Yes. I’m sad to say that I noticed that the trick was to find two numbers both summed up to “b” and produced “c” in college Algebra 50 years ago but, math putz that I am, had no idea how to go on from there. I also noticed that the hypotenuse of a right triangle was the radius of a circle; same story. Failed both classes: end of math studies.
The quadratic equation will turn up to be x² + 4x - 21 = 0 and the best way to solve is factorising: (x + 7) . (x - 3) = 0 So x = - 7 or x = 3 But there are other ways: 2. Symmetry of the parabola: (x+m) . (x+n) = 0 with m + n = 4 and m . n = - 21 (m+n) / 2 = 2 (2 - d) . ( 2 + d) = m . n = -21 so 2² - d² = -21 or d² = 25 so d = 5 m = 2 - d = 2 - 5 = - 3 and n = 2 + d = 2 + 5 = 7 so (x - 3) . (x +7) = 0 x = - 7 or x = 3 3. Using abc-formula: a = 1 b = 4 c = -21 D = b² -4ac = (4)² - 4 . (1) . (-21) = 16 + 84 = 100 (positive so 2 solutions for y = f(x) = 0 ) xtop = -b / 2a = -4 / 2 . (1) = -2 delta = V(D) / 2a = V(100) / 2 . (1) = 5 x1,2 = xtop -/+ delta = -2 -/+ 5 so x = - 7 or x = 3 4. Completing the square no way !
Perhaps I over-complicated things a little. First I simplified and moved over the 3 (X+2)^2 = 22 + 3 Then I went with Ln dividing the result by the exponent which would give me the value of Ln(x+2) Ln25 / 2 = Ln(X+2) Then I reversed the Ln function by using the value of Ln(X+2) as an exponent for e. e^1,609438 = 5 rounded up the 6th decimal to make it more readable, and finally subtracting 2. X = 3 I think this one works(?)
Yeah, right on man. Looked down but didn't see this. My Comment above starts here then adds a bit on this kind of subject, though you might not share my concern past the technicality. Or you may.
@ No matter what, the most important point is that we all are having a good time, getting along with each other nicely and doing Mr John’s math problems! (What else could we wish for happiness, especially when I am an old grandmother?)
It’s a quadratic. I foiled the left side & then subtracted 22 from each side. This gives you x^2+4x-21. If you factor this you get (x+7)(x-3)=0.this gives you x=-7 & x=3.
I think it's time to discuss the Po-Shen Loh method which will factorise all quadric equations. It's interesting that the knowledge in this "new" method goes back 4000 years and was forgotten for 300 years.
The best way to solve this… hmm, let’s see what happens. (x + 2)(x + 2) - 3 = 22 I can quickly tell this is a quadratic equation from the two x terms, but it’s not in a useful form yet. Let’s multiply it out: x^2 + 4x + 4 - 3 = 22 x^2 + 4x + 1 = 22 Next, we need to have this be set to 0, so let’s do that: x^2 + 4x - 21 = 0 Now, we could use the Quadratic Formula, but let’s think for a moment. 21 = 3 * 7 and 4 = 7 - 3 We can factor this! (x + 7)(x - 3) = 0 We press on with our two possibilities: x + 7 = 0 OR x - 3 = 0 x = -7 OR x = 3 We should check these two answers to make sure they work, but I leave that as an exercise for the reader. Rest assured, both are fine. Also, some may want to complete the square on this at some point, but my mind never really latched onto that particular process for some reason. On further review: Oh, now I see a much simpler way to think of this. I think this is along the lines of completing the square, then factor: (x + 2)(x + 2) - 3 = 22 (x + 2)^2 - 3 = 22 (x + 2)^2 - 25 = 0 (x + 2)^2 - 5^2 = 0 Difference of two squares: [(x + 2) + 5][(x + 2) - 5] = 0 (x + 7)(x - 3) = 0 And from here, the answers are clear. Questions?
