20. A person in play lost of his money, and then won 3 shillings; after which he lost of what he then had and then won 2 shillings; lastly he lost of what he then had; and, this done, found he had but 12s. remaining; what had he at first? Ans. 20s.

21. To divide the number 90 into four such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2; the sum, difference, product, and quotient shall be all equal to each other.

Ans. The parts are 18, 22, 10, and 40, respectively. 22. The hour and minute hand of a clock are exactly together at 12 o'clock; when are they next together? Ans. 1 ho. 5 min.

23. A man and his wife usually drank out a cask of beer in 12 days; but when the man was from home, it lasted the woman 30 days; how many days would the man alone be in drinking it? Ans. 20 days.

24. If A and B together can perform a piece of work in 8 days; A and C together in 9 days; and B and C in 10 days; how many days will it take each person to perform the same work alone?

Ans. A 143 days, B 1723, and C 23.

25. If three agents, A, B, and C, can produce the effects a, b, c, in the times e, f, g, respectively; in what time would they jointly produce the effect d?

[99]

QUADRATIC EQUATIONS.

A simple quadratic equation is that which involves the square of the unknown quantity only.

An adfected quadratic equation is that which involves the square of the unknown quantity, together with the product that arises from multiplying it by some known quantity.

Thus ax2b, is a simple quadratic equation,

And ax2 bxc, is an adfected quadratic equation. The rule for a simple quadratic equation has been given already.

All adfected quadratic equations fall under the three following forms:

The rule for finding the value of x, in each of these equations, is as follows:

RULE*.

1. Transpose all the terms which involve the unknown quantity to one side of the equation, and the known terms to the other, and let them be ranged according to their dimensions.

* The square root of any quantity may be either+or-, and therefore all quadratic equations admit of two solutions. Thus the square root of +n2is +n or —n; for (+n) × (+n) or (−n) × (—n) are each equal to n2, but the square root of ―n2 or ✅✔✅n2, is imaginary or impossible.

393350

2. When the square of the unknown quantity has any co-efficient prefixed to it, let all the rest of the terms be divided by that co-efficient.

3. Add the square of half the co-efficient of the second term to both sides of the equation, and that side which involves the unknown quantity will then be a complete square.

4. Extract the square root of both sides of the equation, and the value of the unknown quantity will be de-, termined, as was required.

Note 1. The square root of the first side of the equation is always equal to the unknown quantity, with half the co-efficient of the second term subjoined to it.

2. All equations, in which there are two terms involving the unknown quantity, and which have the index of the one just double that of the other, are solved like quadratics, by completing the square.

Thus, x4+ax2=b, or x2n - ax1=b, are the same as quadratics, and the value of the unknown quantity may be determined accordingly.

So, in the first form (x2+axb), where x+a is found (b+a2), the root may be either + √(b+;a2) or — ✓ (b+÷a2), since either of them, being multiplied by itself, will produce b+4a2. And this ambiguity is expressed by writing the uncertain sign before (b+4a2); thus x±√(b+a2)—a.

In this form, where x±√(b+}a2)—la, the first value of x, viz. x=+ (ba2)—a is always affirmative; for since 4a2+b is greater than 4a2, the greatest square must necessarily have the greatest square root; therefore (b+4a2) will always be greater than 2, or its equal da; and consequently +√(b+}a2)—?a will always be affirmative.

EXAMPLES.

1. Given x2+4x=140, to find x.

First, x2+4x+4=140+4=144, by completing the square.

Then √(x2+4x+4)=√144, by extracting the root; Or, which is the same thing, x+2=12,

And therefore x=12-2=10.

2. Given x2-6x+8=80, to find x.

First, x2-6x=80—8—72, by transposition; Then x2-6x+9=72+9=81, by completing the square, And x-3=81=9, by extracting the root;

The second value, viz. x=(b+4a2)—a, will always be negative, because it is composed of two negative terms. Therefore, when x2+axb, we shall have x=+√(b+a2)—a for the affirmative value of x, and x- == √(b + a2)—a for the negative value of x.

In the second form, where x=t✅✅(b+3a2)+‡a, the first value, viz. x= =+√(b+ a2)+a, is always affirmative, since it is composed of two affirmative terms. The second value, viz. x— +a2)+a will always be negative; for, since b+a2 is greater than a2, (b+a2) will be greater than 2, or its equal a; and consequently --✓(b+↓a2)+a is always a negative quantity.

Therefore, when x2-ax-b, we shall have x+/(b+1a3)+ a for the affirmative value of x, and x- √(b+ a2) + a for the negative value of x; so that, in both the first and second forms, the unknown quantity has always two values, one of which is positive, and the other negative.

3. Given 2x2+8x-20=70, to find x.

First, 2x2+8x=70+20=90, by transposition.
Then x2+4x=45, by dividing by 2,

And x2+4x+4=49, by completing the square;
Whence x+2=✔✅49=7, by extracting the root,
And consequently x=7-2=5.

4. Given 3x2-3x+6=53, to find x.
Here x2-x+2=17, by dividing by 3,

And x2-x=17—2, by transposition ;

9

Also x2-x+1=17—2+1=1, by completing the square;
And x-2=√3, by evolution;
Therefore x=1+1=3, the answer.

5. Given

x2

Here -=423—20=22}, by transposition,

2

And x-444-6, by evolution;
Therefore x=63+}=7, the answer.

In the third form, where x=({a2—b)+}a, both the values of x will be positive, supposing a2 is greater than b. For the first value, viz. x+√({a2—b)+ a will then be affirmative, being composed of two affirmative terms.

The second value, viz. x——✅✔✅(\a2-b)+1a is affirmative; for since a2 is greater than a2b, a2, or a, is greater than (a2b); and consequently-✓✓✅({a2—b)+}a will always be an affirmative quantity. Therefore, when x2-ax-b, we shall have x= + √({a2 —b)+a, and also x=- •√(‡a2—b)+1a, for the

affirmative values of x.