This week in Precalculus 11 we learned about discriminant. A discriminant is a part of the quadratic equation that helps us determine many things such as if the equation is factorable, how many roots it has, or if it has any roots at all before we actually solve the whole thing.

Example

Equation:

Step 1: make sure to take out anything common

Step 2: determine what a, b, and c are.

a = 1

b = -6

c  = 9

Step 3: fill the formula

Step 4: simplify

(36) – (36)

0

Explanation

Step 1: make sure to take out anything common

Taking out things that are common makes figuring out the answer a lot easier because the numbers are smaller to work with, but even if you leave everything in, it comes out with the same answer.

Step 2: determine what a, b, and c are

The formula for finding the discriminant relies on a, b, and c, and they are all in coordination with the equation. a would be equal to the first number (the number that has attached to it. b would be equal to the middle number (the number attached to x). c would be equal to the last number (the number that has no variable attached to it.). Writing out what a, b, and c are equal to makes it a less chance of error.

Step 3: fill the formula

Next, fill the formula according to coordination between letter and number

Step 4: simplify

There’s not much left to do except to find out the answer. The answer for this example was 0. This indicates that this equation has one real root. If the answer was any number bigger than 0 then that would mean that this equation has 2 real roots. If the answer was a number smaller than 0 (any negative number) then the equation has no real roots which means there is no need to solve the rest of the quadratic equation. That’s why it is helpful to find the discriminant, because we don’t need to solve more than we need to if we don’t need to.

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.

This week in PreCalc 11 we learned how to solve radical equations (which is an equation where the variable is under a radical). It was easy if we used our basic knowledge of equations; what we do to one side we do to the other (if I added 25 to one side I would add 25 to the other side of the equal sign)  AND to cancel out an operation you do the opposite of it. (to cancel out -25 I would +25). With this basic knowledge it was pretty straight forward.

Example:

Explanation:

Step 1: apply the opposite operation to get rid of the square root

In the example, square roots were surrounding all the numbers in the equation so not much could be done without them gone. To get rid of the square roots you have to find the opposite operation which in this case will be to square everything. This is because once we square everything and it stays under the square root, the square root will cancel it out.

Step 2: bring like terms together

Now that the square root isn’t in the way of solving the equation we can do it now. First thing’s first, bring like terms together. The 5x and 2x are on opposite sides. So to remove the the 2x from the side it started on originally we have to subtract 2x to cancel it out. Then, since what we do on one side we do to the other, we have to add 2x to the other side. Now the -1 and 5 are on different sides of the equal sign so we have to +1 it started on originally and then +1 on the other side. After all the like terms are on the same side, combine them.

Step 3: SOLVE

All there really is to do is solve. To isolate x we have to divide by 3 and do that to the other side too. Then we are left with the answer which is 2.

Step 4:

Now that we know that x = 2 we need to verify our solution to make sure the root of the equation is a real root. For it to be a real root x has to be greater than or equal to 0. And once we have determined that, we have to put what x equals back into the original equation. In this case when I put 2 back into the original equation, both sides equal 3, therefore; 2 is a real root.