How to use the quadratic formula backwards! How can I find x? Reddit r/askmath

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We will solve a very creative square root equation that is in the quadratic formula form. The equation is -2=(x+-sqrt(x^2+48))/12. I will show you two ways to solve this equation. The first way is to "just solve it" and the second method is to use the quadratic formula backward to write a quadratic equation. This question is from Reddit r/askmath: / bidhkyw8vd
0:00 How can I solve for x?
0:14 "Just solve it"
4:09 Use the quadratic formula backwards
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#math #algebra #mathbasics

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27 июл 2024

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Комментарии : 132

@bprpmathbasics 5 месяцев назад

7:51 好累

@TheLobsterCopter5000 5 месяцев назад

Google says this means "so tired". idk if Google is correct though...

@timetocompose1885 5 месяцев назад

@@TheLobsterCopter5000correct

@shubashuba9209 5 месяцев назад

This guy: deals with complex numbers✅️ Solves difficult geometry problems✅️ Understands power and logarithms✅️ Struggles with 24 × 24 ❌️

@johnmarcusengreso8273 5 месяцев назад

- Every person who loves math ever

@Patrik6920 5 месяцев назад

..happends lol,

@krimbus1236 5 месяцев назад

I’ve figured out a few tricks to square 2 digit numbers easily. First one is using quadratics to find the next square up or down since (x+1)² =x²+2x+1 so using this logic 24²=(25-1)²=625−2(24)+1=576. The next method is doing the same but for all 2 digit numbers, so 24²=(20+4)²=400+2(20×4)+16= 416+160=576

@ishaankumar4587 5 месяцев назад

@@krimbus1236you didnt figure it out its a standard approach for people in small classes

@krimbus1236 5 месяцев назад

@@ishaankumar4587 yeah but I wasn’t taught it. I’m not claiming I discovered it I just figured it out

@YaBoyIsBack 5 месяцев назад

"Let me know which wa-... yeah" -bprp

@ailblentyn 5 месяцев назад

This was a fun problem. I did it by recognizing that it looks like the quadratic formula. I’m happy he saw my way as a possibility!

@user-xm5ku2fv3p 5 месяцев назад

you have taught me maths better than any teacher i have ever encountered!!!! you have blown my head away when i watcher you vid for the derivation of the quadratic formula😂😂😂😂😂😂😂😂

@bprpmathbasics 5 месяцев назад

Watch this video next: This problem deceived me so hard, am I stupid? Reddit r/HomeworkHelp ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-jCIfoKaHBX4.html

@jamescollier3 5 месяцев назад

very cool. I'm thinking there will be a follow up, based on the few comments so far. Please let me add one: How do you know t=-2, maybe it's +12, or some other multiple of +-(12 ×2)? Of course, 12 would change too. Earned a subscription to your "easy" channel too. Lots of BPRP.😂

@F1r1at 5 месяцев назад

You saw the 48 on the right side, so instead of multiply 24*24, you could just write it as 48*12, and just divide everything by 48 at the end, to get x + 12 = 1.

@nitsujri 5 месяцев назад

It's not about short cuts. If the situation is different you want to know the process. He's a good teacher.

@F1r1at 5 месяцев назад

@@nitsujriI believe like everybody at the stage of such equations are able to do simle multiplication. But teaching to see for such shortcuts might be usefull in lot's of other cases

@yurif0331 5 месяцев назад

@@nitsujriare you Dumb?

@monke7566 5 месяцев назад

@@F1r1atnot always, and this is the basic maths channel

@synth3431 5 месяцев назад

@@yurif0331I don't know about them but I am dumb, so I appreciate when he doesn't take shortcuts and just shows the process that can work for any similar question.

@diatonicdissonance 5 месяцев назад

my mans gave up at the end of the video lol

@ytsimontng 5 месяцев назад

Fun question! I made a match using factoring, all the same ofcourse. -2 is the solution to the quadratic a(t+2)(t-p)=0 Matching the equation, a=6 6(t+2)(t-p)=0 6t^2+(12-6p)t-12p=0. Matching the equation, 4ac=-48 so c=-2 -12p=-2 so p=1/6 . 12-6p=11=b=-x so x=-11

@rasnauf 4 месяца назад

This keeps coming up in my video feed, over and over again, never ending. Very peculiar.

