Thank you for the problem and the solution. This was harder. Extending PQ to meet AB at R, as PR will be parallel to CB , AQ:AC is X/(X+6) in similar triangles AQR and ACB and AR:AB is in that proportion. QR/CB = X/(X+6) QR= 5X/(X+6) PA =5 because CB = 5 . They are subtending equal arcs. Using Pythagoras'theorem twice X.X = 25 - PQ.PQ = AR.AR - QR.QR This would come together with a theorem about products of parts of intersecting chords of a circle. (It feels like hard work.) 6X = PQ.QS S is on circle but PS is not a diameter.; that is where it went wrong. -where PQP is a diameter of the full circle-. So R is the centre of the circle. AR = RB = radius. And ........ X^2 = AR.AR - QR.QR = 6X by intersection chords X=6 NO it is not. So R is not the centre. So watching is easier than thinking for myself at the moment.
At about 10:05, Math Booster appears to pull the solution x = 4 out of the air. Let Δ = x³ + 12x² + 11x - 300 and we search for a positive value of x which will produce Δ = 0. If x = 10, Δ = 2010 and, if x = 0, Δ = -300. We also know that Δ is monotonic (in this case, steadily increasing) over the range x = 0 to x = 10, so there is a zero crossing. So, we write a python program to conduct a half interval search: import math x = 5 dx = 2.5 # define function to compute delta = ∆ def compute_delta(x): delta = math.pow(x,3) + 12*math.pow(x,2) + 11*x - 300 return(delta) for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small delta = compute_delta(x) if (delta > 0): x = x - dx else: x = x + dx dx = dx/2 x = round(x , 12) # round x to 9 decimal places formatted = "{:.12f}".format(x) print(f"x = ",formatted) Output: x = 4.000000000000 === Code Execution Successful === We try x = 4 in Δ = x³ + 12x² + 11x - 300 on our own and find that it produces Δ = 0 exactly.
I like your approach. I might add that since the first derivative of the cubic function is easy to calculate, Newton/Raphson converges very quickly. I got lazy so instead of writing some Python code I used an Excel spreadsheet to do Newton. Using an initial guess of x = 4.9 (from the diagram it must be less than 5) it took just 3 iterations to get a value of x = 4.000000417 f(x) f'(x) x0 = 4.9 159.669 200.63 x1 = 4.10416189 16.40661587 160.0323198 x2 = 4.00164125 0.254458393 155.0787881 x3 = 4.000000417
I don't mind that he simply presents the solution. He might have mentioned that to arrive at it he began testing increasing (from zero) factors of 300, or whatever. Since he knew the solution, he might even have contrived a factorization with some educational merit. The real gist of this problem is the geometry, not the numerical value of the solution, and he covered the geometry well.
@@jimlocke9320 thanks for sharing your process for determining X. For all the details provided in getting to the cubic equation, it would be good for the author to mention how to resolve for correct value. I haven’t done this typd of math for 40+ years, and it appeared that four was just selected without stating how. There are a couple methods to use, but neither was specifically stated.
In this question, to avoid the third degree equation, I used a shortcut, I don't know how correct it is, but I reasoned like this: since the hypotenuse of APQ, AP = 5, I supposed that its sides could be the Pythagorean triple 3,4,5. A couple of tests are enough to realize that it is possible. If x = 3 => PQ = 10/3 and this is not good If x = 4 => PQ = 3 and it is ok Once I obtained these values, I verified that with them the triangles PCQ and ABC were still similar, because we know that they are. And it is easy to see that the values obtained are consistent with the similarity between the two triangles (6:3 = (6+4)/5)
@michaelsanders2655 What number satisfies the equation ??? x³+12x²+11x-300=0 4³+12*4²+11*4-300=0 64+192+44-300 = 0 0=0 Other roots are not real numbers!!!! !!! What strange your question is !!!
@@marioalb9726 - no… it’s not strange. You didn’t just pick 4. You knew it couldn’t be zero, nor could it be 10. The number is somewhere between those. You start with a number and work your way up or down. You don’t show an equation that gives you exactly 4. Explain why you chose 4. And, you just did. I’m just looking for a complete answer.
@ - it’s been over 40 years since I’ve done this type of math. I did watch the entire video. At the end, four is chosen. The reason was not elaborated. I’m just asking that be given. Think of it as a student first learning this math. How would they know the answer is 4? You go from that formula to X=4. I don’t think that information was completely provided.
I don’t understand how you just chose 4??? Obviously, the answer isn’t 1 or 2. I figured it would be a whole number/integer from 3 to 5, but why just select four?
