Videos which are actual teaching and not designed for views are getting rarer and rarer. I feel lucky to see people you, sir.
@clue000110 ай бұрын
Totally agree. 👍🏻
@dhobonov10 ай бұрын
Totally + 1 agree.
@mrhtutoring10 ай бұрын
So nice of you. Thank you for the nice comment. 👍
@doloreswillcot632010 ай бұрын
@@clue0001hh on
@doloreswillcot632010 ай бұрын
@@mrhtutoring Thank you for helpful reminders.
@datapro0077 ай бұрын
It's been so long since I was in 9th grade, but it's all coming back to me now.
@theLordsboy9 ай бұрын
Great teacher. Clear explanations. No hype. Thanks!
@escribidormadrid346310 ай бұрын
I watch a few Mr H videos most days. They just reinforce so much. He gets straight to the point, no razzmatazz to distract us, and a specific thing is illustrated. Thank you, Mr H.
@mrhtutoring10 ай бұрын
Thank you for the encouraging words.
@TeguhNoorZaman-uc1xs10 ай бұрын
Mr. H you're teaching Math very well and so clearly. 💡🎓 Nice to meet you on KZbin.👍✨🙏
@mrhtutoring10 ай бұрын
Nice to meet you. Thank you for the nice comment.
@iitb91110 ай бұрын
I love this teacher becoz my little concept gets clear through such question ❤
@Nikioko10 ай бұрын
0:40: What you have to do, is apply the first binomial formula (a + b)² = a² + 2ab + b², with a = √x and b = √3, which gives you (√x + √3)² = x + √12x + 3
@rleroygordon10 ай бұрын
He simply factored out the square root of 12 (i.e. 12 = 4*3): √12x = √(4*3)x = √4 √3x = 2√3x.
@harrymatabal84485 ай бұрын
Your videos are unlike the others. You are brilliant. Thanks
@mrhtutoring5 ай бұрын
Thank you~
@shannonmcdonald75847 ай бұрын
This is one of my practice equations. Every couple days i write it out from memory. Great vid
@Nikioko10 ай бұрын
3:06: The quadratic formula is: x₁,₂ = −p/2 ± √[(p/2)² − q], in the case of x² − 8x + 4 = 0: x₁,₂ = 4 ± √(4² − 4) = 4 ± 2√3.
@mrhtutoring10 ай бұрын
What do p and q represent?
@mealves110 ай бұрын
Dear Professor! Thank you for the awesome lectures. Please, could you provide information about the "electronic blackboard" that you use in your lectures?
@MsHortel7 ай бұрын
Prof H saves me from the pit of mathematics. Thank you Prof H
@DavidMauas10 ай бұрын
Fantastic! Thank you so much!
@mrhtutoring10 ай бұрын
You're most welcome.
@samuelkelly21979 ай бұрын
Mr is a Class Act. His few videos I have watched reminds me of my Additional Maths Tutor in the middle 70s. Mr Malani was his name. God bless Maths Tutors. I do pinch my thighs always NOT taken after these two lovely tutors. I know there still are some wonderful tutors out there.
@titanbs58649 ай бұрын
I am an Indian in High School preparing for the JEE Entrance exams. And this is just one of the topics we are taught in BASIC MATHS. Someone tell me if this is considered hard in the US or somewhere else?
@mrhtutoring9 ай бұрын
This is basic algebra taught to 8th graders in the USA.
@4Chess8 ай бұрын
@titanbs5864 Connect with me for JEE level preparation
@leonakambe10 ай бұрын
Weldon Sir, for your teachings. I'm really learning a great deal from you. Thank you 🙏 once again.
May be mentionned in other comments, a second possible way consists tu replace x by y with y=Root(x). After replacement and squared each side of the equation, and simplification We have y.y -2root(3)y -2=0. Then y=1+-Root(3) Then x=4+-Root(3) Right ?
