In this lecture we looked at “completing the square” algebraically, which helps us to convert our equations from standard form into vertex form.
For your challenge, convert the following equation into vertex, or squared, form;
y = -1/3x^2 + 10/3x - 16/3
Post your answer below and remember to use spoiler tags.
Did it a bit different than in the lecture, instead of multiplying both sides I factored the 1/3 out.
I began with y = (-x² + 10x - 16) / 3
3y = -x² + 10x - 16
-3y = x² -10x + 16
-3y = (x - 5)² - 25 + 16
-3y = (x - 5)² - 9
y = -(1/3)(x-5)² + 3
y = -1/3(x-5)^2 + 3
y = -1/3(x - 5) + 3
Vertex at (5,3)
Oh, right-- it should be -1/3(x - 5)^2
This is giving me flashbacks to math class, lol. Thank god there’s no grades in this course.
Don’t worry @Ajai_Raj. You’re doing great!
Making mistakes is how we learnt and the important thing is that you’re putting in the effort to improve.
Everything will come together with time and practice 
Here goes.
y = -1/3x^2 + 10/3x - 16/3
-3y = x^2-10x+16
-3y = (x-5)^2-25+16
-3y = (x-5)^2-9
y = -1/3(x-5)^2 +3
My biggest problem (and nearly all the time) my pluses and minuses are wrong.
sqrt(2) = 4 = 2 x 2
I meant 2^2 = 4 = 2 x 2
sqrt(-2) = -4, but -2 x -2 = 4
I meant -2^2 = -4, but -2 x -2 = 4 (on my calculator)
so sqrt(-2), should be -1 x sqrt(2) as in -1 x 4 = -4
I meant -2^2, is handled as -1 x 2^2 as in -1 x 4 = -4 ( or in my calulator using -2^2 )
Maybe it is the math semantics … but this makes me mismatching + and - signs …
@FedPete,
I’m a little confused, but I’m assuming that you’re talking about squaring numbers rather than taking the square root of numbers, correct?
If you square 2 you get:
2^2 = 2 x 2 = 4
If you square -2, you also get 4, since the two negatives give a positive result when multiplied, so:
-2^2 = -2 x -2 = 4
Due to this trait of multiplication, there’s no way to square a number and obtain a negative result - It will always be positive.
If you need a negative result then you’d need to do -(2^2) rather than -2^2. Since you’d be calculating the 2^2 first, and then multiplying by -1 to flip the sign.
Now, if you’re asking about square roots, then you have to be very careful when dealing with the square root of negative numbers because the answer isn’t a real number.
I hope that helps, but if I’ve misunderstood your question, please let me know.
Yes, I did confuse you!
I meant #^2, not sqrt(#) !!
Now I’m becoming confused again (back and forth …) 
When I enter this in my mobile phone (android) calculator. Exactly in the way you described, entering keystrokes -2^2=, it returns -4 (minus four). 
But (-2)^2= returns 4. So semantics is an issue here …
Thank you for your understanding.
@FedPete, I just tried it on my phone and you’re absolutely right.
It turns out that the android calculator is a bit rubbish!
If you type in “minus” 2 ^ 2, then calculators will read this as; 0 - 2^2 = 0 - 4 = -4
So the inclusion of that rogue zero at the front is causing a big problem.
Ideally, you need a calculator that has a +/- button, which let’s you flip the sign on your number.
Failing that, you can get around it by just being very careful with your inputs.
In this case, you want to calculate 0 - 2 first and then square the result.
So you can enter your calculation as; 0 - 2 = ^ 2
This will give you:
0 - 2 = -2
-2^2 = 4
If you don’t have an old scientific calculator kicking about in a drawer somewhere, I’d recommend installing one of the many free scientific calculator apps from the Google Play store.
This will make your life a lot easier and will come in handy as you get further into the course.
My scientific Casio calculator of 35 years ago (Yes I still own this one), does this math correctly.
THX!
y=−(1/3) x^2+(10/3)x −(16/3) −Convert to vertex form
−3y=x^2−10x+16
−3y=(x-5)^2−25+16
−3y=(x-5)^2−9
y=−(1/3) (x−5)^2+3
y = -1/3 (x - 5)² + 3
Vertex at (5 , 3)
I am getting the correct results now but it would be helpful if individual steps are not skipped. In this case the signs were throwing me off.
At 4:28 in the video Gary condenses down (-6/2)^2 to -9. At first glance this does not make sense. To explain this away it is my understanding that we are always subtracting that last (b/2)^2 instead of using its square result as the sign.
Slightly confusing as well is that (x + b/2) section preserves the sign of the original b value.
y = (x + b/2)^2 - (b/2)^2
Thank you for your feedback @Jeff_Simon.
Here’s a deeper breakdown of the calculation that will hopefully help to clarify things for you.
Using the equation from the lecture, we have:
y = 2x^2 + 12x + 6
First we try to simplify things to get the x^2 on its own, so in this case we divide everything by 2.
y/2 = x^2 + 6x + 3
Now we can complete the square using the template (x + (b/2))^2 - ((b/2)^2), which gives us:
y/2 = ( x + (6/2) )^2 - ((6/2)^2) + 3
Simplifying the fractions gives us:
y/2 = (x + 3)^2 - (3^2) + 3
Then we convert the -(3^2):
Note: This amount was previously added to the square, so I’ve added an extra set of parentheses here to help make that clear.
y/2 = (x + 3)^2 - 9 + 3
Then we can tidy things up:
y/2 = (x + 3)^2 - 6
And finally multiply everything back by 2 to isolate y:
y = 2(x + 3)^2 - 12
I hope that helps explain things for you but if you have any other questions, please ask.
Thank you for the detail breakdown Gary. Yes those extra parenthesis make things more clear now, also the fact that this amount will be removed since it was previously added to make the square. I had to return to lesson 40 at the 03:10 mark where you show the formula to refresh.
