SOLUTION OF COMPLETE EQUATIONS OF THE SECOND DEGREE. These Equations in their simplest forms contain no other power of the unknown quantity than its square and first power. The value of the unknown quantity will be found by the following RULE XXIII. (260.) For the Solution of a Complete Equation of the Second Degree. 1. Reduce the Equation to the form x2+bx=s; in which x2 must be positive. 2. Aud the square of half the coefficient of x, in the second term, to each ade of the equation:-the first side will then be a perfect square. 3. Eract the square root of each side, and the result will be a Simple Euation,-from which the value of x may readily be found. Clearing the equation of its fractions, transposing, and adding similar terms, we shall find Adding the square of half the coefficient of x, in the second term, to each side,-which is called completing the square, 14x 49 5 49 64 3 + = + = Extracting the square root of each side, Either of these two values of x, 5 or -, will satisfy the given equation; and each of the binomials x- 5 and x+will divide the equation By performing the division it will be found that the left hand side of this equation, is the product of the two binomials; that is, 1. The method of completing the square in the first member, results from the composition of the square of a Binomial. Thus the square of the binomial a+b is a2+2ba+b2, in which b is half the coefficient of a in the second term. 2. If in reducing the equation to the required form, the first term should become -x2, the signs of all the terms in the equation must be changed (117), before completing the square, otherwise the root of the first side would be imaginary, (246). 3. The square root of the first side of the equation-after the completion of the square-will always be the square root of the first term, + or half the coefficient of x in the second term, according as the second term is + or Hence, without completing the square on the first side, we may shorten the solution, by observing that 4. In an equation of the form x2+bx=s, the value of x is half the co-efficient of x in the second term, taken with a contrary sign, + the square root of (the second member of the equation + the square of said half co-efficient). EXERCISES On Quadratic Equations with One Unknown Quantity. 1. Find the value of x in the equation x2-15-45-4x. 2. Find the value of x in the equation 3. Find the value of x in the equation 4. Find the value of x in the equation 5. Find the value of in the equation 5x2+4x-90=114. 6. Find the value of x in the equation x2-x+2=9. 7. Find the value of x in the equation 2x2+12x+36=356. 8. Find the value of x in the equation 36-x 4x-46 Ans. x=5 or -4. Another Method of Solving Quadratic Equations. (261.) A Binomial of the form ax2 + bx will be made a perfect square by multiplying it by 4 times the coefficient of x2, and adding square of the given coefficient of x. the Thus (ax2+bx)4a+b2=4a2x2+4abx+b2=(2ax+b)2, (59). To apply this principle to solution of the Equation 3x2-2x-65. Multiplying both sides of the equation by 4 times the coefficient 3, and adding the square of the coefficient 2 to both sides, we have 36x2—24x+4=780+4=784. Extracting the square root of both sides, we find 6x-2=±28; which gives x= +28+2 =5 or The square root of the first side of the Equation prepared as above, will be x multiplied into twice the given coefficient of x2, + or the given coefficient of x in the second term, according as this term is This is evident from the preceding illustration, (261.) -. + or RULE XXIV. (262.) To Reduce an Equation of the form ax2+bx=s to a Simple Equation. 1. Double the coefficient of x2, and divide the first member by x. 2. Multiply the second member by 4 times the given coefficient of x2, add the square of the given coefficient of x, and extract the square root of the sum. Applying this Rule to the numerical equation we immediately obtain 3x2-5x=50, 6x 550 x 12+25=±√625. When the coefficient of x2 is unity, the multiplier of the second member will be simply 4; and when the coefficient of x is unity, the square to be added after multiplying the second member is 1. This method of solving a Quadratic is preferable to the one first given, whenever the coefficient of x would give rise to a fraction in dividing the Equation by the coefficient of x2, or in completing the square, according to that method. |