Καλημέρα σας από Ελλάδα (Greece). Προφανώς ο καθηγητής βασίστηκε στο θεώρημα: "Αν υπάρχει ακέραια ρίζα της εξίσωσης αυτής, τότε αυτή είναι ένας από τους διαιρέτες του σταθερού όρου" (δηλαδή του -300) Εδώ φυσικά ζητάμε μόνο θετικούς διαιρέτες αφού x>0. Έτσι, αρχίζουμε δοκιμές με το 1, μετά το 2, μετά το 4 και βλέπουμε ότι αν αντικαταστήσουμε στο πολυώνυμο 3ου βαθμού , ρίζα είναι το 4. Έτσι, η εξίσωση x^3+12x^2+11x-300=0 γράφεται: (x-4)(x^2+16x+75)=0x=4 ή x^2+16x+75=0 που όπως εύκολα διαπιστώνεται δεν έχει πραγματικές ρίζες. Άρα μοναδική πραγματική ρίζα το x=4.
I am witnessing your success for years and very glad to see it because your videos helped me to improve my mathematical skill.
Thank you for the problem and the solution.
This was harder.
Extending PQ to meet AB at R, as PR will be parallel to CB , AQ:AC is X/(X+6) in similar triangles AQR and ACB and AR:AB is in that proportion.
QR/CB = X/(X+6) QR= 5X/(X+6) PA =5 because CB = 5 . They are subtending equal arcs.
Using Pythagoras'theorem twice X.X = 25 - PQ.PQ = AR.AR - QR.QR
This would come together with a theorem about products of parts of intersecting chords of a circle.
(It feels like hard work.) 6X = PQ.QS S is on circle but PS is not a diameter.; that is where it went wrong.
-where PQP is a diameter of the full circle-.
So R is the centre of the circle. AR = RB = radius. And ........ X^2 = AR.AR - QR.QR = 6X by intersection chords
X=6 NO it is not. So R is not the centre. So watching is easier than thinking for myself at the moment.
As ⌒AP = ⌒CB, then AP = CB = 5 and PC and AB are parallel. As ∠PCQ and ∠CAB are alternate interior angles, then ∠PCQ and ∠CAB are congruent. As ∠CQP = 90° and, by Thales' Theorem, ∠ACB = 90° (as AB is a diameter and C is on the circumference), then ∆CQP and ∆ACB are similar triangles.
Triangle ∆PQA:
PQ² + QA² = AP²
PQ² + x² = 5²
PQ² = 25 - x²
PQ = √(25-x²)
AC/CQ = CB/QP
(6+x)/6 = 5/√(25-x²)
30 = (6+x)√(25-x²)
30/√(25-x²) = 6 + x
900/(25-x²) = 36 + 12x + x²
900 = (25-x²)(36+12x+x²)
900 = 900 + 300x + 25x² - 36x² - 12x³ - x⁴
x⁴ + 12x³ + 11x² - 300x = 0
x(x³+12x²+11x-300) = 0
x³ + 12x² + 11x - 300 = 0 --- x ≠ 0
By observation, 0 < x < 5:
(1)³ + 12(1)² + 11(1) - 300 =
1 + 12 + 11 - 300 ≠ 0 ❌
(2)³ + 12(2)² + 11(2) - 300 =
8 + 48 + 22 - 300 ≠ 0 ❌
(3)³ + 12(3)² + 11(3) - 300 =
27 + 108 + 33 - 300 ≠ 0 ❌
(4)³ + 12(4)² + 11(4) - 300 =
64 + 192 + 44 - 300 = 0 ✓
x³ + (16x²-4x²) + (75x-64x) - 300 = 0
(x³ - 4x²) + (16x² - 64x) + (75x - 300) = 0
(x-4)x² + (x-4)16x + (x-4)75 = 0
(x-4)(x²+16x+75) = 0
x = 4 | x² + 16x + 75 = 0
√((16)²-4(1)75) = √(256-300) = √(-44)
No additional real solutions
[ x = 4 ]
At about 10:05, Math Booster appears to pull the solution x = 4 out of the air. Let Δ = x³ + 12x² + 11x - 300 and we search for a positive value of x which will produce Δ = 0. If x = 10, Δ = 2010 and, if x = 0, Δ = -300. We also know that Δ is monotonic (in this case, steadily increasing) over the range x = 0 to x = 10, so there is a zero crossing. So, we write a python program to conduct a half interval search:
import math
x = 5
dx = 2.5
# define function to compute delta = ∆
def compute_delta(x):
delta = math.pow(x,3) + 12*math.pow(x,2) + 11*x - 300
return(delta)
for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small
delta = compute_delta(x)
if (delta > 0):
x = x - dx
else:
x = x + dx
dx = dx/2
x = round(x , 12) # round x to 9 decimal places
formatted = "{:.12f}".format(x)
print(f"x = ",formatted)
Output:
x = 4.000000000000
=== Code Execution Successful ===
We try x = 4 in Δ = x³ + 12x² + 11x - 300 on our own and find that it produces Δ = 0 exactly.
