Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

37. There is a certain number, to the sum of whose digits if you add 7, the result will be three times the left-hand digit; and if from the number itself you subtract 18, the digits will be inverted; what is the number?

Ans. 53.

38. A vessel, containing 120 gallons, is filled in 10 minutes by two spouts running successively; the one runs 14 gallons in a minute, the other 9 gallons in a minute ; for what tinie has each spout to run ?

Ans. 14 gallon spout runs 6 minutes.

9 gallon spout runs 4 minutes.

39. In the composition of a certain quantity of gunpowder, twothirds of the whole plus 10 was nitre ; one-sixth of the whole minus four and a half was sulphur; and the charcoal was one-seventh of the nitre minus two; how many pounds of gunpowder were there?

Ans, 60lbs

or QUADRATIC EQUATIONS,

CONTAINING ONLY ONE UNKNOWN QUANTITY.

(68.) There are two kinds of Quadratic Equations with une unknown quantity.

1. When the equation contains only the square of the unknown quantity, it is called a pure quadratic equation.

Thus, r=36; x+5=54 ; 3.ro -4=71; ar-b=c; are pure quadratic equations.

2. When the equation, together with the square of the unknown quantity, contains also the unknown quantity itself, it is then called an adfected quadratic equation.

Thus, xo + 4.r=45 ; 3x*— 2x=21 ; x2 + ax=b; axl+2bx=c+d; are adfected quadratic equations.

(69.) The solution of pure quadratic equations is effected by the following

Rule - Transpose the terms of the equation in such a manner, that the term containing x? may stand on one side of the equation, and the known quantities on the other ; divide both sides of the equation by the co-efficient of xạ, (if it has any co-efficient) and then extract the square root of each side of the equation.

both}

Eramples. 1. Sapposer +5=54, to find s.

Toen, by transposition, t=54-5=49 Extract now the square root of both sides of the equation, and

then x+=v49=7 Ano 2. Let 6r2-8=142, to find x. Then, by transposition, 6x'=142+8=150.

150 Dividing by 6, x =

25.

6
And, extracting the square root, x=25=5 Ans.
3. Let 5.x-27=3x2 +215, to find x.
By transposition, 5x2-3x8=215+27; or 2x2=242

242
Therefore, t?= = 121; or x=W121=11 Ans

2
4. Let ar?-b=c; to find x.

By transposition, ax'=c+b; and x'=

[ocr errors]

C;

[merged small][ocr errors][merged small][merged small]

5. Let ar? -50=bx2—3c+d; to find x.

By transposition, aro-Ur?=50—3c+d;
Or, (ab)x=2c+d;

2c+d

,2c+d Therefore =

and x=N

Ans. a-6

[ocr errors]

a-b

[ocr errors]

Examples for Practice. 6. Given 5r-1=244, to find x..

Ans. x=7. 7. Given 9x2 +9=3x2 +63, to find x...

Ans.x=3. 8. Let 4.x2 + 5 =45, to find x.....

Ans. r=10. 9

C+2 9. Let bx+c+3=26x2 + 1, to find x..... Ans.x=N

b 10. Given 2ax2 +1-4 = cx-5+d-ar", to find x.

d-1-1 Ans. x=v

3a-C (70.) The solution of adfected equations is performed thus :

Rule.- 1. Bring the proposed equation to the form x?Fars+l, hy the rules of simple equations.

2. Add to each side of this equation the square of half tha

co-efficient of its second term ; then extract the square root of each side of the equation; and there will arise a simple equation, by which the value of the unknown quantity may be determined.

[ocr errors]

Examples. 1. Given x + 6x=40 to find the value of x.

Solu.-- We know that ro +6x +9 is the square of 3+3; for (x+3)=r? +6r +9.

If, therefore, we add 9 to each side of the equation, there will arise a new equation x2 + 6x +9=40+9=49, of which one side is the square

of x+3, and the other side is the square of 7. By extracting the square root of each side of this last equation, we have

W x* + 6x+9=v49; or x+3=7;

Therefore x=7-3=4, Ans.
2. Given r? — 10r=ll, to find x.
Solu.-We know that 2* — 10x+25 is the square of 2-5.
Add, therefore, 25 to each side of the oquation,

and we have x2-10x+25=11+25=36. Extracting the square root, we have r-5=136=6;

Therefore, x=6+5=]1. 3. Let xo + 2ax=l, to find x. Solu. —To each side of the equation add a> ;

Then xo + 2axta`=b+a® ;

But x? +2ax+uo is the square

Hence, x+a=vota"; and i. x=v0+ao Fa Note.-1. It will be observed, on reviewing these three examples, that, in each case, the quantity added to both sides of the equation was the square of half the co-efficient of r.