No. The √25 is ONLY +5. Like it or not, stupidity or not. It is ONLY ONLY +5. The proper tricknique (because it gets you stupidly around the stupidity) is the following: (x + 2)^2 - 25 = 0 (x + 2)^2 - 5^2 = 0 {(x + 2) + 5} * {(x + 2) - 5} = 0 Then solving for each factor: x + 2 + 5 = 0 x + 7 = 0 x = -7 and x + 2 - 5 = 0 x - 3 = 0 x = +3 Not adequately teaching students the path in the video and then insisting they get more restrictive with later " 9 - √(x + 3) = 7 " type problems is part of why they get those wrong. (Don't bother me about whatever the shown expression leads to, it's just to evoke the kind of problem, not be a problem in which that ends up mattering. I wrote it on the fly, not over three careful years writing a textbook.) Part of why. Not learning the right idea is another part, but "teaching them wrong" to begin with is more than a wee little bit of it (thanks to early Steve Martin for that). Time to revisit the old shiboleth of "A rectangle has a length three units longer than its width and an area of 42. What are the dimensions?" and planning to totally toss out an answer just because the lengths will be negative numbers. AS IF negative can't just indicate direction on a Cartesian coordinate graph, but must be some intrinsicly good or evil characteristic with only the religion's one meaning of "negative means of a man and a woman only" kind of thing. If the desired answer is 5 by 2, then -2 by -5 is just as real-world-valid because it just means that instead of the rectangle going over 5 units to the right and 2 units up, it goes two units to the left and 5 units down. Gosh, it can't be -2 by anything because nothing in the real world goes left rather than right! And then you want them to have internalized that to the point of absoluteness and somehow instantly shift gears and grasp angles in the Cartesian coodinate graph being "blah-blah plus two pi" and that there are infinitely many of them and so on. Golly gee kidees, you mean that don't compute? I wonder why? First do no harm. Or leave the teaching to doctors who did swear to that principle.
It’s time to discuss how utterly useless this is in the REAL world where people work and earn a living for themselves and their families. This is useful to exactly NOBODY except high school math teachers.
Oh god! You've done it AGAIN!!! You are correct that we need both square roots of 25 here. SO YOU MUST WRITE ±√25 !!!! Without ± in front if it, then √25 equals 5 and only 5. You yourself have made plenty of videos explaining this, and yet over and over again you completely fail to follow your own advice! How do you expect your students to learn from you if you are so hopelessly inconsistent yourself???!!!
Therefore COULD BE MORE THAN TWO SOLUTIONS:: X+2 = square roort of +25 = +5 A POSITIVE INTEGER. OR X+2 = - SQ.ROOT OF 25 (NEGATIVE NUMBER0 = --25 A NEGATIVE INTEGER. THERE ARE TWO MORE P0SSIBILIIES WHICH ARE THE SAME AS THE ABOVE TWO. THE FISAT EQUATION TO OTHER THAN SUM IS NOT EQUAL TO A PERFECT SQUARE MINUS 2 So we mat have two more possibilities involving a complex variables or one that is an irrational number -irrational numbers or a zero. NOW TEST IS THE TOME TO TEST Test each possibility id they make sense or not. Yes, no? Theoreticall is that a this the war to approach it?
Yes. I’m sad to say that I noticed that the trick was to find two numbers both summed up to “b” and produced “c” in college Algebra 50 years ago but, math putz that I am, had no idea how to go on from there. I also noticed that the hypotenuse of a right triangle was the radius of a circle; same story. Failed both classes: end of math studies.
Thanks for the tutorial...I solved the problem by using the difference of two squares. Not sure it was correct method, but got the correct solutions.
Best way? y=x+2; y^2=25; y=±5; x=y-2=3 or -7.
The quadratic equation will turn up to be x² + 4x - 21 = 0 and the best way to solve is factorising: (x + 7) . (x - 3) = 0
So x = - 7 or x = 3
But there are other ways:
2. Symmetry of the parabola: (x+m) . (x+n) = 0 with m + n = 4 and m . n = - 21
(m+n) / 2 = 2
(2 - d) . ( 2 + d) = m . n = -21 so 2² - d² = -21 or d² = 25 so d = 5
m = 2 - d = 2 - 5 = - 3 and n = 2 + d = 2 + 5 = 7
so (x - 3) . (x +7) = 0 x = - 7 or x = 3
3. Using abc-formula: a = 1 b = 4 c = -21
D = b² -4ac = (4)² - 4 . (1) . (-21) = 16 + 84 = 100 (positive so 2 solutions for y = f(x) = 0 )
xtop = -b / 2a = -4 / 2 . (1) = -2
delta = V(D) / 2a = V(100) / 2 . (1) = 5
x1,2 = xtop -/+ delta = -2 -/+ 5 so x = - 7 or x = 3
4. Completing the square no way !
(x + 2)(x + 2) - 3 = 22
(x + 2)(x + 2) = 22 + 3
(x + 2)² = 25
√ (x + 2)² = ±√25
x + 2 = 5 ∨ x + 2 = -5
x = 3 ∨ x = -7
The roots of the Equation are {3, -7}✅
Best and most straightforward way of solving the equation.
