OG详解-OG10 数学1 Q27

The correct answer is 29​3​. Applying the distributive property to the left-hand side of the given equation, x(x+1)-56, yields x² +x-56. Applying the distributive property to the right-hand side of the given equation, 4x(x-7), yields 4x² -28x.  Thus, the equation becomes x² +x-56 = 4x² -28x. Combining like terms on the left- and right-hand sides of this equation yields 0 = (4x² -x²)+(-28x-x)+56,  or 3x² - 29x+56 = 0. For a quadratic equation in the form ax² +bx+c = 0, where a, b, and c are constants, the quadratic formula gives the solutions to the equation in the form x = （−b±√b²−4ac​）2a​ . Substituting 3 for a, -29 for b, and 56 for c from the equation 3x² -29x+56 = 0 into the quadratic formula yields x=(29±√(−29)²−4(3)(56​))2(3)​ , or x=296​±136​​. It follows that the solutions to the given equation are 296​+​136​ and 296​−136​​. Adding these two solutions gives the sum of the solutions:  296​+136​+296​−13​6​ , which is equivalent to 296​+296​​ , or  293​​ . Note that 29/3, 9.666, and 9.667 are examples of ways to enter a correct answer.