Find the constants

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Prime Newtons

7 ай бұрын

In this video, I solved an algebra problem with no special skills. Just algebra!!!

Пікірлер: 77

@chandranshu4892 7 ай бұрын

There is actually a much easier way to solve this. Since we've to find these constants c and d such that they are true for all values of x, set x = 0 to obtain an equation in c and d. Next set, x = c (or -c) and get another equation in c and d. That immediately eliminates d and leaves you with a cubic equation in c with no constant terms. That yields the results pretty quickly with some extraneous values that can be eliminated with simple substitution.

@BartBuzz 7 ай бұрын

Actually, if you set x = 0 , you will get [f(c)]^2 =d and the algebra works out the same as in the video. Except that the algebra is much simpler. There is no cubic equation involved. This approach was easier for me.

@annacerbara4257 7 ай бұрын

if you place numerical values instead of x, for example x=0, x=2, verification is not necessary, which in any case is never a bad thing to do.

@joydeepdas4735 7 ай бұрын

I did the same way and got another pair of solutions: c=0 and d=1

@annacerbara4257 7 ай бұрын

@@joydeepdas4735 I think that you set x=c and x=0 maybe; if you use the undetermined value c you have to verify the solutions because you find the good c=3 and another not acceptable one.

@TheEulerID 7 ай бұрын

@@annacerbara4257 Doing that means taking on trust that there is a value for c where all values of x produces a constant, but whilst the question effectively says there is, it makes me rather queasy. I prefer to prove it (and a well written question would say show that there is a constant c that produces the constant d from that expression).

@hemramachandran5626 15 күн бұрын

You are my favorite teached on the Internet, when I aspire to be a teacher, I would like to be like you!

@dr.santoshshukla1260 7 ай бұрын

Great! Keep it up! I admire your lightness throughout yours solving! God bless!

@weslinpenacamacho1075 7 ай бұрын

I really like your exercies, it works for my to keep (like you say) learning

@mrgamepigeon6586 7 ай бұрын

I really liked your method for finding d at 10:12! When you found c using just absolute values was very cool as well. 11:40 I have never seen someone do it like that before.

@juliovasquezdiaz2432 7 ай бұрын

Excelente, gracias por compartir. He comprobado. Saludos desde Chiclayo Norte del Perú

@user-gi5ix6zr8f 7 ай бұрын

عالی بود سپاسگزارم لذت بردم از شیوه حل سوال

@MDAviaton 7 ай бұрын

Thanks for the video... I guess we learn new things everyday

@vitotozzi1972 5 ай бұрын

Marvellous......

@johnmatthew9253 7 ай бұрын

that was awesome!

@mn-lc7em 7 ай бұрын

Your solution is simple and easy. Question: if expression =d is derived d'=0 and we work only on finding c. Yes is complicated and not elegant but may be in other problems it may be useful. Now a different question: Is any restrictive condition about different of 0 ? And another for c part You have (L)²=(R)² L² -R²=0 (L-R )*(L +R)=0 possible 2 solutions for c. Thank you for the examples

@elmerhuamanpedraza3121 6 ай бұрын

It seems to be difficult. Thanks for showing me the way to solve it.

@surendrakverma555 5 ай бұрын

Very good. Thanks 👍

@carlosangulo2888 7 ай бұрын

Very interesting. Thx.

@andrewhone3346 7 ай бұрын

You can get the answer without any multiplication, by looking at where the rational function can have a zero/pole, and looking at the limit when x goes to infinity.

@sunil.shegaonkar1 7 ай бұрын

Hello Prime Newtons, you have excellent ideas of problem solving.

@PrimeNewtons 7 ай бұрын

Glad you think so!

@user-qy2lx3rt7y 7 ай бұрын

Mind Blown

@earl.c 6 ай бұрын

f(x) = -1/2 + 1/(x-2) f(c+x) f(c-x) = 1/4 + (3-c)/(something with c and x) = d To make the product invariant of x, the second term should be 0. This implies c=3, d=1/4

@abdullahbarish8204 7 ай бұрын

Amazing

@Dark_rider09461 7 ай бұрын

Hello there I love your math videos so much specially the ways you solve them makes me want to watch your videos more❤ You teach even better than my teach😂

@PrimeNewtons 7 ай бұрын

Thanks for such kind words

@johnka5407 7 ай бұрын

How do we get an assumption that 1 = 4d? If it was 1 = 4d + 2 and (c - 4)² = 4d(c - 2)² + 2 the equation would be true, but c and d would be different.

@MightyBiffer 7 ай бұрын

That is where I got stopped as well.

@annacerbara4257 7 ай бұрын

Because x^2 must go away from the equation that has to be not dependent from x.

@johnka5407 7 ай бұрын

@@annacerbara4257 but why 1 = 4d, look at the example I gave.

@BarrieHughes 7 ай бұрын

Good solution

@holyshit922 7 ай бұрын

11:31 yes and thats why i like difference of two squares in this case

@Nathan-ix9ov 7 ай бұрын

Professor can you please suggest some books which helps drastically for IMO preparation ( not SOF)

@blackovich 7 ай бұрын

You said it's easy, but you still had to do quite some work. Nothing good comes easy, my guy.

@salad8176 7 ай бұрын

Just because something is long, doesn't mean its going to be difficult.

