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.
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.
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
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
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 ?
Dear Professor! Thank you for the awesome lectures. Please, could you provide information about the "electronic blackboard" that you use in your lectures?
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 👍
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.
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 :)
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.
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*
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?
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.
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
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 :)
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.
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]
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.
RUclips 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%!
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.
Thanks!
Thank you for your support!
I appreciate it. 🙏
Videos which are actual teaching and not designed for views are getting rarer and rarer. I feel lucky to see people you, sir.
Totally agree. 👍🏻
Totally + 1 agree.
So nice of you. Thank you for the nice comment. 👍
@@clue0001hh on
@@mrhtutoring Thank you for helpful reminders.
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.
Thank you for the encouraging words.
It's been so long since I was in 9th grade, but it's all coming back to me now.
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.
This is one of my practice equations. Every couple days i write it out from memory.
Great vid
Prof H saves me from the pit of mathematics. Thank you Prof H
Great teacher. Clear explanations. No hype. Thanks!
Your videos are unlike the others. You are brilliant. Thanks
Thank you~
I love this teacher becoz my little concept gets clear through such question ❤
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.
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
He simply factored out the square root of 12 (i.e. 12 = 4*3): √12x = √(4*3)x = √4 √3x = 2√3x.
Mr. H you're teaching Math very well and so clearly. 💡🎓
Nice to meet you on RUclips.👍✨🙏
Nice to meet you.
Thank you for the nice comment.
Fantastic! Thank you so much!
You're most welcome.
Thank you!!!
I like your teaching because you are very kind, and very good person 🙌🌹
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
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 ?
Deciphering, filtering, gleaning, applying! Mr. H is committed to teaching.
Which is so helpful and refreshing! Thank you Mr. H sir.
Thank you for the nice comment.
Very clear and easy to follow. Thanks, professor.
As always, masterful lesson.
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.
What do p and q represent?
Dear Professor! Thank you for the awesome lectures. Please, could you provide information about the "electronic blackboard" that you use in your lectures?
please more this kind of this equation sir!
thank you ..
please make videos for integral calculation 😊
Weldon Sir, for your teachings. I'm really learning a great deal from you. Thank you 🙏 once again.
My pleasure ~
Very well done Mr H. Suscribed.
this is pure gold. saving it.
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 👍
This is a great channel
A big thank you.
You're good at showing math, thanks.
Happy to help
Love it😍😍
brilliant teaching, thanks
Great math...
Interesting how we did not know we would end up with a quadratic equation. It sort of looked like though. Thank you sir.
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.
I love this classes im always telling my son to watch lol
very good explanation in good english.
Missing the check if the solution are positive or not when acting in reell space
This is a first class of work. Thank you very much Sir.
Welcome!
This is very informative @mr H .......try open another online platforms for maths questions.....very nice work done sir
Thank you, I will
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 :)
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.
Hello sir ❤❤ from india. I would have really been happy if I had a math teacher like you sir.
👍
Connect with, get free demo and you will sure love it
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*
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?
This is basic algebra taught to 8th graders in the USA.
@titanbs5864 Connect with me for JEE level preparation
Cool beans. 👍
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)
ขอบคุณครับ
How do you know there are no extraneous solutions simply because there is more than one radical in the equation?
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.
Tnx❤❤❤
👍Done! Thanks!
Welcome!
Pourquoi n'as tu pas factorisé pour resoudre l'equation? Merci
God bless you!🕊🌈
how exactly did you simplify the square root of 48?
48 = 2^4 × 3
48^(1/2) = (2^4 × 3)^(1/2)
48^(1/2) = (2^4)^(1/2) × 3^(1/2)
48^(1/2) = 2^(4 × 1/2) × 3^(1/2)
48^(1/2) = 2^2 × 3^(1/2)
48^(1/2) = 4 × 3^(1/2)
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
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 :)
Le plus difficile dans ce problème , c'est de ne pas faire de fautes d'inattention ...... Amitiés à tous .
Connect with me for learning maths from basic to olympiad level, get free demo to decide and you will love it this I assure
Prof. Pls teach us about gausian and row echelon form and reduced row echelon form
Will do~
@@mrhtutoring Thank You Sir🫡
the best
Sorry, read -12 as -16...brain fart. Keep up the fun videos!
No problem. Thanks. 🙏
Can someone please explain how did he simplified the root 48 at 4:04
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.
bro i legit calculated this in my head in like 2 min just from seeing the thumbnail
ok
for some reason, my answer is sqrt(10)/4
How to do that x^2 + 4x +4 = -12x??
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.
The very definition of a theoretical, non-real world problem.
2/3
(sqrt(2x + 5))^4 + (x - 3)^2 = 2 * sqrt(2x + 5)^2 * (x + 3)
(2x + 5)^2 + (x - 3)^2 = 2 * (2x + 5)(x + 3)
4x^2 + 20x + 25 + x^2 - 6x + 9 = 2(2x^2 + 11x + 15)
5x^2 + 14x + 34 = 4x^2 + 22x + 30
0 = x^2 - 8x + 4 = (x - 4)^2 -12.
x = 4 plus or minus sqrt(12) = 4 + or - 2 * sqrt(3).
Plug in and check.
Avg jee questions
I am deep under water.
Stay there forever
@@420sakura1 Rude.
4+2V3; 4-2V3. Пример несложный.
writing a = √x , b = √3 one gets
2 (a^2 + b^2) = ( a + b) ^2 + 1
( a - b) ^2 = 1
√x = √3 -1, √3 +1
x = 4 - 2 √3, 4 + 2 √3
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]
That works! 👍
1
√х + √3 = √2х+5 (х≥0; х≥-2,5)
х+2√3х+3=2х+5
2√3х=х+2
4•3х=х²+4х+4
х²-8х+4=0
D=64-16=48
x=(8±√48):2=(8±4√3):2=4±2√3
I would say that you need to add more details. Why are you doing this and why not. Also the rules as in math teaching rules are more important
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
(x-4)^2=12, how this 12 comes already subtracted to get - 8x
(Rqx+rq3)^2=[rq(2x+5)]^2
x+2(rq3x)+3=2x+5
2rq3x=x+2
(2rq3x)^2=(x+2)^2
12x=x^2+4x+4
x^2-8x+4=0
x=[8+-rq(64-16)]/2=[8+-rq48]/2=
[8+-4rq3]/2=4+-2rq3
X=-2
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.
Should be -12x......not -8x. 4x-16x is -12x......
RUclips 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%!
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.