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10. Required the value of the expression

16+2V5-V6-2V5. 11. Required the value of the expression

14+3V–20+ V4-3V–20. Ans. 6. 12. Required the square root of -3+ -16.

Ans. 1+2V - 1. 13. Required the square root of 8V - 1.

Ans. 2+2V - 1.

Simple Equations containing Radical Quantities. 246. When the unknown quantity is affected by the radical sign, we must first render the terms containing the unknown quantity rational. This may generally be done by successive involutions. For this purpose we first free the equation from fractions. If there is but one radical expression, we bring that to stand alone on one side of the equation, and involve both members to a power denoted by the index of the radical. Ex. 1. Given x+Vx2 – 3x+60=12 to find x. Transposing w and squaring each member, we have

x2 – 3x+60=144–24x+x2; whence


2a Ex. 2. Given Vict Va+x= to find x.

Vata 17-51

Ans. x= Ex. 3. Given

3 to find x.

11 Ex. 4. Given v 4x2 — 72—6=9-22 to find a

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247. If the equation contains two radical expressions combined with other terms which are rational, it is generally best to bring one of the radicals to stand alone on one side of the equation before involution. One of the radicals will thus be made to disappear, and, by repeating the operation, the remain. ing radical may be exterminated.

Ex. 5. Given Vx+19+Vx+10=9 to find x.

By transposition, Vx+19=9-Vx+10.
Squaring, *+19=91+-18Vx+10.
Transposing and reducing,


x+10=16; whence

Ex. 6. Given V36+x=18+ Vī to find me
Ex. 7. Given Vă+4ab=26+ Vāc to find X.
Ex. 8. Given x=Va+XV12+x2 -axta to find 2.





248. When an equation contains a fraction involving radical quantities in both numerator and denominator, it is sometimes best to render the denominator rational by Art. 239; but the best method can only be determined by trial.

VictV2-3 3 Ex. 9. Given

to find x. V-V-32-3 Multiply both terms of the first fraction by Vã+ V-3, and we have (Vic+Vx-3)

3 3-(2-3)

9 (Vät Vã-3)? Extracting the square root,

3 V+V.-3=

VX-3 Clearing of fractions,

Væ2-3+x-3=3; whence

x=4, Ans. 19X+13+1

92 Ex. 10. Given

=13 to find x. 19x+13–V 92 Ex. 11. Given "Vax+b="cx+d to find .

50 Vx+24-9 Ex. 12. Given

=7 to find x. 3+5Vx+24





find x.

32-1 ... Ex. 13. Given =1+}(V34–1) to find 2. ✓ 3x+1

Ans. x=3. Væ+4m_Væ+2m

+ Ex. 14. Given

to find x. Vic+3n Victn

Ans. x=

m Ex. 15. Given (V9x-6) (Vx+25)=(5+3Væ) (Væ+3) to

Ans. 2=9. Ex. 16. Given v2x-3n=3Vn-V23C to find x.

Ans, x=2n. 79x—4 15+ V9X Ex. 17. Given

to find x. Væ+2 Vāc+40

V6x-2 V6x-9 Ex. 18. Given

to find x. V6x+2 V6x+6

5x - 9 752-3 Ex. 19. Given


to find 2. 75x+3

2 Ex. 20. Given V4a+c=2V0+x-Væc to find x.

(a-3)2 Ans. x=


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EQUATIONS OF THE SECOND DEGREE. 249. An equation of the second degree, or a quadratic equation with one unknown quantity, is one in which the highest power of the unknown quantity is a square.

250. Every equation of the second degree containing but one unknown quantity can be reduced to three terms; one con. taining the second power of the unknown quantity, another the first power, and the third a known quantity; that is, it can be reduced to the form

22 +px=4. Suppose we have the equation

3.2 7-6 2-2

-t + +133=(x+3)(x-1).

4 12 3 Clearing of fractions and expanding, we have

932 +73-6+43—8+164=1232 +24x—36. Transposing and uniting similar terms, we have

- 3x2 - 13x=-186. Dividing by the coefficient of x2, that is, by —3, we have

x2 + ,


3 which is of the form above given, p in this case being equal to 1, and q being equal to 62.

251. In order to reduce a quadratic equation to three terms, we must first clear it of fractions, and perform all the operations indicated. We then transpose all the terms which contain the unknown quantity to the first member of the equation, and the known quantities to the second member; unite similar terms, and divide each term of the resulting equation by the coefficient of x.

252. An equation of the form x2 +px=q is called a complete equation of the second degree, because it contains each class of terms of which the general equation is susceptible.

253. The coefficient of the first power of the unknown quantity may reduce to zero, in which case the equation is said to be incomplete.

An incomplete equation of the second degree, when reduced, contains but two terms: one containing the square of the unknown quantity, and the other a known term.

Incomplete Equations of the Second Degree. 254. An incomplete equation of the second degree may be reduced to the form 22=. Extracting the square root of each member, we have

x= . If q be a positive number, either integral or fractional, we can extract its square root, either exactly or approximately, by the rules of Chap. XII. Hence, to solve an incomplete equation of the second degree, we have the


Reduce the equation to the form x2=9, and extract the square root of each member of the equation.

255. Since the square of both + m and —m is + m2, the square of +vq and that of - Vq are both +q. Hence the above equation is susceptible of two solutions, or has two roots ; that is, there are two quantities, which, when substituted for 3 in the original equation, will render the two members identical. These are

+vq and -Vq. Hence, Every incomplete equation of the second degree has two roots, equal in numerical value, but with opposite signs.

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EXAMPLES. Find the values of x in each of the following equations: 1. 4x2-7=3x2 +9.

Ans. x= 4.

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