@vertechua 5 месяцев назад

I liked the second way, since instead of alpha and beta, you just have one variable (t, for case), this suggests that the quadratic so formed must be a Perfect square as well

@mccask 5 месяцев назад

For the 24x24, we have memorized 12x12, so take 144 and multiply by 4 (or double twice).

@AzureKyle 3 месяца назад

The first thing I noticed by looking at the problem was that it was very similar to the quadratic formula (-b+/-sqrt(b^2-4ac))/2a. By working it out a little, you can see that b=-x, because the problem shows x+/-, instead of -x+/-, and negative times negative is positive. And for the b^2 portion, they don't show (-x)^2, because a negative times a negative is positive, so it doesn't matter as much for that part. Then you have the -4ac=48 and 2a=12. With the latter you can figure out that a=6 (12/2=6) and with that you can also figure out c with the former 48/-4=-12. -12/6=-2, so c=-2. If we replace x with -b the original polynomial becomes 6x^2-bx-2=0 And, if we view the problem as the quadratic formula, it shows that the x in the polynomial is equal to -2. So replace the x in the polynomial and you get 6(-2)^2-b(-2)-2=0 Work it out to get 24+2b-2=0 then 2b+22=0. Finally subtract 22 from both sides and you get 2b=-22 and divide both sides by 2 for the solution of b=-11, which results in our original x in the quadratic formula to be -11. And if you plug it back into the original formula, you get (-11+/-sqrt((-11)^2+48))/12. -11^2=121 121+48=169. sqrt(169)=13. And now we have (-11+/-13)/12. So, on one hand, we have (-11+13)/12, and on the other we have (-11-13)/12. The first one becomes -2/12, which can simplify to -1/6 which is not our answer, and the second becomes -24/12 which can simplify to -2, which is our answer.

@jensraab2902 5 месяцев назад

Small suggestion for the first way to solve it. Given that we have 48 under the root, it is better to not calculate the 24² because we can cancel by 24 and make all number much smaller: -(x+24) = √(x²+48) [I also factored out the minus sign to make the calculation more straightforward]] (x+24)² = (x²+48) x² + 48x + 24² = x² + 48 48x + 24² = 48 → this is where we can cancel with 24 and get nice small numbers! 🙂 2x + 24 = 2 x + 12 = 1 x = -11 All that said, the second way to solve it is a lot more exciting! 😁

@subhadeepjana5354 5 месяцев назад

R.H.S. and L.H.S equals to one of the roots of the quadratic equation 6y^2 -xy -2 = 0 As you can see the RHS is in the format of Sridhar Acharya Formula of quadratic Equation root (-b +- √(b^2 -4ac))/2a Anyway, now as the LHS is also one of the roots of the quadratic equation. Just put y= -2 And you can get 6(2^2)+2x -2 = 0 Or, x = -11

@richardhole8429 5 месяцев назад

Since the original equation has +- should we not have two solutions? When you squared both sides you lost one solution. But I don't see how to solve for the other.

@jusfralmedina2979 5 месяцев назад

Remember that in this exercise we are trying to find x (AKA: -b), not both of the solutions of the quadratic equation. But anyway, replacing x with -11, we have: First solution: (-11+sqrt((-11)^2+48))/12 = (-11+sqrt(121+48))/12 = (-11+sqrt(169))/12 = (-11+13)/12 = 2/12 = 1/6 Second solution: (-11-sqrt((-11)^2+48))/12 = (-11-sqrt(121+48))/12 = (-11-sqrt(169))/12 = (-11-13)/12 = -24/12 = -2 So t = {1/6; - 2} are the two solutions of the quadratic equation, though we were looking for x

@rickhole 5 месяцев назад

@@jusfralmedina2979 Thank you. I am thinking that the x value (-b) will have a different value when the + or - radical are used, and thus there will be two values for x. I think you have covered both values.

@robinsparrow1618 5 месяцев назад

THAT'S WHAT I WAS WONDERING! i can't figure out how to get a value of X which returns -2 if you're adding instead of aubtracting

@senseof_outrage9390 5 месяцев назад

Totally unprofessional opinion here... The 'Fundamental Theorm of Algebra' states that there is exactly the same number of solutions/roots as the highest exponent in the equation so: x-2=0 has one solution x²-2=0 has two solutions x³-2=0 has three solutions Etc... Although the highest exponent in the equation is the x² term, it is under the radical, so is really only an x term Hence only one solution The ± is the equation might be there to test the students' knowledge...? And so they have to identify which of the two options provided is correct. Just my 💭

@robinsparrow1618 4 месяца назад

okay, i went through the process of trying to solve for a complex solution and all i got was: a+bi = -11 so i guess it really is the only solution?

