French language. The E, however, often bears the Acute Accent, thus-IE. These Vowels also appear very often in the body of a word with the E accented. In such cases, they do not constitute a Diphthong, and cannot be illustrated by the sound of EE in English word BEE; but each preserves its own distinct Vowel sound Name. IO. Sound. Until the learner has become really familiar with the French language, the surest way to be correct in the use and pronunciation of words commencing with QUA, will be to consult a dictionary. UE. This Diphthong occurs most frequently, as the final letters of French words after the Consonants of G and Q;-in which cases both are silent. When final, and before other Consonants, they have the usual EO. Like the letters IO, in the last syllable of the English sound of French U; examples of which will be given in future word CUR-CU-LIO. lessons. Lui Luee or L'wee Him. Night. Yes. Power. Ruin. The ten Diphthongal Combinations of THREE SUCCESSIVE VOWELS, in the same word, are thus divided to show their pronunciation, which will be further illustrated in future lessons, viz. : Puant Sometimes this Diphthong has the sound of A in the English word FAT, viz. : This last illustration, however, is not strictly correct, because it does not preserve the distinct sound of the French U; which sound, especially in combination, many Frenchmen themselves are not careful to preserve. In common conversation, this Diphthong sounds like an English W. In French words commencing with QUA, the Diphthong UA, has two different sounds. In some the sound of UA would be illustrated by the letters KOUA or K'WA;-but in others, by KA, viz.: QUADRANGLE, is pronounced, KOUAH-DRANH!GL', or Diphthongs of FOUR SUCCESSIVE VOWELS in the same word are thus divided for pronunciation, which latter-the pronunciation-will be amply illustrated in future lessons, viz. : EM "EN is represented by the letters ANH! and is like the sound of the letters AN, in the English words AN-CHOR and CAN-KER, with an effort to speak through the nose, as it is termed. But be particular to avoid the sound of English G IN ALL NASALS. There is, strictly speaking, a real difference between the Nasal sounds of AN, EN and IN;-but it is so slight, and so peculiarly delicate, that scarcely any one not a native Frenchman can detect and describe it intelligibly. In common reading and conversation, these Nas als above-mentioned have but one sound, viz. :-that which has been assigned them in our previous Lessons. It is considered correct enough for all practical purposes. When extraordinary nicety of pronunciation is demanded, as is always the case in using the language of prayer, and in holy and devotional language, the A of the Nasals AM and AN should be pronounced BROADER than the E or I in the Nasals EM, EN, IM and IN. In the former case, let the A Like the sound of E mute. have the sound of AH; in the latter, the sound of A in the word FAT. The sound of OM and There are a few exceptions to the preceding illustrated pro-nunciation, which will be given, viz. :— ENNUI:-according to the rule II, on a preceding page, is represented by the letters the first EN of this word would not be Nasal, because the N is doubled. In this word, however, EN is a Nasal. ONH!, and is like the sound of the letters ON, in the English word CON-QUER, uttered with an effort to speak through the nose, as it is termed. The sound of Ennui Anh!-nuee Tediousness. In the following words, the EN, is a Nasal, viz. :— { Ennuyant Ennuyeusement Anh!-nuee-eeuh!z-manh Ennuyeu Anh!-nuee-eeuh! Anh!-nuee-eeuh! Anh!-nuee-eeuh!z UNH!, and is like the sound of the letters UN, in the English word UN-CLE, uttered with an effort to speak through the nose, as it is termed. Concerning these Nasals, remember these two General Rules, viz.: I. Single M's and N's followed by VOWELS are not Nasals. II. When the M and N are doubled, the Nasality is destroyed. Exceptions to this last Rule will appear in their proper places. We now proceed to illustrate these Nasal sounds. Annoying. Emmancher term Emmancheur Anh!-manh!-shenhr AN. Emmanchure Anh!-manh! shure In the first-AIM,-AI is equivalent in sound to A only; hence, substituting A for AI in the combination AIM, we have simply AM, whose sound has been explained. In the second-AIN, its sound is represented by ANH!, for the same reason. In the third--EIN, EI is equivalent only to A in sound; hence, substituting A in the place of EI in the combination EIN, we have AN, whose sound is represented by ANH! Again-EAN and OAN have each the Nasal sound represented by the letters ANH! AEN in the Proper Name CAEN, have also the sound of AN, represented by the letters ANH!; hence the word CAEN is pronounced KANH! Emmannequiner Anh!-man-kee-nay Emmêler Second syl.) Emmenagement Anh!-may-nazh-manh! Emmener Emménologie Emmuseler Anh!-may-na-zhay Anh!-may-na-gog Anh!-may-nol-o-zhee Anh!