Find all possible values of X | Triangle Inequality Theorem | Important math skills explained
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- Опубліковано 16 вер 2024
- Learn how to find all possible values of X by using the Triangle Inequality Theorem . Important Geometry and Algebra skills are also explained. Step-by-step tutorial by PreMath.com
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Find all possible values of X | Triangle Inequality Theorem | Important math skills explained
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Triangle Inequality Theorem
Find all possible values of X
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Beautifully explained thanks
Glad you liked it
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Case 4: *all side must be positif*
b & c are fine, but 'a' > 0
4x - 5 > 0 ➡ x > 5/4
Luckily x > 5/3 is above 5/4. 😉
Correct. Glad to see someone is paying attention.
Another limit is that x > 5/4. If not, side BC would go negative. However, the x > 5/3 limit overrides this.
Let me make it clear that I love your teaching love your voice and love your channel ... I am a math teacher who love learning new ways to teach math
I wanna add that obviously all sides must be of positive value, so x>-3/2, x>-3, x>5/4 must all be fulfilled too.
But 5/3 is bigger than all of them, so this is redundant.
Good question and good solution. Depiction is superb !
Excellent!
Glad you think so!
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Nicely done. I workred differently. I assigned values to x and went with trial and error (improvement). I got the
No worries!
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Stay blessed 😀
Can you tell me what software you are using ? This software works very well to draw any contents when you are talking.
I thought about using Heron's formula, but I don't think it is necessary. If x = 5/3 then the 3 side lengths are 19/3, 14/3 and 5/3. x must be larger than 5/3 and if it is larger then each of the sides is larger and hence the area is larger. So Heron's formula won't give any values of x that further restrict the possible values of x. What do you reckon?
Love this one, logical rather than mathematical, interesting the inequalities give results.. very interesting, never knew this, thanks for filling in the gaps.... In my head 😃👍🏻
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Many many thanks
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What about saying that a side of the triangle must be greater than the difference of the other two and lower than the sum of the other two and examinating only one case?
Thats nice and useful
Thanks Sir .
Thanks for video.Good luck sir!!!!!!!!!!!!!
Do we use Heron's Formula here?
Nice
What about plugging the values into Heron's formula?
Possible but it's gonna be soo complex
Once you had x>5/3, the x>1 was a non-factor is solving this.
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By using Heron's formula, it turns out that side b is the height. So this is actually a right triangle, where a^2+b^2=c^2. The solution is then x=(23+2*sqrt(51))/13. I didn't consider x=(23-2*sqrt(51))/13 a solution since it's less than 1. BTW, I thought it a bit strange that side b (x+3) looks longer than side c (2x+3).
Wow that's awesome! I'm not sure how to use Heron's formula to solve for this, I only thought it was only used to find the area of a triangle. But it looks like that is a solution for x for a right triangle.
I tried with Pythagorean theorem to find solutions for a right triangle. I found 2 possible solutions but it turns out they didn't work because the value for the hypoteneuse in the equation wasn't the largest value. I wonder why I couldn't get that value with the Pythagorean theorem. It reminds me how sometimes I failed to get an answer with the method I chose that led to nowhere.
Edit: I used the wrong equations the first time around, that's why I ended up with the wrong answers. I ended up with 2 solutions for x that worked that would make it a right triangle. See comment below.
Actually I messed up with Pythagorean theorem earlier that's why I didn't get a working solution. I also got x= (23+2sqrt51)/13 using Pythagorean theorem this time with side AB as the hypoteneuse.
I also got a second solution for a right triangle with side BC as a hypoteneuse with Pythagorean theorem.
x = (29+sqrt764)/11
How do you tell which one is a b or c?
Very easy Math problem
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3x+6>4x-5=>x,11 > x>1
I am first
To
Excellent!
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Hello sir how can i send you more questions?
Or you can write it as x ∈ (5/3,11)
Possible values of x are greater than equal 1and less than 11.
What if the x =0
Hi sir I try to solve it.. What is your real name.. Sir
Could you shout out me please... Sir
Its so sad that I was never exposed to this theorem... I would almost believe that this is not possible
huh?
I didn't understand
Please try to stop saying "go ahead and ...".