Week 6 – Completing the square

Posted on October 23, 2025 by Mia

This week I worked with a new method, completing the square.

This was the equation.

Since there’s an equal sign, I know that I not only factor this if possible but I solve for both x’s.

Since there is a leading coefficient you have to divide by 2, which I did not. The x² must be by itself, or else the method doesn’t work.

When dividing by 2, this leaves me with:

Now we chop the first two terms in half, which is:

Then we add the zero pair. Since it adds to zero it doesn’t change the original equation, but it does make it factorable.

We get the zero pair by raising the 2 to the 2nd power.

2² = 4

So we know we need to add 4 to make the first part factorable. Now, we add the -4.

Then add the like terms to get:

We’re finished factoring. Now we can solve this.

First we move the constant (term without a variable) to the right side.

Now we square both sides.

We move 2 to the right side to isolate x,

to get the final answer of:

Week 5 – Factoring – Grouping method

Posted on October 22, 2025 by Mia

This week I made a mistake with this question:

I remembered factoring 1-2-3, but when its three terms you cant always jump to product and sum, like I did. There needs to be a coefficient of 1 in front of the x².

I thought we had to find two numbers that multiply to 3 and add to 11. Since I couldn’t find an answer, I asked classmates and found out you actually multiply the 6x² with the 3, to get 18. (This is one method, box method also works)

Find factors of 18 that add to 11, like 9 and

Now we put this back, so we are able to factor.

Group and factor:

Important: Both brackets must look the same so we can take one out.

Final answer:

Week Factoring – Factoring Perfect Square

Posted on October 15, 2025 by Mia

My best mistake was when I thought I could factor out -1 to make a perfect square. What I didn’t see is that I can’t do that because 9x is not a perfect square. 9 is, but x isn’t so it can’t be considered perfect.

This was my incorrect work.

Here is the correct way:

What I did first was to move the 4×2 to the left, I did this because I prefer it to be in that order.

Then I factored out x and -1 because I don’t want the coefficient in front of x2 to be negative.

After this I have x=0, and x= 9/4 which are correct.

Remember that everything needs to be a perfect square and subtraction to factor using perfect square method!!

PC Week 4 – Simplifying Radicals

Posted on October 7, 2025 by Mia

This week I worked with this problem.

I multiplied the conjugate:

When you multiply two exact same square radicals, for instance root 12 here, it just becomes 12. My mistake was keeping the radical.

This led me to a mistake, because I thought there was an exponent to be solved.

I got the incorrect answer of:

The right way was:

This leaves

Which leaves me with:

and to simplify further, I got the final answer of:

PC 11- Dividing Radicals

Posted on October 6, 2025 by Mia

This week I worked with dividing radicals. I struggled most with this question.

What I did was divide both by four.

After that I simplified both radicals, and then I divided by 6,

and I kept getting this answer:

However this is the wrong answer, this is how you do it.

First, I realized that the cubed root of 216 is 6.

From there, I multiplied the coefficient 2 by 6.

Which gave me the denominator of 12. After that I divided both the top and bottom by 3.

This cancels out 3 and leaves me with a denominator of 4.

Then I simplified the cube root of 32.

After this I divided the top and bottom by 2.

Leaving me with the final, correct answer:

Week 3 PC 11 – Positive/Negative exponents

Posted on September 18, 2025 by Mia

This week I was stumped with this question.

It looks a lot more complicated than it is, once I understood how negative exponents work it was a lot simpler. (Having to reciprocate)

Here is what I did:

First I switched the x’s, because they both need to reciprocate in order to become positive. I moved the y up as well, next to the other.

From there, I worked with the exponents laws. Division means subtraction, so I subtracted the x’s from each other, leaving x²

Then I multiplies the y’s, which means addition in terms of the exponent. This leaves y³

Now I had to deal with the negative exponent outside the bracket. Since I needed to distribute it to each term, I thought to put everything under 1, so the exponent becomes positive. This makes it easier for me to distribute.

Once that was done, I distributed the three to each term. Remember that the 2 has an invisible exponent of 1, so distributing to that as well.

Once that was done, all I had to do was evaluate 2³

That left me with 8. So, I put it all together,

and this was my final answer.

Thank you for reading.

PC 11 – Sem 1

Posted on September 5, 2025 by Mia

This week I worked with number types, and this worked as a refresher to me since I didn’t remember all of them. What helped me was this diagram, that helped me to visually sort them out.