CHAPTER XXVII. EQUATIONS IN QUADRATIC FORM. = c, n being a 294. An equation in the form ax2 + bxn positive or negative integer or fraction, is in quadratic form. Thus a* + 4x2 = 117, 2+7x=44, and e equations in quadratic form. - a are We give a few examples showing that the ordinary rules for quadratic equations are applicable to those in quadratic form. 295. Any equation which can be thrown into the form ax2 + bx+c+p√ax2 + bx + c = q may be solved as follows. Putting y =√ax2 + bx + c, we obtain y2+py-q=0. Let r, and r2 be the roots of this equation, so that √ax2 + bx + c = r1, √аx2 + bx + c = Taj from these equations we shall obtain four values of x. When no sign is prefixed to a radical, it is usually understood that it is to be taken as positive; hence, if r, and r are both positive, all the four values of x satisfy the original equation. If, however, r, or r2 is negative, the roots found from the resulting quadratic will satisfy the equation ax2 + bx + c−p√ax2 + bx + c = q, but not the original equation. Ex. 1. Solve x2 5x+2V – 5x+3=12. x2 – 5x +3+2V – 5 + 3 = 16. Putting √2-5x+3=y, we obtain y2+2y-150; whence y= 3 or 5. Thus √x25x + 3 =+ 3, or √x2 - 5x + 3 = − 5. Squaring and solving the resulting quadratics, we obtain from the 5±√113 first x 6 or = 1; and from the second x = 2 The first pair of values satisfies the given equation, but the second pair satisfies the equation x25x-2√x2 - 5x + 3 = 12. Ex. 2. Solve 3x2 −7+3√3x2 – 16 x + 21 = 16 x. Transposing, 3 x2 − 16 x − 7 + 3√3 x2 − 16 x + 21 = 0. Thus y2+ 3y = 28; whence y = 4 or — 7. √3x2 - 16x + 21 = 4 or √3x2 – 16 x + 21 = - Squaring and solving, we obtain The values 5 and satisfy the original equation. The other values satisfy the equation 3x2-7-3√3x2 - 16x + 21 = 16x. 296. Occasionally equations of the fourth degree may be arranged in expressions that will be in quadratic form. or Ex. Solve 24 - 8 x3 + 10 x2 + 24 x + 5 = 0. This may be written x4 8x3 + 16 x2-6x2 + 24 x = − 5, by formula, x2-4x= whence 6 ±√36-20 6±4 2 = = = 5 or 1; 2 x = 5, -1, or 2 ± √5. The student will notice that in such examples he should divide the term containing a by twice the square root of the first term and then square the result for the third term. In this case a third term of 16 x2 is required, therefore we write the term 102 of the original equation in the form 16 x2-6x2. 297. Equations like the following are of frequent occur rence. Ex. 2. A person, selling a horse for $72, finds that his loss per cent is one-eighth of the number of dollars that he paid for the horse what was the cost price? Suppose that the cost price of the horse is x dollars; then the loss on $100 is $2. and each of these values will be found to satisfy the conditions of the problem. Thus the cost is either $80, or $720. Ex. 3. A cistern can be filled by two pipes in 33 minutes; if the larger pipe takes 15 minutes less than the smaller to fill the cistern find in what time it will be filled by each pipe singly. Suppose that the two pipes running singly would fill the cistern 15 minutes. When running together they will fill 1 in x and x + ᄒ) x 15 of the cistern in one minute. But they fill of the cistern in one minute; hence Thus the smaller pipe takes 75 minutes, the larger 60 minutes. Ex. 4. The small wheel of a bicycle makes 135 revolutions more than the large wheel in a distance of 260 yards; if the circumference of each were one foot more, the small wheel would make 27.revolutions more than the large wheel in a distance of 70 yards: find the circumference of each wheel. |