Quadratic eqs with squares already factored

[fossil fox]

1. x² = 16

2. t² = 81

3. r² - 100 = 0

4. x² - 144 = 0

5. 2s² = 50

6. 4x² - 225 = 0

7. 3h² - 147 = 0

8. ( x - 4 )² = 169

9. ( k - 7 )² = 289

10. ( 2s - 1 )² = 225

I get the one and two but the rest I don't know how to do

[peak field]

For 3, 4, the equations can be rewritten as ones similar to the equations from 1, 2, simply move the 100 and 144 to the other side

I will continue with the rest once you have completed 3 and 4, so let me know about your progress on those

[fossil fox]

Wait like this? r² = 100

What happens to the 0

[peak field]

Yup

Well, technically what you are doing is rewriting the equation as r² = 0 + 100, but obviously 0 + 100 is just 100, so you can immediately write r² = 100

[fossil fox]

Oh, we have to divide the 100 to itself right

And cancel out the r²

[peak field]

Perhaps you meant taking the square root instead of dividing? There's no division needed for this one

[fossil fox]

Yes

Like 10

For the 100

[peak field]

Yup

So you get r = +-10

What about the 4th one?

[fossil fox]

I suppose the same as 3. x² = 144

Then cancel out x²

Then take the sr

12

[peak field]

Yes, that's it

Now, for 5, the equation can still be rewriting as the equations from 1, 2, this time you just need to divide both sides by the coefficient of s²

[fossil fox]

Huh?

[peak field]

Which bit is confusing? The term "coefficient"?

[fossil fox]

I know the term, I just don't get the dividing both sides by the coefficient of s²

[peak field]

Here the coefficient of s² is 2, so that would mean dividing both sides of the equation by 2, which yields 2s²/2 = 50/2

[fossil fox]

But I can't divide 50 into two

Or is that 2/50

Because then it'll be 25

[peak field]

50 is even and 50/2 = 25

[fossil fox]

Oh

Do I still cancel out the 2s²/2

Or do I get 1

[peak field]

The 2's cancel and you are left with s²

[fossil fox]

Oohh, so s² +_ 25?

[peak field]

Afterwards s² = 25 is obtained, yeah

[fossil fox]

Oh I thought it was the final answer

[peak field]

Not yet, the final step is to now take the square root

[fossil fox]

s² = 25 ?

[peak field]

Yeah, so s = +-5 from that

[fossil fox]

Okay cool, let me try 6

4x² = 225

4/x² 4/225

Am I doing it right

@peak field

[peak field]

Okay so I assume by 4/x² you mean x² divided by 4 and for this conversation I will use the same notation, but keep in mind for future reference: people would write x²/4 instead

Anyway, you got 4x² = 225 correctly, but after that the equation should become 4/4x² = 4/225

[fossil fox]

Ohh I'm sorry

[peak field]

Nothing to be sorry about

[fossil fox]

Is 4/225 56?

[peak field]

No, 225 is not actually divisible by 4 without a remainder, so let's try leaving 4/225 as it is and see if the equation can still be solved regardless

[fossil fox]

So 4+_225

Ohh wait wait

That gives us two answers

x = 4 and x = 225

No?

[peak field]

No, so what we have so far is x² = 4/225

And as before we take the square root of both sides

[fossil fox]

15

[peak field]

So what's left is to determine what the square roots of 4/225 are

[fossil fox]

Then 2

[peak field]

Yes, so a nice property about square roots is that taking the square root of a/b is the same as dividing the square root of a by square root of b

For our equation this means we end up with x = +-2/15

[fossil fox]

We still get two answers no?

[peak field]

Two answers, yes, namely 2/15 and -2/15 rather than 4 and 15

[fossil fox]

I thought the x was supposed to stay

X = 17 then X = 13

[peak field]

I am not sure what you mean by that, are 17 and 13 coming from 15 + 2 and 15 - 2?

[fossil fox]

Yes, that's what we did in class

[peak field]

In that case, 2/15 means the result of dividing 15 by 2 rather than taking 15 and subtracting 2 from it

[fossil fox]

Huh

[peak field]

Then you are either misremembering or what was done in class was wrong, the solutions are just 2/15 and -2/15

[fossil fox]

Oh wait that's a different one

[peak field]

Well that's different because there's no division

[fossil fox]

We also did that

Ya there

[peak field]

Yeah the order of operations that they are using is different from what I mentioned, but it would still be able to solve the equation

Also, there is a pair of mistakes on the board, after taking square root of 9x² and 4x², they should have gotten 3x and 2x

Unless (9x)² and (4x)² were meant, in which case there are parenthesis missing

[fossil fox]

Hold on I'll just go eat then I'll come back

[peak field]

Alright, ping me when you come back

[fossil fox]

@peak field

@peak field

@peak field

@peak field

@peak field

[peak field]

I am here

[fossil fox]

@peak field

@peak field

Hii

[peak field]

Hi

[fossil fox]

Okay I finished eating

Let's continue

Wait let me write down the ones we did

[peak field]

Sure

[fossil fox]

Wait I just want to be sure I did 1. right, I got X = +- 4

Did I

[peak field]

Yes

And t = +-9 for 2.?

