Transposing the 12, and multiplying by x, we have x48x3+19x2-12x=0. 529 16 We have now a Biquadratic Equation, which may be reduced to the form. of a Quadratic by the method just exemplified. We find the first two terms of the square root of the first member, by the common rule. This remainder may be resolved into 3(x2-4x), in which the binomial factor is the same as the part of the square root above found. Then we shall have the Equation The Equation x2-4x=-3 will give x=3, or 1. 54. Find the value of x in the Equation x4—2x3+x=5112. The first two terms of the square root of the first member are x2-x; and the remainder is -x2+x; which may be put under the form The first two terms of the square root of the first member are x2+x; and the remainder is 8x2-8x; which may be expressed by -8(x2+x). We shall then have the Equation (x2+x)2 −8(x2+x)=—12. Completing the square, we find (x2+x)2 −8(x2+x)+16=4. Extracting the square root of each side (x2+x)-4=2; hence x2+x=6, or 2. The Equation x2+x=6 will give x=2, or 3. The Equation x2+x=2 will give x=1, or - 2. 56. Find the value of x in the Equation Transposing, x+10x3+35x2—50x+24=0. x4-10x3+35x2—50x——24. The first two terms of the square root of the first side of the Equa and the remainder is 10x2-50x; which may be expressed by 10( 2–52). Hence, we have the Equation (–5)+10(–5)=−24. Completing the square, (x2−5)+10(x2−5x)+25=1. Extracting the square root, x2-5x+5=±1. From the Equation x2-5x+5=+1, x will be found =4, or 1. From the Equation x2-5x+5=-1, x will be found 3, or 2. 57. Find the value of x in the Equation x4-12x3+44x2-48x=9009. The first two terms of the square root of the first number are x2-6x; and the remainder is 8x2-48x; which may be put under the form 8(x2-6x). Hence, we shall have the Equation (x2 — 6x)2+8(x2 — 6x)=9009. Completing the square, we have (x2-6x)2+8(x2-6x)+16=9025. Extracting the square root of each side (x2-6x)+4=±95. From the Equation x2-6x+4=+95, we shall find x=13, or —7. 58. Find the value of x in the Equation x4-8ax3+8a2x2+32a3x=s. The first two terms of the square root of the first member of the Equation are x2-4ax; and the remainder is -8a2x2+32a3x; which may be expressed by -8a2(x2-4ax). Hence, we shall have (x2 — 4ax)2 — 8a2(x2 — 4ax)=s. Completing the square, we find (x2-4ax)2-8a2(x2-4ax)+16a2=s+16a1. Extracting the square root of both sides, In Pure Equations and Affected Quadratics Containing but One Unknown Quantity. 1. Find two numbers such that their product shall be 750, and the quotient of the greater divided by the less, 31. Let x represent the greater number; then 750 x x2 750 represents the less; and represents the quotient of the greater divided by the less. We shall then have the Equation 750 from which we shall find x=- 50, the greater; ther =15, is the less. 50 2. Find a number such that if and of it be multiplied together, and the product divided by 3, the quotient will be 298. x2 56 Let x represent the number; then represents the product of and of the number; and 3. A mercer bought a piece of silk for £16 4s.; and the number of shillings that he paid per vard, was to the number of yards, as 4 to 9. How many yards did he buy? and what was the price per yard? Let a represent the number of yards he bought; then 4x 9 4x2 9 represents the number of shillings per yard; and represents the number of shillings the whole cost; also £16 4s. is equal to 324 shillings. Hence, we shall have the Equation Then of 27=12 shillings was the price per yard. 4. Find two numbers which shall be to each other as 2 to 3, and the sum of whose squares shall be 208. Let x represent the greater number; then 2x 3 represents the less number; and the Equation will be which will give x=12, the greater number; hence of 12-8, is the less number. 5. A person bought a quantity of cloth for $120; and if he had bought 6 yards more for the same sum, the price per yard would have been $1 less. What was the number of yards? and the price per yard? |