I wish I had you as a teacher back in high school. Everyone had a rough time in our higher math classes because the teacher was a bit out of touch. He had been teaching so long that it seemed he forgot how to explain things.
Very important at the start: if the coefficient of x^2 is not 1, you have to divide everything by that coefficient first. This one has the coefficient as 1, but that has to be checked first!
I didn't learn completing the square in algebra. Either figure out how to factor it or use the quadratic formula. I went my entire college career majoring in math without completing the square. It wasn't until recently when I saw a visual proof of the quadratic formula that I finally understood what was going on.
Of course, in the distant past when completing the square _was_ always geometric, the negative square root part did not arise because areas and so on were always positive. Right up until the 17th century, negative square roots were referred to as "false" (and other epithets).
Solution: To complete the square, when you have an equation of the type "x² + bx ± c", you add AND subtract (b/2)² = b²/4, as the square of (x + b/2)² = x² + 2 * b/2 * x + b²/4 = x² + bx + b²/4 For the whole term x² + bx ± c, you end up with: x² + bx ± c = x² + 2 * b/2 * x + b²/4 - b²/4 ± c = (x + b/2)² - b²/4 ± c So in the given equation x² + 8x - 3, you add and subtract (8/2)² = 4² = 16 x² + 8x + 16 - 16 - 3 = 0 (x + 4)² - 19 = 0 |+19 (x + 4)² = 19 |√ |x + 4| = √19 x + 4 = ±√19 x + 4 = ±√19 |-4 x = -4 ± √19
@@bprpmathbasicsthank you! I made the connection when you showed the geometry. It’s fun when logical connections snap into place in the brain almost autonomously.
I mean if you know that 7^x-7^(x-1)=42 is the correct answer than it isn't wrong. Remember that you can also represent it as: (7-1) * 7^(x-1) = 42 6 * 7^(x-1) = 42 /6 7^(x-1) = 7 7^(x-1) = 7^1 x-1 = 1 +1 x = 2 or use logarithms after you establish that a^b - a^(b-1) = (a-1)(a^(b-1)). IDK what your teacher or someone else wants. And of course this is all assuming that you were told that 7^x-7^(x-1)=42.
@@jamesharmon4994 Given: x^2 + 3*x + 20 = 0 Start with the generalized vertex form of a monic parabola. Note that since the coefficient on x^2 is 1, this means there will be no multiplier in front of the parenthetical term: y = (x + h)^2 + k Expand: y = x^2 + 2*x*h + h^2 + k Match like coefficients: 2*h = 3 h^2 + k = 20 Solve for h & k: h = 3/2 k = 20 - 9/4 = 71/4 Thus, the vertex form is: (x + 3/2)^2 + 71/4 = 0 Now you can rearrange to make x the subject: (x + 3/2)^2 = -71/4 x + 3/2 = sqrt(-71)/2 x = -3/2 +/- sqrt(-71)/2 By definition, i = sqrt(-1). This means we can separate sqrt(-71) as sqrt(-1)*sqrt(71), which means it is i*sqrt(71). Thus: x = -3/2 +/- i*sqrt(71)/2
I challenge you to find and explain all the solutions to x^6 = 1, without saying "x" and without substituting x with another variable. If you mess up, you must start over 😁
Neither does 0,-1 a^x=a^y. But in case of 0 a^x-y=1. You are basically x-y=0. Dividing by 0 x=y In case of 1, it just doesn't because all powers of are equal, it doesn't work for -1 raised to even power, or odd powers. Like - (-1)²=(-1)⁴ - (-1)³=(-1)⁵
I wish I had you as a teacher back in high school. Everyone had a rough time in our higher math classes because the teacher was a bit out of touch. He had been teaching so long that it seemed he forgot how to explain things.
every math teacher I've had just reads off the book, they are getting paid to be text to speech bots and get mad whenever someone asks questions
Very important at the start: if the coefficient of x^2 is not 1, you have to divide everything by that coefficient first. This one has the coefficient as 1, but that has to be checked first!
4:02 ok finally!
great video as always
I didn't learn completing the square in algebra. Either figure out how to factor it or use the quadratic formula. I went my entire college career majoring in math without completing the square. It wasn't until recently when I saw a visual proof of the quadratic formula that I finally understood what was going on.
Of course, in the distant past when completing the square _was_ always geometric, the negative square root part did not arise because areas and so on were always positive. Right up until the 17th century, negative square roots were referred to as "false" (and other epithets).