[fossil fox]

Yes

[peak field]

That’s also correct

[fossil fox]

Awesome, did I skip any steps

[peak field]

Not really, just writing that out immediately should be fine

[fossil fox]

By the way for number three

Why did you remove the exponent

Of r

[peak field]

Let me do a recap as well: So far we have solved the equations 1.-6., all of which can be solved by taking square roots, moving terms to the other side, dividing both sides by a coefficient

[peak field]

You mean the step from r² = 100 to r = +-10?

[fossil fox]

Yes

[peak field]

That’s what taking the square root does, it cancels out the ²

[fossil fox]

I keep forgetting

Ahhhhh

[peak field]

Alright, let me know when you are done with writing down so we can start 7.

[fossil fox]

Um

I got stuck at 5 again

Oh wait nevermind

Okay wait

Wait do I divide the 50 or get its square root

@peak field

[peak field]

50 was first divided by 2, which gets you 25, and then you take the square root of that

[fossil fox]

Does 50 come first or the 2

Like 50/2

Or 2/50

[peak field]

50 on the top and 2 on the bottom, but regarding how it’s texted: normally people write 50/2 for that

But we can stick to using 2/50 if you want just for this conversation

[fossil fox]

@peak field just a question

When do you use the +-

Is it whenever you get a two numbers that can't be divided into each other

[peak field]

When solving the equation of the form x² = a², the +- will always appear, i.e., the solutions of that are x = +-a

Whether the numbers on the right could be divided into each other isn’t relevant to that

[fossil fox]

So for numver three, why s = 25?

Is it not s = +- 25

[peak field]

It was s = +-5 rather than that, I am pretty sure I wrote it with +- before

Perhaps you are confusing s² = 25 and s = +-5?

[fossil fox]

You said s² = 25

And I thought the exponent is called no?

Cancelled*

[peak field]

The exponent on the left is cancelled when you take the square root, yes, and you get s = +-5 as a result

[fossil fox]

But I already divided the 50 to 2 how'd you get +-5

[peak field]

Dividing 50 by 2 was a step earlier

Alright, I will write how the equation is solved entirely

[fossil fox]

Wait wait let me do it

Let me write it down the way I did and tell me if it's correcr

So 2s² = 50 right

Cancel out the ²

[peak field]

No, first divide both sides by 2

[fossil fox]

Then divided 2s/2 = 50/2

Oh

[fossil fox]

Is this correct

[peak field]

If you add the ² to s, it becomes correct

[fossil fox]

Then cancel out the two 2s

[peak field]

Your first step is to just divide both sides by 2, so the exponent stays

[fossil fox]

But you have to cancel it out

[peak field]

Yes

You mean the exponent cancels out with the /2?

[fossil fox]

2s²

[peak field]

Yes, dividing that by 2 gives you just s²

[fossil fox]

You would get s = 25 yes?

Then get its square root

So the final answer would be s = +- 5

[peak field]

s² = 25, the exponent of s stays until you take the square root

[fossil fox]

I'll try to do the 7 so I won't take much of your time, however the last three though

That's where I don't know what to do now

[peak field]

I am not in a hurry anywhere though so you can take your time

Just for reference, here are the steps for 5.:
2s² = 50

Divide both sides by 2:
2s²/2 = 50/2

Simplify:
s² = 25

Take square roots:
s = +-5

[fossil fox]

Oh, I think I did it right, I did cancel when dividing the two 2s though

Leaving only the s

Okay let's move to 8

Wait I haven't wrote down the 6 yet, hold on

@peak field

[peak field]

Done with 6.?

[fossil fox]

I think I did it wrong

4x² = 225 right

[peak field]

Yes

[fossil fox]

Then 4x²/4 = 225/4

Cancel out 4

[peak field]

Yes

[fossil fox]

Leaves us with x

X²

[peak field]

squared

Yeah

[fossil fox]

What's for 225/4

[peak field]

It's fine to leave it as it is there

225 and 4 are perfect squares, so when you take the square root there will be no trouble

[fossil fox]

So the answer will be x² = 225/4?