2. And in perforniing operations of this description, the learner should collect, that, after completing thie squaro ai corling to the rule, the squ root of that side of the equation containing the anknown qnantity, is always x, with half the co-efficient of the second term annered with its proper sig

Thus, the square root of x2 + 6x + 9 was x + 3 that of gr - 100+ 25 was x- -5; and that of x+ 2ax +a? was x* ta.

of x+a;

2

Examples for Practice. 4. Given 8+8x=65, to find X....... Ans, X 5. 5. Let x?—4x=45, to find x....

. Ans. x = 9. 6. Let x+125=108, to find x. Ans. X 6. 7. Let re – 14.051, to find x..... Ans. x = 17. 8. Given x2 +6bc=c%, to find ........ Ans. x=v0+919–36.

,

[ocr errors]
[merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

=5, Ans

8

[ocr errors]

с

2a

37 3 40 Therefore, x= +

8

8 3. Given 2aro-br=c, to find x.

b Solu. First, dividing by 20, x

2a Ն

12

12 Then, add the square of and x2 x+

4a

2a 16a22a' 16a?

Bac+1 Vyac + 12
Extract the root, and x-

42
16u

4a

Bac+62 +6
Therefore, x =N

4a

с

+

[ocr errors]
[ocr errors]
[ocr errors]

Examples for Practice. 4. Given 3x + 2r=161, to find x....... Ans. x= 7. 5. Let 2x2 - 5x=117, to find x...

Ans. x= 9. 6. Given 33 2r=280, to find x.. Ans. x=10 7. Let

7x_201= 32, to find x... Ans, x= 4. 3. Given 5r? + 4x=273, to find x.. Ans. r= 7.

Aac+/-6 axl + bx=c, to find

Ans. x=

[ocr errors]

9. Let

[ocr errors]

ах

S

(71.) OnE value only, of the unknown quantity, hath been found in the preceding questions; but every quadratic equation contains two values of the unknown quantity, arising from the following circumstance.

Since - a gives + a2, as well as tax ta; when we take the square root of + a', we may obtain for a result, either + a, or – a; consequently, in extracting the square root of each side of the equation, after the square has been completed, there will arise two simple equations to determine the value of the unknown quantity.

(72.) For Example. Let x + Px=Q, represent a quadratic equation in its most general form, wherein P and Q are any numbers integral or fractional, positive or negative.

Р Solu. To each side of the equation, add the square of

2' P2 P2 P+4Q. And it becomes 2? ++Px+*=+Q=-* x

4

+

[blocks in formation]
[ocr errors]

then r

x+ Ž

P: +4Q tv P?+4Q
4

2

[ocr errors]

2

P +P+4Q

P -P2 +4Q Hence, x+3 , and x +

are two 2 2

2 simple equations for determining the value of x;

+P+4Q-P VP:+4Q-P From which it appears, that x=

-sand 2

2

Note.—If Q be a negative quantity, and P be less than 4Q, then the quantity P + 4Q is negative; and because the square root of a negative quantity hath no real representation, the two values of the unknown quantity, containing the radical pe + 4Q, are said to be imaginary or impossible.

Eramples. 1. Let x2 + Ox=112, to find the two values of x. Solu. First, add 9 to each side, then x2 +6x+9=112+9=121 ;

Therefore, extracting the square root, x+3=+V121=+11; and, consequently, x=+11—3=8, or 14 Ans.

2. Given ?- 16r=-63, to find, in this equation, the two values of x.

Solu. Add 64 to each side, then r?— 16x+64=64–63=1; consequently, x–8=+Vī=+1, or x=8+1=9 or 7.

3. Let x + 8x=-31, to find the two values of x in this equation.
Solu. Add 10 to each side, then x2 +8/+16=-31+16=-15.
Now extract the square root, and x+4=+V-15;
Wherefore, the two values of x are 4+1=15, and 4-V15;
But these quantities, containing -15, are both impossible.

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
« ΠροηγούμενηΣυνέχεια »