@sleha4106 5 месяцев назад

Daniel, The Cooler Daniel as mathematical methods abridged version

@themammoth67 5 месяцев назад

Is methods this easy? Lets gooo

@IsYitzach 5 месяцев назад

I immediately jumped to the second solution because I knew the origin, but I knew there was no way I could do it my head. It requires too much working memory for that. I just wanted to see it done.

@PedroHenrique-vs3mf 5 месяцев назад

Bro the goat of math lmao

@Mike__B 5 месяцев назад

It's probably because it's early and I haven't been properly caffeinated yet, but that second method where you recognize the quadratic equation (which I did at the beginning), it seems to be that a,c & t could have an infinite combination of values. For instance 'a' could equal 12 instead, which would make c half as large, as well as t. I mean I guess that doesn't matter because any solution for t,a,c that you use will get you to the right b value, but it's just something that bugs me when it probably shouldn't... ok I'm off to get caffeine

@darranrowe174 5 месяцев назад

Because of the /2a matching the /12, this means that a could only ever be 6.

@Mike__B 5 месяцев назад

@@darranrowe174 I think you're missing what I was getting at. You can "simplify" the given equation by multiplying or dividing both sides by any value you want, as such there are an infinite number of combinations of t,a,c that are perfectly valid. e.g. I could multiply both sides of that quadratic by 2, then 12t²+11t-1 = 0 is a perfectly valid expression as well to give x = -11 (b=-x), then for t = 1, a = 12 & c = -1 Like I said, no caffeine when I wrote that, it doesn't bug me as much now.

@phiefer3 5 месяцев назад

@@Mike__B While true, this really doesn't matter. It doesn't change the relationship between the variables and still results in the same solution. Your complaint is literally the equivalent of being given a quadratic equation such as: 12x^2 + 3x - 14 = 0 Seeing someone break out the quadratic formula for it using a=12, b=3 and c=-14; and then complaining about their arbitrary choices for a,b and c, because they could have used 24, 6 and -28 instead. Yes, with any equation you can introduce whatever factors you want, it doesn't change anything. That's sort of the whole point of equations.

@blacksnow7106 5 месяцев назад

The professor watched that video of flammablemath and thought: "what if we start there and work backward lol"

@duegia44 5 месяцев назад

Careful when you trying to square both sides of an equation to obtain a solution, sometime it will introduce an extra solution that is not a solution of the original equation. In this case, if you multiply both the negative and positive expression on the right side with each other, then compare it to its supposed value, you will see that it has no solution. Basically -11 is the extra solution that was introduced by squaring both sides.

@ethohalfslab 5 месяцев назад

The reason it has "no solution" is because the original equation can never be 'always' true. The LHS is just a single number, and the RHS has a +/-. So the LHS can only ever be 1 thing, and the RHS can be two different things. So they can never be always equal. The assumption made in the video was the +/- sign was just going to resolve into one of the two being correct, not both.

@aditaggarwal3486 5 месяцев назад

@@ethohalfslab+/- does mean 'plus or minus' doesn't it?

@ethohalfslab 5 месяцев назад

@@aditaggarwal3486 Yes it does

@duegia44 5 месяцев назад

@@ethohalfslab was that assumption ever made? I don't think I saw it in the video.

@ethohalfslab 5 месяцев назад

@@duegia44 The assumption was made by attempting to solve the problem. Otherwise he would have just said "There's a +/- on one side but not on the other, so there is no solution."

@iAzazelHD 5 месяцев назад

Since "-x-24" = +- sqrt(), Wouldnt x+24 = +- sqrt() as well?

@abysslight2490 4 месяца назад

So, here’s the thing. If you plug -11 back in for x, you end up with the statement “-2=-2 or -2=1/6”. Which, while technically true, isn’t really useful. You don’t need the plus or minus there, only the minus, and it would still be recognizable as something derived from the quadratic formula.

