ond term, taken with a contrary sign, plus or minus the square root of the second member, increased by the square of half the coeffi cient of the second term. Ex. 1. Let it be required to solve the equation x2-10x=-16. Completing the square by adding to each member the square of half the coefficient of the second term, we have To verify these values of x, substitute them in the original equation, and we shall have Also, 82-10x8-64-80--16. 22-10x2= 4-20=-16. Ans. x +5 or -9. Ex. 2. Solve the equation 2x2+8x-20-70. Ex. 3. Solve the equation 3x2-3x+6=51. Completing the square, x2-x+1=1—3=38· Hence x== or, Ans. Second Method of completing the Square. 260. The preceding method of completing the square is always applicable; nevertheless, it sometimes gives rise to inconvenient fractions. In such cases the following method may be preferred. Let the equation be reduced to the form ax2+ bx=c, in which a and b are whole numbers, and prime to each other, but c may be either entire or fractional. Multiply each member of this equation by 4a, and it becomes 4a2x2+4abx=4ac. Adding b2 to each member, we have 4a2x2+4abx+b2=4ac+b2, where the first member is a complete square, and its terms are entire. Extracting the square root, we have 2ax+b=±√4ac+b2. which is the same result as would be obtained by the former rule; but by this method we have avoided the introduction of fractions in completing the square. If b is an even number, will be an entire number; and it 2 would have been sufficient to multiply each member by a, and b2 add to each member. Hence we have the following 4 RULE. 1st. Reduce the equation to the form ax2+bx=c, where a and b are prime to each other. 2d. Ifb is an odd number, multiply the equation by four times the coefficient of x2, and add to each member the square of the coeffi cient of x. 3d. If b is an even number, multiply the equation by the coefficient of x2, and add to each member the square of half the coefficient of x. Ex. 4. Solve the equation 6x2-13x=-6. Multiplying by 4×6, and adding 132 to each member, we 144x2-312x+169-169-144=25. have Ex. 5. Solve the equation 110x2—21x=−1. Multiplying by 440, and adding 212 to each member, we have 48400x2-9240x+441=1. Ex. 6. Solve the equation 7x2-3x=160. Ans. x-5 or -37. 261. Modification of the preceding Method. The preceding method sometimes gives rise to numbers which are unneces sarily large. When the equation has been reduced to the form ax2+bx=c, it is sufficient to multiply it by any number which will render the first term a perfect square. Let the resulting equation be m2x2+nx=q. The first member will become a complete square by the addition of n and the equation will then be Having reduced the equation to the form ax2+bx=c, multiply the equation by any quantity (the least possible) which will render the first term a perfect square. Divide the coefficient of x in this new equation by twice the square root of the coefficient of x2, and add the square of this result to both members. Ex. 7. Solve the equation 8x2+9x=99. Ex. 8. Solve the equation 16x2-15x=34. Ex. 9. Solve the equation 12x2=21+x. Solve the following equations: Ex. 10. x2-x+20=42. Ex. 11. x2-x-40-170. Ex. 12. 3x2+2x-9=76. Ans. x 7 or -61. Ans. x 15 or -14. This equation reduces to x2-15x=-46. Ex. 23. (x-1)(x-2)+(x-2)(x-4)=6(2x-5). 170 170 51 5 Ans. x=8 or 2 Ex. 24. = Ans. x 4 or -1. х x+1x+2° Equations which may be solved like Quadratics. 262. There are many equations of a higher degree than the second, which may be solved by methods similar to those employed for quadratics. To this class belong all equations which contain only two powers of the unknown quantity, and in which the greater exponent is double the less. Such equations are of the form where n may be either integral or fractional. For if we assume y=x", then y=x, and this equation y2+py=q; becomes Extracting the nth root of each member, we have Ex. 1. Solve the equation x-13x2——36. Assuming x2=y, the above becomes Thus x has four values, viz., +3, −3, +2, −2. To verify these values: 1st value, (+3)-13(+3)2= −36, i. e., 81-117-36. 2d value, (-3)—13 (-3)2=—36, i. e., 81-117—36. 3d value, (+2)1 — 13 (+2)2 = −36, i. e., 16— 52=—36. 4th value, (-2)-13 (−2)2 ——36, i. e., 16— 52——36. Ex. 2. Solve the equation x6-351⁄23——216. Assuming =y, the above becomes This equation has four other roots which can not be de termined by this process. Ex. 3. Solve the equation x+4√x=21. Assuming √x=y, we have y2+4y=21; |