Before the break, we got introduced to quadratics. When I first learned about them, I made a very big mistake. I didn’t understand the concept of finding an x1 and an x2. This is mainly because I didn’t get that quadratic formulas are where two points intersect on a graph, so I just didn’t get why we had to find two points. Now I feel that I understand the questions much better, as I’m normally used to solving for one x, but cancelling the other out makes it very easy.

for example, with questions like (x-2)(x+7)=0 I thought I was supposed to expand the equations then solve for x but because of the zero I can just swap the factors for ones that make zero with the other numbers. in this case x1 would be 2 and x2 would be -7.

This is a simple concept I was just very confused at first.

This week in Pre Cal 11 we learned more about factoring. We did a lot with factoring in grade 10 and it was one of my best units so I already feel pretty confident in my skills but there were some new topics introduced that I think I could use some more practice on.

One of those topics is using substitution. This wasn’t a very complicated concept it was similar to most things we’ve done in algebra but it was very useful when factoring equations with more complex variables and made It much easier.

This was the first question I tried, and although it looks like a very complicated factoring question, it is incredibly easy to use substitution.

Here is my work:

Pretty much you choose a different variable to replace the parts in brackets. I chose A for this. Write the sentence let (x+3) = a and factor your equation like normal after. Once you’ve factored your substituted equation, replace your variable for its original form and you’re done.

This seemed very complicated but it wasn’t and I think it’s a very helpful way of factoring bigger equations.

Last week in Pre-Calculus I struggled with restrictions. This was challenging for me as I didn’t really understand it on the first test, and I needed to work on making sure I fully complete the questions given by doing restrictions, as they are considered incomplete if not done.

This question wasn’t hard to solve, but I didn’t understand how to figure out its restriction. To me it seems so much easier just to solve the question for X than trying to figure out the value of X, but now I understand that restrictions are important for many reasons.

For this question, the X variable is under a square root sign. This means that x has to be greater than or equal to 0. This is because anything under a square root sign cannot be a negative as a negative times a negative is a positive. If this root was cubed instead of squared, it would be possible, but it’s not.

Now you’ve narrowed down your possible answers so it’s much easier when solving the equation.

I was making the mistake of just saying x is equal to all real numbers as I didn’t understand what a restriction was there for, but now I do and feel much more confident in my variable finding skills.