@thegreatbambino3358 5 месяцев назад

This is wrong. There is not solution, plug your answer back in. You get -2 = -11 +/- sqrt((-11^2) + 48) / 12. Given that the value of the radicand is sqrt(121 + 48) = 13 you have -2 = (-11 - 13) / 12 or -2 = (-11 + 13)/12. The left side is always -2 but the right side can be either -2 or 1/6, thus you can't put an equal sign there. Your problem came as it always does, when blasting away the plus or minus with squaring and retaining the = sign.

@mikeyk730 5 месяцев назад

Completely agree. There is no solution to the original equation, - 2 = (x +- sqrt(x^2 + 48)) / 12, since there is no value of x that satisfies both the plus and minus case

@thegreatbambino3358 5 месяцев назад

@730 I realized this when I saw the initial problem statement in which I reasoned that for this to be true, the radicand had to be 0, i.e. only a single solution since + or - 0 is just 0, meaning x would have to be sqrt(-48), which would obviously not work with the x outside the radicand. I was curious how he would solve this problem, and instead he just got it wrong.

@pacattack2586 5 месяцев назад

Shouldn't there be two answers? This answer only seems to work on the negative side of the +/- but doesn't work on the positive side. Of course I also suspect the second answer may involve imaginary numbers, but even those I can't seem to figure out how to work. Edit On second thought the + side requires the square root to be *less* than 0, so this is absolutely a second answer that's not a real number, but even imaginary doesn't work because you get left with a bit of error as there is at least one i you just won't be able to get rid of if you try standard complex numbers I'm now asking if there's something that maybe involves i^3 for the positive answer?

@bvssmouq6gamingofficialyt 5 месяцев назад

I mean positive*positive and negative*negative both equal a positive. So there would only be one

@bvssmouq6gamingofficialyt 5 месяцев назад

Keep in mind we aren't solving for t here.

@bvssmouq6gamingofficialyt 5 месяцев назад

So the +/- doesn't really apply to the a b(x) and c values

@hhhhhh0175 5 месяцев назад

i think it's because the square root is multi-valued. you can subtract the positive branch or add the negative branch

@captianmorgan7627 5 месяцев назад

3:08 So why do you first subtract 576 from each side instead of dividing by 48? The answer is obviously different but it seems more simple and there is nothing that I know of that would keep you from doing it....?

@Thepotatoz. 5 месяцев назад

so it would just be the same answer because you would get x+12 = 1 then x = -11 but its just the way to keep things simple by just isolating x and only diving 1 number instead of two (your way would divide 576 and 48 by 48 instead of just -528)

@captianmorgan7627 5 месяцев назад

I suppose this is why math and I don't get along. I see (48x+576)/48 as = to x+576 not x+12.

@Thepotatoz. 5 месяцев назад

@@captianmorgan7627 well your original question was still correct it's just a bit simpler, and the reason why we do it his way is exactly what you just mentioned, you see it as x+576 and not x+12, by doing it first you avoid that confusion as a whole. Don't worry about these small mistakes, being able to ask questions and listen to answers shows that you and math do get along and that you are actually really smart

@alejomdp 5 месяцев назад

The only problem I find is that x=-11 only works when using the negative sign before the root in the original equation. When I replace x=-11 but using the positive sign before the root, it doesn't verify. It results in -2=1/6, which is false. I can't understand why that is, it's like x=-11 is like a half-answer. I know that the absolute value definition must have something to do with this.

@epikherolol8189 5 месяцев назад

That's because -2

@chitlitlah 5 месяцев назад

There's no number that will work when doing the positive version. The square root function will always give a positive result no matter what x is, so x will have to be negative since two positives can't add up to -2, but the number given by the square root will always be larger than the absolute value of x and a large positive number plus a small negative number will be positive, hence it can't be -2.

@alejomdp 5 месяцев назад

@@chitlitlah @epikherolol8189 So, the problem should've been divided into TWO problems, one with the negative sign and giving an answer of x=-11, and the other with the positive sign giving no solutions (I tried complex ones but with no success). That's why I considered x=-11 to be a half-answer, I didn't want to say there were more solutions, no, I wanted to say that x=-11 only solves one half of the problem. That's why I consider there are originally TWO equations, not one. One with a solution, the other one with no solutions.

