If x = 6, then x 2 - 5x = 6 becomesĬhecking your solutions is a sure way to tell if you have solved the equation correctly. Step 4 Check the solution in the original equation. This applies the above theorem, which says that at least one of the factors must have a value of zero. Since we have (x - 6)(x + 1) = 0, we know that x - 6 = 0 or x + 1 = 0, in which case x = 6 or x = - 1. Step 3 Set each factor equal to zero and solve for x. Solution Step 1 Put the equation in standard form. Of course, both of the numbers can be zero since (0)(0) = 0. We can never multiply two numbers and obtain an answer of zero unless at least one of the numbers is zero. We will not attempt to prove this theorem but note carefully what it states. In other words, if the product of two factors is zero, then at least one of the factors is zero. The method of solving by factoring is based on a simple theorem. This method cannot always be used, because not all polynomials are factorable, but it is used whenever factoring is possible. The simplest method of solving quadratics is by factoring. It is possible that the two solutions are equal.Ī quadratic equation will have two solutions because it is of degree two. This theorem is proved in most college algebra books.Īn important theorem, which cannot be proved at the level of this text, states "Every polynomial equation of degree n has exactly n roots." Using this fact tells us that quadratic equations will always have two solutions. The solution to an equation is sometimes referred to as the root of the equation. In other words, the standard form represents all quadratic equations. The standard form of a quadratic equation is ax 2 + bx + c = 0 when a ≠ 0 and a, b, and c are real numbers.Īll quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation. Solve a quadratic equation by factoring.Ī quadratic equation is a polynomial equation that contains the second degree, but no higher degree, of the variable.Place a quadratic equation in standard form.Upon completing this section you should be able to: QUADRATICS SOLVED BY FACTORING OBJECTIVES You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations. In previous chapters we have solved equations of the first degree. All skills learned lead eventually to the ability to solve equations and simplify the solutions. The obvious root we would be interested in for most chemical applications is the positive number, 3.00 x 10 -2.Solving equations is the central theme of algebra. QUIZ: Solve for x using the quadratic equation:Īnswer: x = 3.00 x 10 -2 and -3.11 x 10 -2. Therefore, A and B both lost 0.099 M and the equilibrium concentrations of both C and D are 0.099 M. Since it is impossible to have a negative concentration remaining, the 0.309 number is extraneous (meaningless) and the other, x = 0.099 is the root we are interested in. If 0.309 M of one reactant was lost, that would leave behind (0.300 - 0.309) = -0.009 M of one reactant and (0.100 - 0.309) = -0.209 M of the other reactant. The value of x represents the concentration of these reactants that were converted into products. In this particular problem, the initial concentrations of two reactants were 0.300 M and 0.100 M - these numbers appeared in the denominator of the original problem. If 0.300 mol of A and 0.100 mol of B are mixed in a 1.00 liter container and allowed to reach equilibrium, what concentrations of A and B will react and what concentrations of C and D will be formed? Let me give you the original problem:Ĭonsider the following equilibrium having an equilibrium constant = 49.0 at a certain temperature: Now, plug the numbers into the quadratic formula,Ĭhemical Equilibrium Application: At this point, it may be difficult for you to see which root (answer) is useful and which one is not. If we then subtract x 2 from both sides, we can rearrange the equation to get a quadratic equation: If we cross-multiply (See the review on Algebraic Manipulation), we get: To expand the denominator, multiply the two terms together: Let's work through a typical quadratic calculation that you might find in equilibrium problems. Since nothing can exist as a negative concentration, the other answer must be the RIGHT one. This will be obvious! Usually when the WRONG answer is plugged in, it will lead to a negative concentration or amount. This means that one answer will make sense, the other answer won't. In most chemistry problems, only one answer will be meaningful and have physical significance. There are two roots (answers) to a quadratic equation, because of the in the equation. A quadratic equation can always be solved by using the quadratic formula: It has the general form:Įach of the constant terms (a, b, and c) may be positive or negative numbers. when solving equilibrium problems, a quadratic equation results.
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