Solving Modulus Equations and Inequalities
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- Опубликовано: 11 окт 2024
- How to solve Modulus inequalities and equations. Also know as solving absolute value inequalities. Modulus inequalities revision questions / handout. Absolute Value Equations. Leaving Cert Maths and a levels.
Thank you!!! Finding this every where, the drawing graph part totally answered my question ❤
I'm glad you found it helpful. Yes the sketching clarifies it easier.
Overall good explanation but u could've used (a+b)²= a²+b²+2ab
It's easy of course but I'm alitle bit confused here (x-2)² ....was it not supposed to be difference of two square?
difference of two squares is when squaring each part separately, such as (x)² -(2)² . Hope this clarifies.
@@tullamaths much appreciation...it's clear now
@@HaroldMussah best of luck with your studies
thank you!!!
it’s very helpful to me!
You are very welcome
Thank you sir love from Pakistan
Great Tulla Maths is reaching Pakistan
Thank you
You're very welcome
thank you very much sir
No problem at all - glad to of helped!
Thankyou sir.. 💌in part 12:17 how did you solve the eqn into (3x+ 7)(x+5) i mean which were the numbers that you considered?
it is factorising quadratics. Please check this video if you need support. Any queries, let me know. ruclips.net/video/UekDsivwxYo/видео.html
It's great but lengthy process!
Thxs broo nice explanation tq soo much
no problem at all - glad it helped
bro,
what to do if equation in form of
2|x+1|>x+4
p.s. understood all the concepts, just had a doubt in this type of question. Keep up the good content👍💪
let me check this further
@@tullamaths hello there
but can we square the rhs without knowing that it might be negative.
since squaring might effect the inequality for eg
2 > -3
after squaring
4>9
which is wrong
how do you solve inequalities when there is a - or + outside of the abs(x)? e.g abs(x-1)>4-abs(x+1)?
that's great sir
very welcome
Thank you...
Im glad it helped
Thank you ❤
you're welcome!
Thank you Sir
no problem - glad you found it helpful
Explanation is good , TNX
I'm glad it helped !
thank u from kashmir
Glad to help !
taught me faster than my teacher
great that it helped you
Just question cus Im struggling with factoring. Where did you get the 15 (FIFTEEN) to add to (7+5) in the part 12:00 ?
Factorising can always be though. The fifteen is coming from multiplying the 3x by 5 = 15x. Then multiplying the x by 7= 7x. Add 15x and 7x makes the 22x. Perhaps watch this video and it could help better
ruclips.net/video/UekDsivwxYo/видео.html
Polynomial
Thank you
you're welcome
How did get to conclusion of example 1 and 2. In example 1,6 is correct however when you minus 2 from -2 it doesn't give you 4. In example 2 is it not suppose to be -25😢😢. I'm already confused ain't planning to go further 💔
-2 -2 is = -4 modulus of -4 = 4 as per my video. Be careful with your rules of subtraction. Example 2 the answer is x=2 as per video (not -25 as you state).
Him : Twenty-five
My ears: Turkey-five
The answer to question 4 is incorrect. You want the outside areas not the negative section under graph. You can easily see this by subbing in x=zero and it fails to be correct since 1 is not greater than 6!
How do you solve |x-2| = |x| +1? Does anyone have any ideas?
I have found out a way
If inequality is in form :
1) |a|≤/≥|b| then square both side snd find the domain.
2) |a|±|b|≥/≤(any constant)
Then use identiy
|Z1|+|Z2|≥|Z1 + Z2|
|Z1|-|Z2|≤|Z1-Z2|
Crispy
How to draw modular graph in Y = 4 - |2x - 8|
Your method is too lengthy+complicated... infact the first math you have done is way too lengthy and on the other hand I completed it within 2 lines only.
Please try to make it easy
There are many methods to solving mathematical problems. My tutorial video aims to explain just one of these methods.
You can't square both sides in inequalities 🤦 because x is a variable it can be a negative value too
You have murdered all the laws in mathematics
sguares are always positive. aise karega iit?
@@atharvsoni5308
-4 < 3✅
Squaring both sides
(-4)² < (3)²
16
In this case you can because both sides are positive, plus |x| = sqrt(Re(x)^2 + Im(x)^2) which is sqrt(x^2) for all real numbers.
Both sides can be squared when both sides of the modulus inequality are absolute values, because -4 < 3, but |-4|>|3|, so both equations can be squared
Im more confused now than before i watched this video
I'm sorry to hear that, hopefully you can find a video elsewhere that can help.
misleading, make easy to complicate!
Agreed,this isnt hard but overcomplicated for no reason
It's not that hard if u understand polynomial 😌
@@Carrymejane doesn't matter there are easier ways
@@Uchiha...Itachi how? can u suggest?
Some of your working are wrong
13 minutes of absolute no use
Worst explanation
its three not tree
If you know better maths then y tf u watching this 🤭🤓
If you know better maths then why tf u watching this video ? 🤓
🤓 @@adw8s941
I said about english accent not maths dummy