Although the method of finding the value of x is the same in each of these forms, it is convenient to distinguish between them. See Art. 215. From the preceding we derive the RULE, FOR THE SOLUTION OF A COMPLETE EQUATION OF THE SECOND DEGREE. 1st. Reduce the equation, by clearing of fractions and transposition (if necessary), to the form ax2+bx=c. 2d. Divide each side of the equation by the coefficient of x2, and add to each member the square of half the coefficient of the first power of x. 3d. Extract the square root of both sides, and transpose the known term to the second member. EXAMPLES. 1. Find the roots of the equation x2+8x=33. Completing the square by taking half the coëfficient of x(§); squaring it, and adding the square to each member, we have x2+8x+16-33+16=49 9+24=33. Vrification. (3)2+8(3)=33, that is, Or (-11)2+8(-11)=33, that is, 121-88-33. In verifying these values of x, it is to be noticed, that the square of -11, is 121, and that 8 multiplied by -11, gives -88. 2. Solve the equation x2-6x=16. Completing the square, Both of which will be found to verify the equation. 3. Solve the equation x2+6x=—5. Completing the square, 4. Find the values of x, in the equation x2-10x=-24. Completing the square, Extracting the root, x2-10x+25=25-24-1 Transposing, Whence And The preceding examples, illustrate the four different forms, when the equation is already reduced. Equations of the second degree, however, generally occur in a more complicated form, and require to be reduced before completing the square. 5. Find the values of x, in the equation 3x-5= 3x2-5x=7x+36 7x+36 x Clearing of fractions, Transposing, 3x2-12x-36 Here the coefficient of x is 3, the half of which is square of this is, which being added to both sides, we have the EXAMPLES FOR PRACTICE. NOTE.-The first sixteen of the following Examples, are arranged to illustrate the four forms, to one of which every complete equation of the second degree may be reduced. REVIEW.-212. To what form may every complete equation of the second degree be reduced? What are the four forms that this gives, depending on the signs of 2p and q? Explain the principle, by means of which the first member of the equation x2+2px=q may be made a perfect square. What is the rule for the solution of a complete equation of the second degree? 41. x2-4x=-1.". . . Ans. x=2/3=3.732+, or .268-. ART. 213.-THE HINDO0 METHOD OF SOLVING QUADRATICS.— When an equation is brought to the form ax2+bx=c, it may be reduced to a simple equation, without dividing by the coefficient of x2; thus avoiding fractions. If we multiply both sides of the equation ax2+bx=c, by a, the coefficient of x2, it becomes a2x2+abx=ac.. Now, if we regard a2x2+abx, as the first and second terms of the square of a binomial, a2x2 must be the square of the first term, and abx the double product of the first term by the second. Hence, the first term of the binomial is va2x2-ax; and the second term, the quotient derived from dividing abx by the double of ax, the abx first term; that is, b. 2ax2 Adding the square of b2 the equation becomes a2x2+abx+4=ac+4· Now, the left side is a perfect square; but it will still be a perfect square, if we multiply both sides by 4, which will clear it of fractions. Thus, 4a2x2+4abx+b2=4ac+b2 Extracting the square root, Whence 2ax+b=1/4ac+b2 -b±√4ac+b2 Now, it is evident, that the equation 4a2x2+4abx+b2=4ac+b2, may be derived directly from the equation ax2+bx=c, by multiplying both sides by 4a, the coefficient of x2, and then adding to each member, the square of b, the coefficient of the first power of x. This gives the following RULE, FOR THE SOLUTION OF A COMPLETE EQUATION OF THE SECOND DEGREE. Reduce the equation to the form ax2+bx=c, and multiply both sides, by four times the coëfficient of x2. Add the square of the coëfficient of x to each side, and then extract the square root. This will give a simple equation, from which x is easily found. 1. Given 3x2-5x-28, to find the values of x. Multiplying both sides by 12, which is 4 times the coëfficient of x2, 36x2-60x=336 Adding to each member 25, the square of 5, the coefficient of x, Extracting the root, 36x2-60x+25=361 6x-5=19 x=5±19=24, or —14 By the same rule, find the values of the unknown quantity in each of the following examples. If further exercises are desired, the examples in the preceding article may be solved by this rule. REVIEW.-213. Explain the Hindoo method of completing the square |