-mea-lay Handle-maker. To put into a basket. Sea-hardened. To entangle. * To handcuff. To consecrate a bishop. Banked with earth. Employment. Enivrer English. Meanwhile. To charm. Pronunciation. and all deri-y (ved from it) Anh!-nee-vray Enorgueillir, Anh!-or-guaygl-yeer IM. Pronunciation. English. To intoxicate. French. Anh!-bay-sil Impénitence Anh!-pay-ne-tanh!s Moment. Impératoire Anh!-pay-ra-toahr or t'wahr Appointed place. Impossible Anh!-po-sibl’ Limbe Lanhb To surprise. The French word MONSIEUR is pronounced by foreigners all sorts of ways, except the right way, in common conversation. The author knows of no one French word so much in use by those who speak the English language, as this; and yet pronounced so variously and incorrectly. Let us analyze this word, and, if possible, set forth its correct sound. Remember, then, that the N and R of the word MONSIEUR are always silent;-the N is silent by the Rule of custom, and the R is silent according to the General Rule which obtains concerning final Consonants. Take out of the word the letters N and R, and we have MOSIEU. Divide it now into syllables, and we have MO and SIEU. In the first syllable, the O is short like the letter O in the English word NOT,-therefore the_pronunciation of the first syllable MO, is easily ascertained. But in the second and last syllable SIEU, we have a Diphthong of THREE SUCCESSIVE VOWELS, viz.: IEU divided thus, I-EU, but pronounced as one syllable, preserving the sounds of both divisions. The sound of I is short, like I in the English word FIG;-and the sound of EU is exactly like E Mute or unaccented. These are the elements of the different sounds in the French word MONSIEUR, and are thus pronounced, viz. :-MOSIEU or MO-SIUH! Sometimes it is pronounced MOS-SIEU, but incorrectly, because the Parisian critic and scholar gives it but one S, and that, at the beginning of the second syllable. Hence it will be perceived, that it is simply ridiculous to pronounce this word Mong-seer or Mon-secuh. The ON in this word is not a Nasal, because the N is silent. The I is not long, and cannot be illustrated by EE, but is short, as above explained. LESSONS IN ALGEBRA.-No. XXV. METHOD FOR COMPLETING THE SQUARE.—Continued. Before completing the square, the known and unknown quantities must be brought on opposite sides of the equation by transposition; the square of the unknown quantity must also be positive, and it is preferable to make it the first or leading term. If the square of the unknown quantity is in several terms, the equation must be divided by all the co-efficients of this square. 16. Reduce the equation Dividing by +d, 17. Reduce the equation Given ax + bx=d, to find x. If this equation is multiplied by 4a, and if b2 is added to both sides, it will become, 4a2x2 + 4abx +b2 = 4ad + ba ; 1. The object of multiplying the equation by the co-efficient of the highest power, is to render the first term a perfect square without removing its co-efficient, and at the same time to obtain the middle term of the square of a binomial. But we must multiply all the terms of the equation by this quantity to preserve the equality of its members. The equation above, when multiplied by a, becomes a2x2 2 +abx=ad. That the first term will, in all cases, be rendered a complete square when multiplied by its co-efficient, is evident from the fact, that it will then consist of two factors, each of which is a square, viz. x2, and the square of its co-efficient. But the product of the squares of two or more factors is equal to the square of their product. 2. It will be seen that one term is still wanting in the first member, in order to make it the square of a binomial, viz. the square of the last term. This deficiency may be supplied by adding to both sides the square of half the co-efficient of the lowest power, as in the first method of completing the square. But in taking half of this co-efficient, the learner will often be encumbered with fractions which it is desirable to avoid. Thus in the equation above, half of the co-efficient of the lowest power is the b 2' Take the equation Multiplying by 16, etc. 16x248x+36=100 By evolution and transposition x=4. Ans. Mult. by n2, etc. n2a2x2 + n3acx+ =n2ad+n2c2; the first member of which is the square of the binomial, Adding this to both sides, the equa- may then be completed. tion will become a2x2+abx+ 62 4 62 =ad+ the first member 4' Now it is obvious that multiplying the equation by 4 is the same as removing the denominator 4 from the third term. Hence multiplying the equation by 4 will avoid the introduction of fractions, and also leave the square of the whole of the co-efficient of the lowest power to be added to both sides according to the rule. The first term evidently continues to be a square after it is multiplied by 4, for it is still the product of the powers of certain factors. 