Something ive noticed is in the USA it seems they use the quadratic formula right away, while where i live we first calculate b^2 - 4ac to check if its a negative or a positive number (since if its a negative the equation has no solutions unless youre working with complexe numbers) Is there any reason the USA (or wherever) do this or is it just to make it easier for students to understand?
for p-q formula you have to break it down to a lonely x² (plusminus px plusminus q) and you are allowed to divide to reach it. the common rule is to abstain from divisions of additions or substractions; p-q is a rare excusion :)
Is there a real reason why you would use the PQ formula over the ABC-formula? I mean it can do everything the abc-formula can but it's less capable. If you don't have 1x^2 you have to put in more work to rewrite your expression so it works with the pq. Additionally the abc formula has (-b+-sqrt(delta))/2a if delta is 0 you have -b/2a which I think is a good reason why you shouldn't use the pq as your default formula.
•The extra work you put in by using the pq formula is saved later as you wont end up with big numbers. Thats because it’s basically the same as the abc: assume a is 1, shorten the abc formula and you‘ll end up with pq …Meaning the equivalent of saying its extra work to divide by x^2‘s coefficient would be criticizing to have to divide by 2a in the abc‘s case. •-p/2 +- sqrt(0) will also result inthe result just being -p/2 so this isn’t a reason to not use pq either => I personally think you have way less to write down when using pq and it looks better and more structured too… But thats just me and i assume you like better what you learned first and they are both equally useful.
The pq-formula can be rewritten as x = [ -p ± √(p² - 4q) ]/2 or x = -½p ± ½√(p² - 4q) I don't see why it should give any problems when "delta" equals 0. Furthermore, in general mathematical analysis, there is some usefulness to (re)writing polynomials as _monic_ polynomials (i.e. with leading coefficient equal to 1). However, I myself have just the "abc-formula" memorised (although I rarely use it; instead, I tend to use factoring or "completing-the-square").
@@yurenchu I didn't mean to say that a delta of 0 causes problems in the pq. I was simply pointing out that if delta is 0 in the abc formula it gives you x=-b/2a. I like that because x=-b/2a is the formula to calculate the highest/lowest point of a parabola and you simply add/substract the delta to obtain the 0 points of set polynomial. I think it's pretty because you can easily see the how the abc formula works in this regard.
Question : Is it possible to find a system to find the terms in an ap of certain digits? For eg - If a is a single digit number, and d is a 2 digit number, is it possible to find a formula that allows us to get the, say first 3 digit number? Like is it possible to find a formula for situation like these, except for every numbers? I guess any nonnatural numbers cn be discarded as AP can only hold natural terms (not numbers, terms)
Sometimes, if you allow the use of the ceiling function. First determine the formula for the arithmetic progression. Second, _find the inverse of this formula._ Arithmetic progression formulae are linear functions so they have well-defined inverses. Apply that inverse to 10^(k-1) where k is the number of digits you require. 10^(k-1) is the smallest number with k digits, so we use this as starting point. Applying the inverse will therefore get us somewhere close. _Take the ceiling of this value,_ call it n. Find the nth value of the arithmetic progression with the original formula. You are done. Edit: (And why I changed the first word from "Yes" to "Sometimes") This might not work if the common difference is greater than the number of digits we're looking for. e.g. try finding the 2-digit element of the progression 1, 101, 201, etc. The same trick can be used wherever a formula or function has a well defined inverse over an interval.
Why would nonnatural numbers be discarded? AP can contain any numbers. First (positive) n-digit number is a + d*ceil((10^n - a)/d) Works for any a and positive d. EDIT: "n-digit number" here interpreted as "a number with n digits before the decimal point". EDIT 2: I see I made some errors. Just as the other poster said, sometimes it won't work, for example, if d has n digits or more than n digits, then it's likely not going to work (although the formula would work to find the (positive) value with _at least_ n digits). Secondly, my formula above finds the first (n+1)-digit number, not the first n-digit number.
When I read this comment, I thought "Well, just multiply the trinomial by 4 or some other suitable even number", but as it turns out, even that would never work.
Quadratic formula and Completing the Square same thing. Because derivation of Quadratic After done by completing the Square so don't worry But you are general the root or solution By x=(-b(+, -) √{b^2-4ac}) /2a Of course a not be Zero(0) Your pq formula are irrating when p and q are fraction
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Teim
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What are the name of other
Are there any else?
Thank you!
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Decimal? Ti-84 put x^2+7x-11 into y1 and zero into y2 and use the intersect function to find where the parabola crosses the y-axis.
Something ive noticed is in the USA it seems they use the quadratic formula right away, while where i live we first calculate b^2 - 4ac to check if its a negative or a positive number (since if its a negative the equation has no solutions unless youre working with complexe numbers)
Is there any reason the USA (or wherever) do this or is it just to make it easier for students to understand?
for me, i just use factorization unless question states to give answer in decimal places
for p-q formula you have to break it down to a lonely x² (plusminus px plusminus q) and you are allowed to divide to reach it.
the common rule is to abstain from divisions of additions or substractions; p-q is a rare excusion :)
I have this in school atm so these videos really help
creo pfp based
@@evilsmileyface aura my beloved
@@evilsmileyface and sphere, just haven't updated it yet on here lmfao
@@RemecArt in my case i’ve never gotten bored of crazy
@@evilsmileyface that's also a really good one
Is there a real reason why you would use the PQ formula over the ABC-formula? I mean it can do everything the abc-formula can but it's less capable. If you don't have 1x^2 you have to put in more work to rewrite your expression so it works with the pq. Additionally the abc formula has (-b+-sqrt(delta))/2a if delta is 0 you have -b/2a which I think is a good reason why you shouldn't use the pq as your default formula.
