Here’s how I do this with my students. I tell them I’m thinking of a number and when you multiply it by itself, you get 64. Most will say it’s 8, but then I tell them it’s -8. That lets them know you have no way of knowing which unique number I started with.
Dear Sir.... Please tell me which model of Wacom tablet and pen are you currently using for this nice looking lecture..... I want to know as i am interested in making lectures in my local dialect as i am inspired by your initiative.....
Agree with other comment. Any Wacom tablet will do the job. Choosing one depends on your budget. The key to make high quality "Sal Style" videos is setting up the canvas to the same resolution as the final video. For example, this one was made at 1280 x 720 pixels. If canvas and final video resolution match it will render beautiful. Hope this helps. :)
@@lafudge2929 No, it can't, but I understand your confusion. There are two possible answers to an equation like 81=x^2 in that case we think of what your thinking of the positive and the negative square root of 81. So our answer is -sqrt(81) and +sqrt(81) which would be -9 and +9. However, the sqrt(9) will always be a positive number because "sqrt" is an implicit "positive sqrt". Also because y=sqrt(x) is a function and by definition 1-to-1. So that means that the same input value cannot have two different output values, aka needs to pass the vertical line test. So it can't be true and isn't true that the sqrt(x)=y has both (9,3) and (9,-3) as solutions. It only has one and it is (9,3). However, if you plotted -sqrt(x)=y then you would have (9,-3) as a solution.
that's exactly what I was thinking...I don't have a problem understanding the concept of extraneous solutions, but I fail to wrap my head around the deeper issue; You do everything algebraically correct and yet arrive to a wrong solution....it also got me thinking whether this means algebra itself is flawed. Did you do any further research and if so, what did you find?
@@simeonstan5843 It's not flawed. You're just not including the logic behind it. The moment you squared both sides you included a solution that doesn't support the initial equation. So, the operation is not proper. Though you can easily eliminate that solution by checking if it supports the initial equation or not.
@@RandomDudeOnTheInternet-lm4bn But WHY doesn't it support the initial equation? His explanation doesn't make sense. Squaring a value IS reversible operation. You simply square the value lmao. Anyways, I guess I'll be tripping out over this phenomenon for the rest of the day.
7:34 Bro you can't write it as an equation: 0×3=0×4 Because when 0 of L.H.S goes to R.H.S it is just like this: 3=0×4/0 Which undefined because 0 cannot be in denominator.
multiplication of an equation by a variable is also not a reversible operation. That is, when the range of the variable includes zero. xa=xb is true for x = 0. It is not true from xa=xb that a=b. For example 3*0=4*0 and 3 is not equal to 4.
Here’s how I do this with my students. I tell them I’m thinking of a number and when you multiply it by itself, you get 64. Most will say it’s 8, but then I tell them it’s -8. That lets them know you have no way of knowing which unique number I started with.
Beautiful. Thanks Sal!
Dear Sir.... Please tell me which model of Wacom tablet and pen are you currently using for this nice looking lecture..... I want to know as i am interested in making lectures in my local dialect as i am inspired by your initiative.....
Any Wacom tablet should work fine.
Agree with other comment. Any Wacom tablet will do the job. Choosing one depends on your budget.
The key to make high quality "Sal Style" videos is setting up the canvas to the same resolution as the final video. For example, this one was made at 1280 x 720 pixels. If canvas and final video resolution match it will render beautiful. Hope this helps. :)
Benito Estrada thanks
4:50 To get to the point
Sal! Do you have any on restrictions on radicands??
I thought that the square route of 9 can equal 3 and -3, since -3 ×-3 = 9, so wouldn't -3 = square route of 9 be a correct answer?
Archie Brew lol, I’m lost too
@@amypalacios6314 The square root of a number CAN have a negative solution. I don't know what you're on about...
@@lafudge2929 No, it can't, but I understand your confusion. There are two possible answers to an equation like 81=x^2 in that case we think of what your thinking of the positive and the negative square root of 81. So our answer is -sqrt(81) and +sqrt(81) which would be -9 and +9. However, the sqrt(9) will always be a positive number because "sqrt" is an implicit "positive sqrt". Also because y=sqrt(x) is a function and by definition 1-to-1. So that means that the same input value cannot have two different output values, aka needs to pass the vertical line test. So it can't be true and isn't true that the sqrt(x)=y has both (9,3) and (9,-3) as solutions. It only has one and it is (9,3). However, if you plotted -sqrt(x)=y then you would have (9,-3) as a solution.
@@amyp7067 That makes sense.
The square root is defined to output only positive numbers
What if both are extraneous
😂
doesn't this mean algebra itself is flawed??
that's exactly what I was thinking...I don't have a problem understanding the concept of extraneous solutions, but I fail to wrap my head around the deeper issue; You do everything algebraically correct and yet arrive to a wrong solution....it also got me thinking whether this means algebra itself is flawed.
Did you do any further research and if so, what did you find?
@@simeonstan5843
It's not flawed. You're just not including the logic behind it. The moment you squared both sides you included a solution that doesn't support the initial equation. So, the operation is not proper. Though you can easily eliminate that solution by checking if it supports the initial equation or not.
@@RandomDudeOnTheInternet-lm4bn But WHY doesn't it support the initial equation? His explanation doesn't make sense. Squaring a value IS reversible operation. You simply square the value lmao. Anyways, I guess I'll be tripping out over this phenomenon for the rest of the day.
What happens if there is no solution? How do we call it?
tysm literally needed this rn
thank you :) !
What do I do if I can’t factor out a (x+1)?
Why do I have to do this 😭 it makes no sense
7:34 Bro you can't write it as an equation:
0×3=0×4
Because when 0 of L.H.S goes to R.H.S it is just like this:
3=0×4/0
Which undefined because 0 cannot be in denominator.
multiplication of an equation by a variable is also not a reversible operation. That is, when the range of the variable includes zero. xa=xb is true for x = 0. It is not true from xa=xb that a=b. For example 3*0=4*0 and 3 is not equal to 4.
this should be taught in highschool maths :/
sogood