I like your approach. I might add that since the first derivative of the cubic function is easy to calculate, Newton/Raphson converges very quickly. I got lazy so instead of writing some Python code I used an Excel spreadsheet to do Newton. Using an initial guess of x = 4.9 (from the diagram it must be less than 5) it took just 3 iterations to get a value of
x = 4.000000417
f(x) f'(x)
x0 = 4.9 159.669 200.63
x1 = 4.10416189 16.40661587 160.0323198
x2 = 4.00164125 0.254458393 155.0787881
x3 = 4.000000417
I don't mind that he simply presents the solution. He might have mentioned that to arrive at it he began testing increasing (from zero) factors of 300, or whatever. Since he knew the solution, he might even have contrived a factorization with some educational merit. The real gist of this problem is the geometry, not the numerical value of the solution, and he covered the geometry well.
@@jimlocke9320 thanks for sharing your process for determining X. For all the details provided in getting to the cubic equation, it would be good for the author to mention how to resolve for correct value. I haven’t done this typd of math for 40+ years, and it appeared that four was just selected without stating how. There are a couple methods to use, but neither was specifically stated.
Why do all that detailed work and not show the last part where x = 4?
Nice problem, well solved.
In this question, to avoid the third degree equation, I used a shortcut, I don't know how correct it is, but I reasoned like this: since the hypotenuse of APQ, AP = 5, I supposed that its sides could be the Pythagorean triple 3,4,5. A couple of tests are enough to realize that it is possible.
If x = 3 => PQ = 10/3 and this is not good
If x = 4 => PQ = 3 and it is ok
Once I obtained these values, I verified that with them the triangles PCQ and ABC were still similar, because we know that they are. And it is easy to see that the values obtained are consistent with the similarity between the two triangles (6:3 = (6+4)/5)
That is a literal example of easier than it looks. And there is only one positive real solution and that solution is x = 4.
PA=5...PC=√(61-x^2)..gli angoli opposti del trapezio APCB sono supplementari...arcsin(x/5)+arcsin(6/√(61-x^2))+arctg(6+x)/5=180....calcoli x=4
Why PC = √(61-x²) ?????
Similarity of triangles:
y/6 = 5/(6+x) --> y = 30/(6+x)
Pytagorean theorem:
y² + x² = 5²
[30/(6+x)]² + x² = 5²
30² / (6+x)² = 5² - x²
(5²-x²)(6+x)² = 30²
(x²-5²)(x²+12x+6²) = -30²
x⁴+12x³+6²x²-5²x²-12.5²x-30²=-30²
x³+12x²+11x-300=0
x= 4 cm ( Solved √ )
And, how did you arrive at 4? I know that’s the answer, but you don’t explain how you got 4.
@michaelsanders2655
What number satisfies the equation ???
x³+12x²+11x-300=0
4³+12*4²+11*4-300=0
64+192+44-300 = 0
0=0
Other roots are not real numbers!!!!
!!! What strange your question is !!!
@@marioalb9726 - no… it’s not strange. You didn’t just pick 4. You knew it couldn’t be zero, nor could it be 10. The number is somewhere between those. You start with a number and work your way up or down. You don’t show an equation that gives you exactly 4. Explain why you chose 4. And, you just did.
I’m just looking for a complete answer.
@@michaelsanders2655
Please, review your school notes about solving third degree equations
or see video, almost the same as I wrote !!!
@ - it’s been over 40 years since I’ve done this type of math.
I did watch the entire video. At the end, four is chosen. The reason was not elaborated. I’m just asking that be given.
Think of it as a student first learning this math. How would they know the answer is 4? You go from that formula to X=4. I don’t think that information was completely provided.
I don’t understand how you just chose 4??? Obviously, the answer isn’t 1 or 2. I figured it would be a whole number/integer from 3 to 5, but why just select four?
Καλημέρα σας από Ελλάδα (Greece). Προφανώς ο καθηγητής βασίστηκε στο θεώρημα: "Αν υπάρχει ακέραια ρίζα της εξίσωσης αυτής, τότε αυτή είναι ένας από τους διαιρέτες του σταθερού όρου" (δηλαδή του -300) Εδώ φυσικά ζητάμε μόνο θετικούς διαιρέτες αφού x>0. Έτσι, αρχίζουμε δοκιμές με το 1, μετά το 2, μετά το 4 και βλέπουμε ότι αν αντικαταστήσουμε στο πολυώνυμο 3ου βαθμού , ρίζα είναι το 4. Έτσι, η εξίσωση x^3+12x^2+11x-300=0 γράφεται: (x-4)(x^2+16x+75)=0x=4 ή x^2+16x+75=0 που όπως εύκολα διαπιστώνεται δεν έχει πραγματικές ρίζες. Άρα μοναδική πραγματική ρίζα το x=4.
@@solomou146 - no translation available. I know how/why X is chosen as 4. It wasn’t properly explained.
All fine up to 9:09 which is where I got to, then you just start guessing. Unimpressive.
(5)^2 (6)^2={25 +36}=45 180ABCPX/45=4ABCPX 2^2 (ABCPX ➖ 2ABCPX+2).
Lame.