This week in Pre-Calculus 11 we reviewed the different ways to factor, we learned how to find out what x can equal in a quadratic equation when we can factor it, and we learned how to find out what x can equal when a quadratic equation isn’t factorable.  The ways we can do that is by completing the square or filling in the quadratic equation. The most difficult thing I think we learned is completing the square.

Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that one side is a perfect square trinomial. This makes the equation ‘kind of’ factorable and able to solve for x. Without the completing the square method, there would be no way to solve for x for a non factorable trinomial.

Example:

Explanation:

Step 1- find out if it’s a quadratic equation and if it’s factorable

To find out if it’s a quadratic equation, look for an exponent in the equation that has a degree of 2. To check if it’s factorable, multiply the coefficient attached to x2 and the number with no variable attached to it. In this case it would be 26. Now, I would check if two numbers that multiply two 26 also add up to the middle term then it’s factorable. In this case, the equation is non factorable, but it is quadratic.

Step 2- make sure x2 has a coefficient of one

If it doesn’t have a coefficient of one then make sure the number can be divided by the middle term. If it does, then divide the x2 coefficient and the middle term coefficient by the x2 coefficient to make it a 1.

Step 3- divide the middle term by 2 and square it.

At this point, we are at an equation that cannot be solved for x. We can make it solvable by completing the square. The middle term in this example is 6. So if we divide that by 2, which is 3, and square that, which is 9, then we just completed the square.

Step 4- add the # to the equation with a zero pair.

So if we add 9 to the equation (after the 6x), then we have to make it a zero pair by adding a negative 9. If we don’t add the negative 9 then the whole equation is changed, but adding a negative 9 makes a zero pair so it basically cancels each other out, but is very significant.

Step 5- multiply the 2 with the 9

After adding the zero pair, then brackets go around the first three terms. In this example it’s 2(x2 – 6x +9), and the negative 9 goes on the outside of it with the +13. In this case, the 2 on the outside of the bracket is multiplying the positive 9 that we added. To keep the zero pair we have to multiply the negative 9 by the 2 as well.

Step 6- factor and solve

The reason the brackets went around the terms before is because it factors perfectly into two factors that are exactly the same. Here it’s (x-3)(x-3). To put it in simpler terms its (x-3) squared. Now there’s not much left to do except to solve for x.

First move the -5 to the other side of the equal sign. Now to get rid of the two, divide both sides by it. Then to get rid of the square, square root both sides. When you add a square root sign to the fraction, you must put a ± sign before it because the outcome of the square root can be either a negative or a positive. Last, move the three over to the side with the fraction.

Step 7- don’t leave the denominator as a radical

As we learned last unit, radicals can’t be left in the denominator. So to get rid of it, multiply the top and bottom of the fraction by the square root of the denominator which is two. This makes only the top a radical now because the denominator will be equal square root of 4 which is 2.

Step 8- simplify where necessary.

This example didn’t need any because 10 can’t be broken broken down anymore.