@chitlitlah 5 месяцев назад

@@alejomdp I see what you mean. It appears that it didn't split into two problems when he was solving it algebraically because when he squared both sides, he got a couple of x^2 terms that canceled each other out leaving a linear equation. If b had been positive x instead of negative x, he would have ended up with a 2x^2 that could have resulted in two answers. It's like if you had the equation (x + 1)^2 = (x - 1)^2, you could just square everything and then most of the terms would cancel out so you just have 4x = 0, no complex answers or DNE answers. If you started with (2x + 1)^2 on the left, it wouldn't cancel so nicely and you might end up with two answers.

@chitlitlah 5 месяцев назад

I take back what I said above about if b was +x, you could end up with two answers. Of course the sign is going to go away when you square it anyway, so it's always going to cancel out nicely. It just seems like using the quadratic formula in this way, it's guaranteed to work out into a single answer.

@albertreynespuig4076 5 месяцев назад

when you have 48x+576=48 why dont you divide the whole expression by 48

@mathmachine4266 5 месяцев назад

You subtract x and THEN you square it. Isolate the square root at all costs.

@shaind 5 месяцев назад

but x=-11 doesn't work for (x+sqrt(x^2+48))/12=-2😢

@MarioFanGamer659 5 месяцев назад

But it does work for (x-sqrt(x^2+48))/12=-2. Notice that the equation has a ± init, not just a +.

@darranrowe174 5 месяцев назад

I think it is easy to miss, but what you are calculating is the coefficient of the x term. The way this works is that you use one of the two roots as a valid value. It would have worked just as well with 1/6, that would have the + side of the sqrt that you mentioned.

@mikeyk730 5 месяцев назад

@@MarioFanGamer659The ± is what makes the original equation have no solution. There is no value of x that satisfies both criteria

@basedhutao775 4 месяца назад

Beautiful

@amoghsod2212 5 месяцев назад

Well if you try , putting x = -11 in the equation, and you first open the root with +ve sign you get -2 = 1/6 😂

@Kd-FunBruh 5 месяцев назад

Exactly bruh...

@K0nam3Two 5 месяцев назад

Quadratic equations can have 2 answer and sometimes only one is correct...

@AkitoLite 5 месяцев назад

(x±√x²+48)/12 needs to have 2 answers, because x/12 is the mid point and (√x²+48)/12 is the distance. There cannot be an answer so that it is only =-2 since there is ADDITION in the sqrt, and x² cannot be negative unless imaginary number, however then the x outside would be imaginary aswell, unable to obtain -2. What I said is basically proof by contradiction.

@forbidden-cyrillic-handle 5 месяцев назад

Yep. There's no solution. The equality sign doesn't hold fir both cases. Basically the square root should be zero so it makes any sense. But then x is complex and left side isn't complex, so I don't know what to say...

@K0nam3Two 5 месяцев назад

@@forbidden-cyrillic-handle X is not complex

@Ahmad-yi6d 5 месяцев назад

👍

@ChavoMysterio 5 месяцев назад

This means the original problem would be x²+22x+121=0

@sans1331 5 месяцев назад

third

@Lovuschka 5 месяцев назад

I think x is that thing on the right side of = and also again on the left side of ². I hope this helps you to find x.

@thejaegerbomber99 5 месяцев назад

There are two issues with that problem. First the equation doesn't even have the proper form of a quadratic formula. Second, that equation should only represent one root, so the +- doesn't make any sense. If I use the equation as is, the the associated quadratic equation is 6z^2 + xz - 2 = 0, and for x = -11, z = -2 isn't a root for the aformentioned quadratic equation. Now, if I use the proper form of the quadratic formula, then the quadratic equation is 6z^2 - xz - 2 = 0, and for x = -11, z = - 2 IS a root for the quadratic equation. But keep in mind x = -11 is only valid for only one root.

@darranrowe174 5 месяцев назад

What do you mean by "doesn't even have the proper form." Starting from z = (-b +- sqrt(b^2 - 4ac) /2a) with a = 6, b = - x and c = - 2, I calculate this to be z = (- (- x) +- sqrt((- x)^2 - 4(6)(- 2)))/2(6) => z = (x +- sqrt(x^2 + 48))/12. What other form of the equation are you using? Edit: Just to go all the way. Using the values of a, b and c in the general quadratic equation, we get y = 6z^2 - xz - 2. If we let z = - 2 when y = 0, we get 0 = 6(- 2)^2 - (- 2)x - 2 => 0 = 24 + 2x - 2 => 0 = 22 + 2x => -22 = 2x => -22/2 = x => x = -11. If we substitute this for x in our quadratic equation, we will get y = 6z^2 - (-11)z - 2 => y = 6z^2 +11z - 2. Both -2 and 1/6 are roots for this.

