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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 time 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. 69lbs

OF QUADRATIC EQUATIONS,

CONTAINING ONLY ONE UNKNOWN QUANTITY.

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

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

Thus, 36; x2+5=54; 3x2-4-71; ax2-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, x2+4x=45; 3x-2x=21; x2+ax=b; ax2+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 2 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 x, (if it has any co-efficient) and then extract the square root of each side of the equation.

Examples.

1. Suppose x+5=54, to find r.

Toen, by transposition, x2=54-5=49.

Extract now the square root of both sides of the equation, and

both

then x+49=7 An.

2. Let 6x-8=142, to find x.

Then, by transposition, 6x=142+8=150.

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And, extracting the square root, x=/25=5 Ans.

3. Let 5x2-27=3x2+215, to find x.

By transposition, 5x2-3x2=215+27; or 2x2=242

242
2

Therefore, x2= =121; or x=/121=11 Ans

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5. Let ax2-5c=bx2-3c+d; to find x.
By transposition, ax2-bx2-5c-3c+d;
Or, (a-b) x2=2c+d;

a

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c+2

b

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9. Let bx2+c+3=2bx2+1, to find x.... .. Ans. x=.

10. Given 2ax2+b−4= cx2-5+d-ax3, to find x.

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(70.) The solution of adfected equations is performed thus : Rule.-1. Bring the proposed equation to the form a+ax=+l, by the rules of simple equations.

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

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.

Examples.

1. Given x2+6x=40 to find the value of x.

Solu. We know that x2+6x+9 is the square of x+3; for (x+3)2=x2+6x+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

√x2+6x+9=49; or x+3=7;

Therefore x=7-3=4, Ans.

2. Given x2-10x=11, to find r.

Solu. We know that x2-10x+25 is the square of x-5.
Add, therefore, 25 to each side of the oquation,
and we have a2—10x+25=11+25=36.

Extracting the square root, we have x-5=√36=6;
Therefore, x=6+5=11.

3. Let x+2ax=b, to find x.

Solu. To each side of the equation add a2;

Then x2+2ax+a2=b+a2;

But x2+2ax+a is the square of x+a;

Hence, x+a=√b+aa; and :.x=√/b+a2Fa

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 x.

2. And in performing operations of this description, the learner should collect, that, after completing the square according to the rule, the squ root of that side of the equation containing the anknown quantity, is always x, with half the co-efficient of the second term annexed with its proper sig

Thus, the square root of x2 + 6x + 9 was x + 3 ; that of r2-10x+25 was x-5; and that of r+2ax+a2 was x2+a.

Examples for Practice.

4. Given x2+8x=65, to find x....

5. Let x2-4x-45, to find x....

6. Let x2+12x=108, to find x...

Ans. x 5.

9.

Ans. x =
Ans. x= 6.

7. Let x2-14x51, to find x........ Ans. x = 17.

8. Given x2+6bx=c2, to find x.. • Ans. x=√c2+9b2—3b.

....

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Examples for Practice.

4. Given 3r+ 2x=161, to find x.........

Ans. x= 7.
Ans. x

9.

5. Let 2x2- 5x=117, to find x.......
6. Given 3x2- 2x=280, to find x....... Ans. x=10
7. Let 7x2-20x= 32, to find x...
3. Given 5x2+ 4x=273, to find x..

9. Let ax2+ bx=c, to find x

....

....

Ans. x 4.

Ans. x= 7.

Ans. x

√4ac+b2-b

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(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.

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Since -ax a gives + a2, as well as a x +a; when we take the square root of a2, we may obtain for a result, either + a, 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.

or

(72.) For Example. Let x2 + PrQ, represent a quadratic equation in its most general form, wherein P and Q are any numbers integral or fractional, positive or negative.

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

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P
2'

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Note.-If Q be a negative quantity, and P be less than 4 Q, then the quantity P + 4Q is negative; and because the square root of a hegative quantity hath no real representation, the two values of the unknown quantity, containing the radical ✔P2 + 4Q, are said to be imaginary or impossible.

Examples.

1. Let x2+6x=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=√/121=+11; and, consequently, x=+11-3-8, or 14 Ans.

2. Given x2-16x=-63, to find, in this equation, the two values of x.

Solu. Add 64 to each side, then x2-16x+64—64—63=1; consequently, x-8=±√I=+1, or x=8+1=9 or 7.

3. Let x2+8x=-31, to find the two values of a in this equation. Solu. Add 16 to each side, then x2+8x+16=−31+16=-15. Now extract the square root, and x+4= +√−15;

Wherefore, the two values of x are-4+√15, and-4-√√—15; But these quantities, containing 15, are both impossible.

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