@aura_beast2678 7 ай бұрын

Dude this comment suggests that you are still immature

@theupson 7 ай бұрын

i used the same reasoning as euclids lemma to infer that (x+c-2) divides either (x+c-4) or (x-c+4). 2nd edit: fundamental theorem of algebra does this for us.

@ramunasstulga8264 7 ай бұрын

How to evaluate logarithm inside a logarithm? For example log_0.8(log_144(288×3^[1÷2]))? _0.8 is the base and _144 is also the base. Using calculator the outcome is -1 but I don't want to always use a calculator for this kind of questions 😭

@Jari1973 7 ай бұрын

Really great.. Also suggestions for answers.. But isn't the solution to this based on memory, you Only remember how to do the solution??

@user-if5do8md6n 7 ай бұрын

thanks sir

@user-zv6fk4ed1o 5 ай бұрын

Вы молодчина.

@wildwillberry 7 ай бұрын

I tried this using variable substitutions for the two function inputs, t = c - x and u = c + x. This results in t = -u. I multiplied f(x) f(-x) and then solved for d and got d = 1/4. This is only works if c = 0 is also a solution. Am I missing something?

@user-yd4ky5vb3w 7 ай бұрын

Thanks for an other video....

@SuperAnouri 5 ай бұрын

don't need expand at bottom of video. rule is c-4 = c - 2 or c - 4 = -(c-2)

@TheEulerID 7 ай бұрын

I had two ways of doing it. First was to note that the expansion could be written as the the difference of two squares at the top and bottom. Hence you get ((4-c)^2-x^2)/((2c-4)^2-4x^2). Now note that we have a dividend of the form A-x^2 and a divisor of the form B-4x^2 where A & B are constants. In order for that to be invariant with the value of X, then we need to pull out a common factor 4 from the divisor so it is of the form 4(B/4-x^2). As the dividend is of the form A-x^2, it can be seen that this division will only produce a constant when B/4 = A or, re-arranged, that B = 4A. As A=(4-c^2) and B=(2c-4)^2, then (2c-4)^2 =4(4-c)^2. Expanding gives us 4c^2-16c+16=64-32C+4c^2. Simplify, and we get 16c=48, or c=3. Substitute that into the formula earlier and you find d=1/4. The other way is to differentiate the expanded expression with respect to x and you find that the gradient will be 0 when 2(c-3)=0, or c=3. Which amounts to much the same thing.

@richardbraakman7469 7 ай бұрын

I made a transcription error from one line to the next and ended up deep in very complicated expressions. Do you have any advice for avoiding such errors?

@isaiaholaru5013 7 ай бұрын

Read the question, write down the first line of your working, then read it again, making sure to match your signs and values. With enough practice, this only takes a few seconds with the back-and-forth

@theupson 7 ай бұрын

take smaller steps. if your expressions completely changes appearance between lines, because you're doing a bunch of simplifying in your head as you go, errors are both likelier to happen and harder to notice. if you for instance do distributing in one step and combining like terms in another, as you are contemplating the _next_ step it's easy to quickly check for bugs in the _last_ step.

@thiagodonascimento7926 7 ай бұрын

I don't get how you found d. Where does this x^2 = 1 come from, and why can you compare those coefficients?

@harrymetu2746 7 ай бұрын

First!!❤

@zakinaqvi998 4 ай бұрын

The extreme right side is entirely invisible in the video.

@playgirl7305 7 ай бұрын

What if f(x) male and f(c+x) *f(c-x) =d when c is female? Would 'd' be male or female child.

@iMvJ27 7 ай бұрын

Q.) A value of x satisfying 85x ≡ 45(mod 15) is 35 10 25 15 How to solve such question sir ??

@annacerbara4257 7 ай бұрын

x=3k k=1,2,3 ... so the solution is x=15.

@cliffordabrahamonyedikachi8175 6 ай бұрын

C equals to 3. D equals to 1/4.

@nanamacapagal8342 7 ай бұрын

12:06 For those who want to see how it would look like with absolute values, here: (c-4)^2 = (c-2)^2 |c - 4| = |c - 2| Now since there are two absolute values, there's four cases to worry about: 1) +(c - 4) = +(c - 2): no solution. 2) +(c - 4) = -(c - 2): c = 3. 3) -(c - 4) = +(c - 2): c = 3 (same as #2) 4) -(c - 4) + -(c - 2): no solution (same as #1) The only way this would work is if c = 3

@theupson 7 ай бұрын

going from a trivial polynomial to a multiply-piecewise abomination takes a special kind of hate :). difference of perfect squares ftw

@Shirobanasumire 7 ай бұрын

But what happens if ( C - 4 )² is not equal to ( C - 2 ) ² when you compare coefficients ?

@blackovich 7 ай бұрын

It cannot happen if the original equation was true.

@ahnafhasankhan2781 7 ай бұрын

2nd

@rescyy2235 7 ай бұрын

Once reaching (c-4)^2=(c-2)^2, I recommend solving this in the following way, a^2=b^2 => a=b or a=-b; same way c-4=c-2 leading to no solutions, and c-4=2-c giving us c=3. It's the same result but it's easier because you don't have to expand the parantheses

@user-gi5ix6zr8f 7 ай бұрын

عالی بود