[peak field]

No, you take the square root now

[fossil fox]

X = 225/4

@peak field

[peak field]

Square root of 225/4 is not 225/4

Here you can use the fact that 225 = 15² and 4 = 2², meaning you would have x = +-15/2 at the end

[fossil fox]

You said leave it as it is though

Oh wait I get it

[peak field]

I meant before taking square root, it's fine to not simplify 225/4 any further

[fossil fox]

Okay done

Can we do 8

Please

I don't get it

Because there's parentheses now

[peak field]

Alright, I assume you can handle 7, let's start 8

The given equation may look different from what we've been dealing with so far, but the logic stays the same: We can start by taking the square root since the left hand side is the square of something and on the right hand side 169 = 13²

[fossil fox]

Do we have to apply the foil method

[peak field]

And similar to before we get x - 4 = +-13

No, here that would make things more difficult

Simply take advantage of the fact that the left hand side is already written as a square

Andtake the square root

[fossil fox]

What do we do with the left side

Or do we transpose

[peak field]

So the exponent cancels out with ² as usual and we are left with x - 4 = +-13

All that's left to do is to just move 4 to the other side of the equation

[fossil fox]

Oh so x = +-13/4?

[peak field]

Looks like you tried to divide both sides by 4, which would rather rewrite the equation as (x - 4)/4 = +-13

By moving 4 to the other side I meant x = 4 +- 13

[fossil fox]

Wait so just cancel the exponent, remove the parentheses, and just move the number inside the parentheses on the right?

[peak field]

Yeah

[fossil fox]

Then the final answer would be x = 4 +- 13?

[peak field]

Yeah

[fossil fox]

OHHH

@peak field

I finished 8 and 9

But the 10

It comes with a number now

Not just the letter itself

[peak field]

It's fine, you can still use the same strategy, the only difference is that you will need to divide both sides by 2 at the end

Take square root, move the constant to the other side, divide both sides by 2, what do you get?

[fossil fox]

112

And 1

[peak field]

Can you show your steps?

[fossil fox]

2x - 1 = 225 (cancel the exponent)

Then 2x = 1 +- 225

[peak field]

You forgot to take the square root of 225

[fossil fox]

Wait huh

[peak field]

There's no such step as just cancelling the exponent, cancelling the exponent is the result of taking the square root of both sides

[fossil fox]

How come I cancelled the exponent for 8 and 9

[peak field]

Back then you would also be taking square root of the number on the right

And similarly here you need to take the square root of 225

15² = 225, so you get 2x - 1 = +-15

[fossil fox]

Wait I didn't need to cancel the square on 8 and 9?

[peak field]

No, it was needed; Taking the square root resulted in the exponent cancelling on the left and it also affected the right hand side

[fossil fox]

I got k = 7 +- 17 for 9

Tell me if that's correct

[peak field]

Yes

[fossil fox]

Okay great

[fossil fox]

Could you please write the step by step on how to do the 10.

[peak field]

(2x - 1)² = 225

Take square roots:
2x - 1 = +-15

Move the constant to the other side:
2x = 1 +- 15

Divide both sides by 2:
2x/2 = (1 +- 15)/2

Cancel out:
x = (1 +- 15)/2

That's the solution, but if you want you could also write it as the solutions being x = 8 and x = -7

[fossil fox]

2s?

[peak field]

Forgot to mention that step, hold on

[fossil fox]

Wait it's 2x not 2s

[peak field]

Fixed

Okay I will rename that to x but the name of the variables doesn't matter much

[fossil fox]

I found a problem at number 7

I don't know the squared for 3h

²

[peak field]

Simply do the same thing as you did in 6: Move 147 to the other side and divide both sides by 3, this time 147 is divisible by 3 and 147/3 = 49

[fossil fox]

So instead of getting the squared we do division?

[peak field]

Yeah

[fossil fox]

Okay my final answer is h = 49

[peak field]

Should be h² = 49

What I meant is that you get right of 3 using division, but the exponent of h stays

3h² = 147
3h²/3 = 147/3
h² = 49
h = +-7

[fossil fox]

Huh?

Which is it h² = 49 or h = +-7

[peak field]

The final answer is h = +-7

So when the problem says "solve for x", they want you to end up with an equation of the form "x = ..." where ... is some number that doesn't contain x in it

[fossil fox]

Is the final answer for 10 1+-15/2

[peak field]

Yeah

Another thing about notation in text messages: When reading 1 +- 15/2, someone could interpret it as 1 +- (15/2) rather than (1 +- 15)/2, which is what you meant, so make sure to put parenthesis in the future

[fossil fox]

Oh thank you so much for your help, it's late and I still have to do two more assignments from my other subjects, I'm gonna go finish them now so I can sleep, only my math assignment gave me trouble

Again thank you

Much love ❤

[peak field]

You're welcome

[fossil fox]

Bless

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Quadratic eqs with squares already factored

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