3. It will be perceived at once, that the second term is composed of twice the root of the first term multiplied into the co-efficient of the last term, which constitutes the middle term of a binomial square. Obs. It is manifest from the preceding demonstration, that multiplying by 4 is not a necessary step in completing the square, but is resorted to as an expedient to prevent the occurrence of fractions. When, therefore, the co-efficient of the lowest power is an even number, so that half of it can be taken without a remainder, we may simplify the operation by multiplying by the co-efficient of the highest power alone, and adding to both sides the square of half the co-efficient of the lowest power of the unknown quantity. 22. Reduce the equation 4x-x2-32. In a quadratic equation the first term 2 is the square of a single letter. But a binomial quantity may consist of terms, one or both of which are already powers. Thus 3a is a binomial, and its square is 6+2ax3 + a2, where the index of a in the first term is twice as great as in the second. When the third term is deficient, the square may be completed in the same manner as that of any other binomial. For the middle term is twice the product of the roots of the two others. So the square of x+a, is x2+2ax" + a2. 1 And the square of x+a, is x + 2ax + a2. Therefore Any equation which contains only two different powers or roots of the unknown quantity, the index of one of which is twice that of the other, may be solved in the same manner as a quadratic equation, by completing the square. N.B. It must be observed, that in the binomial root, the letter expressing the unknown quantity may still have a fractional or integral index, so that a further operation may be necessary. 23. Reduce the equation Completing the square Extracting and transposing Extracting again 24. Reduce the equation 25. Reduce the equation 26. Reduce the equation The solution of a quadratic equation, whether pure or adfected, gives two results. For after the equation is reduced, it contains an ambiguous root. In a pure quadratic, this root is the whole value of the unknown quantity. Thus the equation Becomes, when reduced, x2=64, z=±√64. That is, the value of x is either +8 or -8, for each of these is a root of 64. Here both the values of x are the same, except that they have contrary signs. This will be the case in every pure quadratic equation, because the whole of the second member is under the radical sign. The two values of the unknown quantity will be alike, except that one will be positive, and the other negative. But in adfected quadratics, a part only of one side of the reduced equation is under the radical sign. When this part is added to, or subtracted from, that which is without the radical sign; the two results will differ in quantity, and will have their signs in some cases alike, and in others unlike. 31. Divide the number 30 into two such parts that their product may be equal to 8 times their difference. If the less, then 30 - the greater part. By the supposition, xX (30 — x) = 8 × (30 — 2x). This reduced, gives x = 23 ± 1740, or 6, the less part But as 40 cannot be a part of 30, the problem can have but one real solution, making the less part 6, and the greater part 24. The preceding principles in quadratic equations may be summed up in the following 1. Remove the co-efficient of the second power of the unknown quantity, and add the square of half of the co-efficient of the first power of the unknown quantity to both sides of the equation. 2. Or multiply the equation by four times the co-efficient of the highest power of the unknown quantity, and add to both sides the square of the co-efficient of the first power of the unknown quantity. IV. Reduce the equation by extracting the square root of both sides; and transpose the known part of the binomial root thus obtained to the opposite side. When one of the values of the unknown quantity in a quadratic equation is imaginary, the other is so also. For both are equally affected by the imaginary root. Thus in the example above, The first value of x is x-4 =2x+ 100-9x 4.x2 x +1=10 = 10 — — 2. 4x+7 10.x2+1 x2-6x+9 3 =-3. And the second is contains the imaginary quantity An equation which, when reduced, contains an imaginary root, is often of use to enable us to determine whether a proposed question admits of an answer, or involves an absurdity. 30. Suppose it is required to divide 8 into two such parts that the product will be 20. If x is one of the parts, the other will be 8-x. By the conditions proposed This becomes, when reduced, (8x) Xx 20 x=4±√ −4. Here the imaginary expression -4 shows that an answer is impossible; and that there is an absurdity in supposing that 8 may be divided into two such parts that their product shall be 20. Although a quadratic equation gives two results, yet both these may not always be applicable to the subject proposed. The quantity under the radical sign may be produced either from a positive or a negative root. But both these roots may not, in every instance, belong to the problem to be solved. 11. Reduce x+1 3x x+2 a |