, as shall be fhewed further on.

SCHOLI U M.

From the Work of the three laft Examples, it may be observed ; that the Sum of both the Roots will always be equal 10 the Co-ëficiens of their respective Equations, with a contrary Sign. Thus. In Example 1. à®+32a=4644

Here a =

54 And a=

86

20=32 In Example 2.

aa-70=948,75 Here a = 34,5

And a=-27,5} Add

2a=+7
In the last Example 360-aa= 243
Which was changed into a a— 36 a=-243

Here a = 9
And a 27S

2a = 36
Dd

Hence

Hence it is evident, that if either of the Roots be found, the other may be easily had without Divifions.

If the Contents of this Section be well understood, it will be easy to give a Numerical Solution to any Quadratick Equation, that happens to arise in resolving of Questions, &c. And as for giving a Geometrical Construction of them, I think it not proper in this place; because I here suppose the Learner wholly ignorant of the first Principles of Geometry, therefore I shall refer that Work to the next Part.

CH A P. IX.

Of analytis, or the Method of resolving Poblenis exem

plified by Variety of Numerical Düestions. N.B.HERE I advise the Learner to make life always of the fami

sy in Viz. {

any Number And e reprefent a les pubert }or other Quantity

, rates their sum: Led

eir Difference.

aep their Product. Then let

=q their Quotient.
aatee=z the Sum of their Squares.
iaa-re=* the Difference of their Squares.

e

Any two of the fix (s, d, p, q, 2, *) being given, thence to find the rest; which admits of fifteen Variations, or Questions.

ate=s} and suppose

a

Question 1. Suppose s and d were given, and it were required by them to find a .e.p.q.z. and x. Le

-e=d).
I +21312a=std=432

std
3 • 24

= 216, here a is found. 1 - 21512c=s-d=48.

id=192 } Then

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2

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