@Psykolord19899 ай бұрын
Tackling this problem before watching fully, want to see how rusty I am... Alright so, √x + √3 = √(2x+5)? I have two ways in mind to potentially solve this. One is to simply square both sides and work from there, and the other is to multiply by √x - √3 on both sides so the left hand side is all in terms of x^1. I'm going to try the first one Squaring a+b = a^2 + 2ab + b^2, so squaring both sides gives us: x + 3 + 2(√3)(√x) = 2x+5. (NOTE: I was stuck on this part for a while. I kept thinking the 2ab term on the left would be 2√3 and not 2(√x)(√3) Now, we want all of the X on one side. This is going to be a tall order, but I have a strange idea. Subtract 2x + 5 from both sides... -x -2 + (2√3)(√x) = 0. How do we deal with this? We're going to pretend it's a quadratic equation! We will engage in a process called "U-substitution," where we define a variable in terms of another arbitrary variable, which is traditionally labeled "U." For example, if you give me an equation like x^4 -4x^2 + 3, the problem can be put into the quadratic equation by making the substitution U = x^2 and solving for U. Once you solve for U, *YOU MUST put things back in terms of your initial variable* . I, and many other students, have messed up problems because we forgot this last step. In this case, we define the variable u = √x. Then x = u^2, meaning... -u^2 + u(2√3)-2 = 0. Quadratic equation: First check the discriminant, b^2 -4ac. b = 2√3 so b^2 = (4*3) = 12, 4 ac = 4(-1)(-2) = 8. 12-8 = 4, so we will in fact have two real roots for "u" Using the quadratic equation (-b±√(b^2 - 4ac))/2a... -> ((2√3±2))/ -2 = √3 ±1. Now recall, these are the values for U. To get X, we must square these roots, and thanks to binomial expansion arrive at: x = 3 ±2√3 +1 = *4±2√3*
@ichdu67109 ай бұрын
very good explanation in good english. Missing the check if the solution are positive or not when acting in reell space
@arminyt3979 ай бұрын
in this case sqrt(3)sqrt(x) = sqrt(3x) is correct when we multiply two square roots, we always have sqrt(a)sqrt(b) = sqrt(ab) no matter what the following case is a > 0, b> 0 a > 0, b < 0 a < 0, b > 0 but when a
@radoslavvrabel48667 ай бұрын
I wish for many videos and comments like this because some don't even know what they are doing and teach math and don't even know the basic rules and only calculate with a calculator 👍
@davedeatherage49029 ай бұрын
Deciphering, filtering, gleaning, applying! Mr. H is committed to teaching. Which is so helpful and refreshing! Thank you Mr. H sir.
@mrhtutoring9 ай бұрын
Thank you for the nice comment.
@jmich710 ай бұрын
Interesting how we did not know we would end up with a quadratic equation. It sort of looked like though. Thank you sir.
@carultch4 ай бұрын
You don't necessarily. In special cases of combinations of square root terms like this, you may not end up even thinking about the quadratic formula. Those are usually cases where a quadratic would either have a repeated root, or two equal and opposite roots, if you did eventually take it to the point of a quadratic formula. In the general case, you do. You've got two instances of your unknown variable, and one is equal to the other squared. As an example, had you been given: sqrt(x) + sqrt(3) = sqrt(2*x + 2*sqrt(3*x) + 2) Upon squaring both sides and expanding, you end up with: x + 2*sqrt(3)*sqrt(x) + 3 = 2*x + 2*sqrt(3)*sqrt(x) + 2 The middle terms are eliminated: x + 3 = 2*x + 2 Solve for x: x = 1 This is a case where the b-term of the quadratic was eliminated, and it turned into a linear equation.
@redheadedmoos120410 ай бұрын
I squared both sides and got x + 3 + 2 root x = 2x + 5 I i subtracted both sides by x, 3 and 2 root 3x and got x + 2 - 2 root 3x then equated the equation to 0 After taking 2 common from equation, I got x + 2(1-root 3x) =0 and by dividing both sides by 1 - root 3x I got x = -2 I know the answer is wrong but want to know what I did wrong
@GreggRomaine7 ай бұрын
Your mistake occurred here: "dividing both sides by 1 - root 3x". When you divide both sides of an equation, you must divide the ENTIRE side of the equation. So, dividing the left side of the equation: x + 2(1-root 3x) = 0 would result in: x/(1-root 3x) + 2 = 0 NOT simply: x + 2 = 0 You forgot about that poor, little x :)
@gilbertgosset570810 ай бұрын
Le plus difficile dans ce problème , c'est de ne pas faire de fautes d'inattention ...... Amitiés à tous .
@leaDR3569 ай бұрын
for some reason, my answer is sqrt(10)/4
@Lemonsieur-m4m7 ай бұрын
Very well done Mr H. Suscribed.
@sistemaassonnato-889210 ай бұрын
I wanted to ask a question, isn't it necessary to establish the conditions for the existence of radicals? before squaring. thank you professor for the work you do :)
@carultch9 ай бұрын
Yes. When you square both sides to clear a radical, you lose information from the original equation. Square roots inherently have a restricted range by only giving the positive square root by convention. Squaring both sides opens up that restricted range, and introduces the possibility of extraneous solutions. It's common that the extraneous solution may be a solution that doesn't make sense in the application of the equation, and you don't really need to think about this problem.
@ekbalmokhammad86207 ай бұрын
Thank you!!! I like your teaching because you are very kind, and very good person 🙌🌹
@gleysonsantos39607 ай бұрын
this is pure gold. saving it.
@mathsfamily67669 ай бұрын
please more this kind of this equation sir!
@naderhumood119910 ай бұрын
This is a first class of work. Thank you very much Sir.
@mrhtutoring10 ай бұрын
Welcome!
@kookyta10 ай бұрын
thank you .. please make videos for integral calculation 😊
@name19277 ай бұрын
This is a great channel
@Cheerup-l7j10 ай бұрын
Hello sir ❤❤ from india. I would have really been happy if I had a math teacher like you sir.
@mrhtutoring10 ай бұрын
👍
@4Chess8 ай бұрын
Connect with, get free demo and you will sure love it
@devansh56449 ай бұрын
Avg jee questions
@johnmaulkin-ur8lm8 ай бұрын
A big thank you.