(X+2)(x+2)=22+3==25, x+2=5, x=3.
Perhaps I over-complicated things a little.
First I simplified and moved over the 3
(X+2)^2 = 22 + 3
Then I went with Ln dividing the result by the exponent which would give me the value of Ln(x+2)
Ln25 / 2 = Ln(X+2)
Then I reversed the Ln function by using the value of Ln(X+2) as an exponent for e.
e^1,609438 = 5
rounded up the 6th decimal to make it more readable, and finally subtracting 2.
X = 3
I think this one works(?)
(x+2)(x+2)=(x+2)sqt=25, x+2=+/-5. x(1)=3, x(2)=-7.
Answer: X = -7 or X=3
----------
(X+2)(X+2)-3 =22
X^2+4X+4-3 = 22
X^+4X-21=0
(X+7)(X-3)=0
Either X+7=0
OR
X-3=0
If X+7=0,
X= -7
If X-3 =0,
X=3
Yeah, right on man. Looked down but didn't see this. My Comment above starts here then adds a bit on this kind of subject, though you might not share my concern past the technicality. Or you may.
@
No matter what, the most important point is that we all are having a good time, getting along with each other nicely and doing Mr John’s math problems!
(What else could we wish for happiness, especially when I am an old grandmother?)
got -7, 3 expand simplify solve thanks for the fun QE with 2 answers (the Sq proves it)
nice SR option
It’s a quadratic. I foiled the left side & then subtracted 22 from each side. This gives you x^2+4x-21. If you factor this you get (x+7)(x-3)=0.this gives you x=-7 & x=3.
Best way to solve simple ones like this is in my head
X=3,-7
Rewrite , refactor, set to zero. It becomes X^2+4X-21=0. (X+7)(X-3) . X+7 =0. X=-7. X-3=0 X=3. Voila
Its a quadratic equation and the answer is x = -7 or x = +3 by fractorization method
x²+4×−21=0= (x+7)(x-3) hence x=3 x=-7
I think it's time to discuss the Po-Shen Loh method which will factorise all quadric equations. It's interesting that the knowledge in this "new" method goes back 4000 years and was forgotten for 300 years.
(x+2)(x+2)-3=22
(x+2)^2-25=0
(x+2+5)(x+2-5)=0
(x+7)(x-3)=0
x=-7,3
3 and -7 by inspection
(x + 2)^2 = 25
x + 2 = +/- 5
x = -7, 3
The best way to solve this… hmm, let’s see what happens.
(x + 2)(x + 2) - 3 = 22
I can quickly tell this is a quadratic equation from the two x terms, but it’s not in a useful form yet. Let’s multiply it out:
x^2 + 4x + 4 - 3 = 22
x^2 + 4x + 1 = 22
Next, we need to have this be set to 0, so let’s do that:
x^2 + 4x - 21 = 0
Now, we could use the Quadratic Formula, but let’s think for a moment.
21 = 3 * 7 and 4 = 7 - 3
We can factor this!
(x + 7)(x - 3) = 0
We press on with our two possibilities:
x + 7 = 0 OR x - 3 = 0
x = -7 OR x = 3
We should check these two answers to make sure they work, but I leave that as an exercise for the reader. Rest assured, both are fine.
Also, some may want to complete the square on this at some point, but my mind never really latched onto that particular process for some reason.
On further review: Oh, now I see a much simpler way to think of this. I think this is along the lines of completing the square, then factor:
(x + 2)(x + 2) - 3 = 22
(x + 2)^2 - 3 = 22
(x + 2)^2 - 25 = 0
(x + 2)^2 - 5^2 = 0
Difference of two squares: [(x + 2) + 5][(x + 2) - 5] = 0
(x + 7)(x - 3) = 0
And from here, the answers are clear.
