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. And we got it right.Solving equations is the central theme of algebra. So first I'll do the vertexĪt 2 comma negative 5, which is right there. Which is equal to- let's see, this is equal to 2 squared is 4. When x equals 2, y is going toīe equal to 5 times 2 squared minus 20 times 2 plus 15, To substitute back in to figure out its y-coordinate. Sits exactly smack dab between the roots, I want to figure out, is this point right This is true, and you canĪdd 3 to both sides of this. And so this will be true ifĮither one of these is 0. X's will make this expression 0, and if they make Side, we still have that being equal to 0. On factoring quadratics if this is not so fresh- isĪ negative 3 and negative 1 seem to work. And whose sum is negativeĤ, which tells you well they both must be negative. Whose product is positive 3? The fact that their And now we can attempt toįactor this left-hand side. Plus 15 over 5 is 3 isĮqual to 0 over 5 is just 0. Me- these cancel out and I'm left with x squared The x squared term that's not a 1, is to see if I canĭivide everything by that term to try to simplify I like to do whenever I see a coefficient out here on We're going to try to solve the equation 5x Those three points then I should be all set with And then I also want toįigure out the point exactly in between, which is the vertex. Minus 20x plus 15, when does this equal 0? So I want to figure Seen, intersecting the x-axis is the same thingĪs saying when it does this when does y equal I want to first figure out whereĭoes this parabola intersect the x-axis. You can just take threeĬorresponding values for y are and just graph The following equation y equals 5x squared Therefore, 3 and 1 are the only possible x values. So -3 doesn't work as an x, and the same thing would happen with -1:īoth of these solved binomial equations show that -3 and -1 cannot be x values for the parabola. X cannot equal -3 or -1 because if x was -3 then this would happen: The reason that Sal made the x values positive 3 and 1 is because they are the only two x values that would make his equation equal zero. (this basically explains how to get the 3 and 1 for the x values and it explains why -3 and -1 do not work for x values.) I think that was the question you were asking, I hope that helped, if not, hopefully this will: you can check them out if you are still confused. The -3 and -1 were not x values, they were just what he used to factor in a way that he teaches in previous factoring videos. In order to find the x values he used that binomial and made it equal 0, and his x values that he found were 3 and 1. the -3 and -1 were numbers that he got when he factored the equation into a binomial. the -3 and the -1 that he got were not his x values. He didn't exactly switch his x values to positive.
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