I have a quiz on this today lol what a coincidence
Solution:
To complete the square, when you have an equation of the type "x² + bx ± c",
you add AND subtract (b/2)² = b²/4, as the square of (x + b/2)² = x² + 2 * b/2 * x + b²/4 = x² + bx + b²/4
For the whole term x² + bx ± c, you end up with:
x² + bx ± c
= x² + 2 * b/2 * x + b²/4 - b²/4 ± c
= (x + b/2)² - b²/4 ± c
So in the given equation x² + 8x - 3, you add and subtract (8/2)² = 4² = 16
x² + 8x + 16 - 16 - 3 = 0
(x + 4)² - 19 = 0 |+19
(x + 4)² = 19 |√
|x + 4| = √19
x + 4 = ±√19
x + 4 = ±√19 |-4
x = -4 ± √19
Is this a geometric method to derive the quadratic formula? Curious.
Yes, here’s the proof th-cam.com/video/AD58TWGIcsQ/w-d-xo.htmlsi=jN71IKiyQlNy8edk
@@bprpmathbasicsthank you! I made the connection when you showed the geometry. It’s fun when logical connections snap into place in the brain almost autonomously.
I wish that you had taught diff eq’s back in the middle of the last century !
I did. I made tons of videos back in 2017 on my main channel blackpenredpen
Can you please explain this (why isn't the right way of solving):
7^x-7^(x-1)=42
7^x-7^(x-1)=49-7
7^x-7^(x-1)=7²-7¹
7^x-7^(x-1)=7²-7²‐¹
x=2
Thanks.
I mean if you know that 7^x-7^(x-1)=42 is the correct answer than it isn't wrong. Remember that you can also represent it as:
(7-1) * 7^(x-1) = 42
6 * 7^(x-1) = 42 /6
7^(x-1) = 7
7^(x-1) = 7^1
x-1 = 1 +1
x = 2
or use logarithms after you establish that a^b - a^(b-1) = (a-1)(a^(b-1)).
IDK what your teacher or someone else wants. And of course this is all assuming that you were told that 7^x-7^(x-1)=42.
How about an example where, if you use the quadratic formula, the discriminate is negative? Like:
x^2 + 3x + 20 = 0
Ok!
@bprpmathbasics I suggest this because so MANY videos abandon looking for x if the discriminate is negative.
@@jamesharmon4994 Given: x^2 + 3*x + 20 = 0
Start with the generalized vertex form of a monic parabola. Note that since the coefficient on x^2 is 1, this means there will be no multiplier in front of the parenthetical term:
y = (x + h)^2 + k
Expand:
y = x^2 + 2*x*h + h^2 + k
Match like coefficients:
2*h = 3
h^2 + k = 20
Solve for h & k:
h = 3/2
k = 20 - 9/4 = 71/4
Thus, the vertex form is:
(x + 3/2)^2 + 71/4 = 0
Now you can rearrange to make x the subject:
(x + 3/2)^2 = -71/4
x + 3/2 = sqrt(-71)/2
x = -3/2 +/- sqrt(-71)/2
By definition, i = sqrt(-1). This means we can separate sqrt(-71) as sqrt(-1)*sqrt(71), which means it is i*sqrt(71). Thus:
x = -3/2 +/- i*sqrt(71)/2
Why doesnt this work if the coefficient of x² is a? And thanks.
I challenge you to find and explain all the solutions to x^6 = 1, without saying "x" and without substituting x with another variable. If you mess up, you must start over 😁
Remember me 😅,
take my doubt:
1^4 = 1^5, bases are same then by comparing powers we can say,
4 = 5, or
2+2 = 5 How?
Doesn't work when base 1
Neither does 0,-1
a^x=a^y. But in case of 0
a^x-y=1. You are basically
x-y=0. Dividing by 0
x=y
In case of 1, it just doesn't because all powers of are equal, it doesn't work for -1 raised to even power, or odd powers.
Like - (-1)²=(-1)⁴
- (-1)³=(-1)⁵
you complete me😅
Bad timing with the T-shirt, the text is hard to read, some people may get confused just because of that hahahaha Great video anyway
But the cat is crying tho 😆
x = -4 +/- ✓21
” (x+y) squared = x squared + y squared ” That was on your T-shirt 👕 😢
Bottom text
'my students did this again...'
Hence why the cat is crying.