@deepeshsinghthakur6427 3 месяца назад

When a Westerner gives equation, it is known by his name But when it is given by an Indian🇮🇳, you just say Quadratic Formula It was actually given by Shri Dharacharya Ji And we call it Shri Dharacharya Formula

@MightyBiffer 5 месяцев назад

There is something wrong with your solution method. If we started with the equation minus 2 = (x + (x^2 + 48)^.5)/12 - which is just looking at the + of the +/- and then solved it exactly as you did above: -24 = x + (x^2 +48)^.5 Mult by 12 -x -24 = (x^2 +48)^.5 Sub x x^2 +48x +576 = x^2+48 Squaring both sides 48x +576 = 48 Sub x^2 48x = -528 x = -11 Dividing by 48 Well, we get the same solution. But if we check the solution with the plus side it fails. in fact, the limit of (x + (x^2 + 48)^.5)/12 as x goes to negative infinity is about 0. So there should be no solution to the equation with the plus sign out of the +/- part. But your solution method finds one.

@OMGclueless 5 месяцев назад

When you square both sides of an equation, you might add an extra solution. That's what happened here, we added a solution to the positive equation that doesn't actually work. Not sure why BPRP didn't go back and check his solution with the original equations, it's important to do that whenever you square both sides of an equation because it's possible the result you get doesn't actually work with *either*.

@AndreChaosweapon 5 месяцев назад

@@OMGclueless the equation where this one solution is 2 is 6x^2-11x-2=0 and if you look for the solutions you'll see is 2 and -1/6

@OMGclueless 5 месяцев назад

@@AndreChaosweapon The x in your equation there is not the x in the original equation (this is why BPRP named it "t" while solving). t has multiple solutions because it comes from a quadratic, but the only one that's relevant is t=2 as the problem writer has chosen that one to write down in the original equation.

@darranrowe174 5 месяцев назад

@@AndreChaosweapon Actually, the quadratic equation you get from this is y = 6x^2 + 11x - 2. What you are calculating is -b, not b.

@darranrowe174 5 месяцев назад

@MightyBiffer @OMGclueless That is because the + side of the square root was unused. The usage of the quadratic formula was used as a compare the coefficient method of giving two of the three coefficients. The -2 (which is the negative root) was then used to give the value of b. Try doing this with 1/6 instead of -2. You should notice two interesting things. First, it will give you the same quadratic equation back y = 6x^2 + 11x -2. Second, this will work with the positive root but not the negative.

@builderdog3875 5 месяцев назад

So he, Struggled with subtraction and basic multiplication, But completed a challenging division problem within a matter of seconds?

@sangram4627 5 месяцев назад

a + b + c = 6 a² + b²+ c² = 14 a³ + b³ + c³ = 36 ab + bc + ac = 11 abc =6 Values of a , b and c are ? Help me, solve this

@devanshthepro9865 5 месяцев назад

a=2,b=1,c=3

@OscarMorales-wn7ql 5 месяцев назад

Good morning, Mr. presenter of this block! Firstly, in this presentation it is seen that the modern mathematics introduced since the end of the 60s of the last century has not arrived. For a start: The concept of an equation that you have is wrong. Second: What you have written does not express anything in Mathematics; You have to know how to use the symbolic language of mathematics, if you think and affirm that it is similar to the formula to find the solutions of a second degree equation, you are wrong, I suggest you see a high school book where the topic of quadratic equations and you will see the error you make. Do you know which one it is? It is a consequence of poor learning of the most basic things that methodically and didactically should be avoided. Look at this exposition that makes it something that for students to be mathematics teachers would serve them as conceptual errors are made in mathematics, which in turn brings pedagogical and methodical errors in the teaching of mathematics at a basic level. Excuse the harshness of my comments, it is for you to review what you are doing in this case, it is a meaningless digression, but it carries an increased value of how. You should not do mathematics without having the basic concepts of mathematics well defined. We must avoid generating more illiterate people in these times. Good day! Another thing, a formula is not an equation.