Questions?
Let y = x+2
So...
y² - 3 = 22
y² = 25
y = ±5
x + 2 = ±5
x = ±5 - 2
x = 3, x = -7
(x+2)^2=25
x+2=+/-5
x=-7
x=3
x={3-7}. A quadratic in x has 2 values of x .ans
I was hoping that you would have graphed this equation so that visually oriented students could see the solution
Thank you. x=-7, 3. Edited to correct accidental sign swap
(x+2)(x+2)=22+3=25=>(x+2+5)(x+2-5)=0=>(x+7)(x-3)=0 gives x={3,-7}ans.way i solved.
No. The √25 is ONLY +5. Like it or not, stupidity or not. It is ONLY ONLY +5.
The proper tricknique (because it gets you stupidly around the stupidity) is the following:
(x + 2)^2 - 25 = 0
(x + 2)^2 - 5^2 = 0
{(x + 2) + 5} * {(x + 2) - 5} = 0
Then solving for each factor:
x + 2 + 5 = 0
x + 7 = 0
x = -7
and
x + 2 - 5 = 0
x - 3 = 0
x = +3
Not adequately teaching students the path in the video and then insisting they get more restrictive with later " 9 - √(x + 3) = 7 " type problems is part of why they get those wrong. (Don't bother me about whatever the shown expression leads to, it's just to evoke the kind of problem, not be a problem in which that ends up mattering. I wrote it on the fly, not over three careful years writing a textbook.) Part of why. Not learning the right idea is another part, but "teaching them wrong" to begin with is more than a wee little bit of it (thanks to early Steve Martin for that).
Time to revisit the old shiboleth of "A rectangle has a length three units longer than its width and an area of 42. What are the dimensions?" and planning to totally toss out an answer just because the lengths will be negative numbers. AS IF negative can't just indicate direction on a Cartesian coordinate graph, but must be some intrinsicly good or evil characteristic with only the religion's one meaning of "negative means of a man and a woman only" kind of thing. If the desired answer is 5 by 2, then -2 by -5 is just as real-world-valid because it just means that instead of the rectangle going over 5 units to the right and 2 units up, it goes two units to the left and 5 units down. Gosh, it can't be -2 by anything because nothing in the real world goes left rather than right!
And then you want them to have internalized that to the point of absoluteness and somehow instantly shift gears and grasp angles in the Cartesian coodinate graph being "blah-blah plus two pi" and that there are infinitely many of them and so on. Golly gee kidees, you mean that don't compute? I wonder why?
First do no harm. Or leave the teaching to doctors who did swear to that principle.
(X + 7) • (X - 3) = 0
X = -7, X = +3
Jeez it’s been a while
X=3
X=-7 x=3
X=4,x=-7
It’s time to discuss how utterly useless this is in the REAL world where people work and earn a living for themselves and their families. This is useful to exactly NOBODY except high school math teachers.
Oh god! You've done it AGAIN!!!
You are correct that we need both square roots of 25 here. SO YOU MUST WRITE ±√25 !!!!
Without ± in front if it, then √25 equals 5 and only 5. You yourself have made plenty of videos explaining this, and yet over and over again you completely fail to follow your own advice! How do you expect your students to learn from you if you are so hopelessly inconsistent yourself???!!!
Well, you kind of gave up the nature of the equation in your title. ;)
Therefore COULD BE MORE THAN TWO SOLUTIONS::
X+2 = square roort of +25 = +5
A POSITIVE INTEGER.
OR
X+2 = - SQ.ROOT OF 25 (NEGATIVE NUMBER0 = --25
A NEGATIVE INTEGER.
THERE ARE TWO MORE P0SSIBILIIES WHICH ARE THE SAME AS THE ABOVE TWO.
THE FISAT EQUATION TO OTHER THAN SUM IS NOT EQUAL TO A PERFECT SQUARE MINUS 2
So we mat have two more possibilities involving a
complex variables or
one that is an irrational number -irrational numbers or a zero.
NOW TEST IS THE TOME TO TEST
Test each possibility id they make sense or not. Yes, no? Theoreticall is that a this the war to approach it?