The sum of the first two integers is equal to one-fourth the product of the second and third. We can represent them as x, x +2 and x + 4FACTS: The sum of the first two is represented by: x + (x + 2) The product of the second and the third is: (x + 2)(x + 4) One fourth of that value is: FORMULA: From the facts we can create an equation: x + x + 2 =Ĭonsecutive Integer ProblemSOLVE: x + x + 2 = 2x + 2 = combine like terms 4(2x + 2) = (x + 2)(x + 4) multiply both sides by 4 8x + 8 = x2 + 6x + 8 left: distributive property / right: FOIL 8x + 8 – 8 = x2 + 6x + 8 – 8 subtract 8 from both sidesĘx = x2 + 6x simplify 0 = x2 + 6x – 8x subtract 8x from both sides 0 = x2 – 2x combine like termsx2 – 2x = 0 switch sides (symmetry principle) x(x - 2)= 0 factor out common factor of x x = 0 OR x – 2 = 0 Set factors equal to zero (zero x = 2 product rule), and solveANSWER:If x is 0, the second integer is x+2 = 0+2 = 2, the third integer is x+4 = 0+4 = 4If x is 2, the second integer is x+2 = 2+2 = 4, the third integer is x+4 = 2+4 = 6So we express the solutions as 0, 2, 4 OR 2, 4, 6Ĭheck the SolutionConsecutive Integer ProblemLet’s see if our answers fit into the original problem:We have three consecutive even integers. Find all possible solutions for the three integers.Using FFFSA method:FIND: We need to find three consecutive even integers. According to the zero product rule, we know that x + 2 must be equal to 0, so we set it to zero and solve.3x2 + 12x + 12 = 03(x2+ 4x + 4) = 0 3(x + 2)2 = 0(x + 2)2 = 0x + 2 = 0 x = -2ĬheckSubstitute the solution into the original equationfor x = -2 3x2 + 12x + 12 = 03(-2)2 + 12(-2) + 12 = 0 3(4) + 12 (-2) + 12 = 0 12 – 24 + 12 = 0 0 = 0The solution is validĪpplications with Quadratic equations Consecutive Integer ProblemWe have three consecutive even integers. We recognize it as a perfect square.We can divide both sides of the equation by 3 and are left with two identical factors of (x + 2). The resulting expression can be factored further. First we find that there is a common factor of 3, so we factor it out. Solve by Factoring – perfect squareThis equation is in standard form, so we need to determine whether the expression on the right side of the equation can be factored. We find that we have a common factor of 3x in each of the terms of the expression.Using the zero product rule, we set each of the resulting factors equal to zero and solve to find the two solutions.ĬheckSubstitute each of the solutions into the original equationfor x = 09 x2 – 5 = 12x – 59(0)2 – 5 = 12(0) – 5 9(0) – 5 = 12(0) – 5 0 – 5 = 0 – 5 -5 = -5for x = 9 x2 – 5 = 12x – 59( )2 – 5 = 12( ) – 5 9( ) – 5 = 12( ) – 5 16 – 5 = 16 - 5 11 = 11Both solutions are valid. The constant term cancels out in our resulting equation.Next, we factor the expression on the right side of the equation. To do so, we move the terms from the right to the left side of the equation. Solve by Factoring – common factor 9 x2 – 5 = 12x – 5 9x2 = 12x 9x2– 12x = 0 9x2– 12x = 03x(3x – 4) = 0 3x = 0 OR 3x - 4 = 0 x = 0 3x = 4 x = In this example, the equation is not in standard form, so the first step is to express the equation in standard form. ![]() Solve by FactoringHere is an example of an equation in Standard Form:y2 – 6y + 5 = 0Can we factor the expression on the left side of the equation? Yes, we can express it as the product of two polynomial factors.Since the product of those two factors is zero, then according to the Zero Product Rule, one of the factors must be equal to zero.So we set each of the factors to zero and solve to determine the two possible solutions.y2 – 6y + 5 = 0 (y – 5) (y – 1) = 0 y – 5 = 0 OR y – 1 = 0y = 5 y = 1ĬheckSubstitute each of the solutions into the original equationfor y = 5 y2 – 6y + 5 = 0(5)2 – 6(5) + 5 = 0 25 – 30 + 5 = 0-5 + 5 = 00 = 0 for y = 1 y2 – 6y + 5 = 0(1)2– 6(1) + 5 = 0 1 – 6 + 5 = 0-5 + 5 = 00 = 0 Both solutions are valid. (factor2) = 0, then either factor1 = 0, or factor2 = 0.If so, then the product of those factors is 0, and since(factor1) Then determine whether the expression on the left side of the equation can be factored. ![]() Zero Product RuleSolving a quadratic equation by factoring is based upon the Zero Product Rule which states: if ab = 0, then either a = 0 or b = 0To apply this rule to solving a quadratic equation:We first must ensure that the equation is in Standard form: Ax2 + Bx + C = 0 This lesson involves those that can be solved by factoring. There are several methods for solving them. They may have zero, one or two solutions. Solving Quadratic Equations by FactoringQuadratic Equations are also known as Second Degree Equations because the highest power of the variable is 2.
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