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 2.

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:

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!!

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:

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.

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.

These are the symbols for each type from my notes.

Real Numbers – R (1, 2, 3,…)

Natural – N (0, 1, 2, 3,…)

Whole – W

Rational Q – (Quotient) (1/2,  3.23)

Irrational – (π or √2)