@Lemonsieur-m4m4 ай бұрын
A complicate answer to a simple question. Thank you for taking the time to walk us through it. And underlining what not to do. Which is a must IMHO.
@ockham19639 ай бұрын
brilliant teaching, thanks
@BINIYAMAZMACH10 ай бұрын
Tnx❤❤❤
@4Chess8 ай бұрын
Connect with me for learning maths from basic to olympiad level, get free demo to decide and you will love it this I assure
@jacobgoldman57809 ай бұрын
I think the main reason we know no extraneous solutions is that both 4+2sqrt(3) and 4-2sqrt(3) are positive since 1< sqrt(3)
How do you know there are no extraneous solutions simply because there is more than one radical in the equation?
@Mycroft61610 ай бұрын
Since both sides use the square root function, we know neither side can output a negative value. Squaring creates extraneous solutions only when applied to a function that can generate negative outputs since squaring a negative generates a positive. When we are not turning negative outputs into positives, squaring does not cause extraneous solutions. We can also check this case-by-case by setting each radical x-term greater than or equal to 0 and solving for our minimum x values. Any solutions above the highest minimum is valid. In this case, both solutions are positive, so they are both valid.
@BruceLee-io9by2 ай бұрын
As always, masterful lesson.
@อืซี่โมบาย-ฉ9ษ10 ай бұрын
ขอบคุณครับ
@amann028 ай бұрын
Can someone please explain how did he simplified the root 48 at 4:04
@neillawrence41988 ай бұрын
48 can be factored as 3 times 16. Remove the square root of 16, which is of course 4, to the outside the radical, leaving the 3 inside.
@anchorskid6 ай бұрын
I am deep under water.
@420sakura12 ай бұрын
Stay there forever
@anchorskid2 ай бұрын
@@420sakura1 Rude.
@rob8769 ай бұрын
x + 3 + 2√(3x) = 2x + 5 2√(3x) = x + 2 12x = x^2 + 4x + 4 x^2 - 8x + 4 = 0 (x - 4)^2 = 12 x - 4 = ±√12 x = 4 ± √12 x = 4 ± 2√3
@saileshkumarsaila52419 ай бұрын
(x-4)^2=12, how this 12 comes already subtracted to get - 8x
@Rooney11845 ай бұрын
Prof. Pls teach us about gausian and row echelon form and reduced row echelon form
@mrhtutoring5 ай бұрын
Will do~
@Rooney11845 ай бұрын
@@mrhtutoring Thank You Sir🫡
@jozitro45549 ай бұрын
God bless you!🕊🌈
@jim23769 ай бұрын
Cool beans. 👍
@toxiqdog101510 ай бұрын
bro i legit calculated this in my head in like 2 min just from seeing the thumbnail
@Samanthak_Prajwal10 ай бұрын
ok
@ramonbina-oro57218 ай бұрын
the best
@JulesMoyaert_photo10 ай бұрын
👍Done! Thanks!
@mrhtutoring10 ай бұрын
Welcome!
@asmaaftab44979 ай бұрын
X=-2
@margaretwelsh64457 ай бұрын
1
@আলোরপথিক-থ৫ত8 ай бұрын
How to do that x^2 + 4x +4 = -12x??
@carultch7 ай бұрын
Given: x^2 + 4*x + 4 = -12*x Shuffle everything to one side: x^2 + 16*x + 4 = 0 Use the mean-product formula to solve. The mean m, will be the middle coefficient divided by -2, and the product p will be the final constant term. x = m +/- sqrt(m^2 - p) x = -8 +/- sqrt(8^2 - 4) x = -8 +/- sqrt(60) x = -8 +/- 4*sqrt(15) Then you need to interpret which one applies.
@vansin00110 ай бұрын
2/3
@Brid72710 ай бұрын
When I got to x^2-8x+4, I first thought of finding the value of x by using the quadratic formula, but that’s too straightforward for me which in turn is boring So instead I used the “complete the square” method x^2-8x+4=0 x^2-8x=-4 x^2-8x+16=-4+16(take half of b then square it meaning (-4)^2=16 and add on both sides) (x-4)^2=12 x-4=+-sqrt(12) x=4+-2sqrt(3) x=2{2+-sqrt(3)}[extra step but yeah]
@mrhtutoring10 ай бұрын
That works! 👍
@venkybabu814010 ай бұрын
When you use psychology you should know how to control people. Like FB which makes you work 24×7. Some ask you to fill timesheet. Mostly orphanage kids are hired for such work. Because orphans are in high demand throughout the world. For cooking cleaning road side canteen. Sabarimala canteen. Government posters sticking. Movie wall paper. CA means street cleaning jobs. Etc.
@kristopherkerry752410 ай бұрын
Should be -12x......not -8x. 4x-16x is -12x......
@spasticnapjerk8 ай бұрын
KZbin just presented an ad for a homework ai where the announcer said that it spit out 1000 words on global warming, and it solved a hairy equation, and said look at my grades now, 94%!
@carultch4 ай бұрын
AI undermines the entire point of education. It makes it stupidly simple to cheat your way to a degree, and makes the degree worthless when you graduate.