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3 + 2 v/7

.=—- - =2,123, exact, to within 0,001.

O v V'4 — 0

Remark. Expressions of this kind might be calculated by approximating to the value of each of the radicals which enter the numerator and denominator. But as the value of the denominator would not be exact, we could not form a precise idea of the degree of approximation which would be obtained, whereas by the method just indicated, the denominator becomes rational, and we always know to what degree the approximation is made.

The principles for the extraction of the square root of particular numbers and of algebraic quantities, being established, we will proceed to the resolution of problems of the second degree.

Examples in the Calculus of Radicals.

1. Reduce V 125 to its most simple terms.

Ans. 5 V!,.

. / 50

2. Reduce v t0 'ts most Slmple terms.

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3. Reduce V 98a2x to its most simple terms.

Ans. la Vix.

4. Reduce V(n?—aV) to its most simple terms.

5. Required the sum of Vl2 and V 128 .

Ans. 14 VY.

6. Required the sum of VZ7 and V 147.

Ans. 10 VY.

. . 2 /27 7. Required the sum of \/ Tt and v

a 50

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8. Required the sum of 2 Y aab and 3 V646a;4.

9. Required the sum of 9 V 243 and 10 V 363 .

/T /~5~

10. Required the difference of "v — and "v ^>

Ans. VTB. 45

11. Required the product of 5 V* 8 and 3 V 5 .

Ans. 30 YlO.

2 . /T 3 . /Y

12. Required the product of — v -g- and — v —.

Ans. Yli5. 40

13. Divide 6 VTo by 3 a/If. Ans. 2 A/y.

Of Equations of the Second Degree.

137. When the enunciation of a problem leads to an equation of the form a^—b, in which the unknown quantity is multiplied by itself, the equation is said to be of the second degree, and the principles established in the two preceding chapters are not sufficient for the resolution of it; but since by dividing the two members by a, it

becomes x?=—, we see that the question is reduced to finding the b

square root of —.


138. Equations of the second degree are of two kinds', viz. equations involving two terms, or incomplete equations, and equations involving three terms, or complete equations.

The first are those which contain only terms involving the square of the unknown quantity, and known terms; such are the equa tions,

1 5 7 299

3*^5 ; -^_3+-^=--x2+

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These are called equations involving two terms because they may be reduced to the form ax'=h, by means of the two general transformations (Art. 90 & 91). For, let us consider the second equation, which is the most complicated; by clearing the fractions it becomes

8x?—72 +1Ox2=7 -- 24JC2 + 299, or transposing and reducing

42^= 378.

Equations involving three terms, or complete equations, are those which contain the square, aod also the first power of the unknown quantity, together with a known term; such are the equations

5 1 3 2 273

5*2-7*=34; -*-Tx+-=S--x-*+—.

They can always be reduced to the form axs+bx=c, by the two transformations already cited.

Of Equations involving two terms.

139. There is no difficulty in the resolution of the equation

b /b ax'—b. We deduce from it x2=—, whence x—\/ —.

a v a

b .... When — is a particular number, either entire or fractional, a

we can obtain the square root of it exactly, or by approximation. If — is algebraic, we apply the rules established for algebraic quantities.

But as the square of +?n or — m, is +m2, it follows that

/ / b \2 b

J is equal to .—. Therefore, x is susceptible of two

values, viz. x=+\/—, and x= sV—. For, substituting

either of these values in the equation ax2b, it becomes

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3 '12 24 '24

We have already seen (Art. 138.), that this equation reduces to


42x2=378, and dividing by 42, ^=-^-=9; hence x—±3. Lastly, from the equation 3x3=5; we find

x=±v/i.lN/ is.

As 15 is not a perfect square, the values of x can only be determined by approximation.

Of complete Equations of the Second Degree. .

140. In order to resolve the general equation ax>+bx=-.c.

we begin by dividing both members by the co-efficient of x3, which gives,

b c

x3-\ x=—, or x3+px^q

b c by making ~a=p an<* ~a~=^'

Now, if we could make the first member (x2+^x) the square of a binomial, the equation might be reduced to one of the first degree, by simply extracting the square root. By comparing this member with the square of the binomial (x+a), that is, with x3+2ax+a>, it is plain that a^-j-ya; is composed of the square of a first term x,

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