A modified version is very useful to quickly calculate Eigenvalues of 2x2 Matrices
•The extra work you put in by using the pq formula is saved later as you wont end up with big numbers. Thats because it’s basically the same as the abc: assume a is 1, shorten the abc formula and you‘ll end up with pq …Meaning the equivalent of saying its extra work to divide by x^2‘s coefficient would be criticizing to have to divide by 2a in the abc‘s case.
•-p/2 +- sqrt(0) will also result inthe result just being -p/2 so this isn’t a reason to not use pq either
=> I personally think you have way less to write down when using pq and it looks better and more structured too… But thats just me and i assume you like better what you learned first and they are both equally useful.
Pq is super easy and is really easy to simplify 2.
The pq-formula can be rewritten as
x = [ -p ± √(p² - 4q) ]/2
or
x = -½p ± ½√(p² - 4q)
I don't see why it should give any problems when "delta" equals 0.
Furthermore, in general mathematical analysis, there is some usefulness to (re)writing polynomials as _monic_ polynomials (i.e. with leading coefficient equal to 1).
However, I myself have just the "abc-formula" memorised (although I rarely use it; instead, I tend to use factoring or "completing-the-square").
@@yurenchu I didn't mean to say that a delta of 0 causes problems in the pq. I was simply pointing out that if delta is 0 in the abc formula it gives you x=-b/2a. I like that because x=-b/2a is the formula to calculate the highest/lowest point of a parabola and you simply add/substract the delta to obtain the 0 points of set polynomial. I think it's pretty because you can easily see the how the abc formula works in this regard.
Instead of factoring, use the method of "completing the square":
x² + 7x - 11 = 0
x² + 2*(7/2)*x = 11
x² + 2*(7/2)*x + (7/2)² = 11 + 49/4
(x + 7/2)² = 93/4
x + 7/2 = ±√(93/4)
x = -7/2 ± (1/2)√93
Done!
Alternatively:
...
(x + 7/2)² = 93/4
(x + 7/2)² - (√(93/4))² = 0
... recall a² - b² = (a+b)(a-b) ...
(x + 7/2 + √(93/4)) * (x + 7/2 - √(93/4)) = 0
(x + 7/2 + √(93/4)) = 0 OR (x + 7/2 - √(93/4)) = 0
x = -7/2 - √(93/4) OR x = -7/2 + √(93/4)
x = -7/2 - (1/2)*√93 OR x = -7/2 + (1/2)*√93
Done (again)!
Question : Is it possible to find a system to find the terms in an ap of certain digits?
For eg - If a is a single digit number, and d is a 2 digit number, is it possible to find a formula that allows us to get the, say first 3 digit number? Like is it possible to find a formula for situation like these, except for every numbers? I guess any nonnatural numbers cn be discarded as AP can only hold natural terms (not numbers, terms)
Sometimes, if you allow the use of the ceiling function. First determine the formula for the arithmetic progression. Second, _find the inverse of this formula._ Arithmetic progression formulae are linear functions so they have well-defined inverses. Apply that inverse to 10^(k-1) where k is the number of digits you require. 10^(k-1) is the smallest number with k digits, so we use this as starting point. Applying the inverse will therefore get us somewhere close. _Take the ceiling of this value,_ call it n. Find the nth value of the arithmetic progression with the original formula. You are done.
Edit: (And why I changed the first word from "Yes" to "Sometimes") This might not work if the common difference is greater than the number of digits we're looking for. e.g. try finding the 2-digit element of the progression 1, 101, 201, etc.
The same trick can be used wherever a formula or function has a well defined inverse over an interval.
Why would nonnatural numbers be discarded? AP can contain any numbers.
First (positive) n-digit number is
a + d*ceil((10^n - a)/d)
Works for any a and positive d.
EDIT: "n-digit number" here interpreted as "a number with n digits before the decimal point".
EDIT 2: I see I made some errors. Just as the other poster said, sometimes it won't work, for example, if d has n digits or more than n digits, then it's likely not going to work (although the formula would work to find the (positive) value with _at least_ n digits). Secondly, my formula above finds the first (n+1)-digit number, not the first n-digit number.
There is a generalization of this, no trinomial with only odd coefficients factors because they necessarily sum to an odd number and 0 is even.
When I read this comment, I thought "Well, just multiply the trinomial by 4 or some other suitable even number", but as it turns out, even that would never work.
I always prefer to have the 2c on the top and the -b +-sqrt(b^2-4ac) on the bottom of the formula.
great so you'll get 1/x
@@uggupuggu No, I will get x .
But what if c = 0 ?
😉
@@yurenchu You get 0 and -c/b as the answers.
@@stevencarr4002 No, you don't.
Consider the equation (3x² - 21x + 0) = 0 , for example.
x²+7x-11 = 0
x²+7x+12,25 = 23,25
(x+3,5)² = 23,25
x= -3,5±√23,25
What is this formula known as internationally?
quadratic formula right? its the same name but the real name is- sridharacharya's formula(saint grom india)
I think people call it the A-B-C-Formula
@@markerguy ok
@@dd-di3mz yes it can be
@@homagninandy1222 welcome and the p/q formula is nothing but b=p,c=q and a =1
Quadratic formula and Completing the Square same thing.
Because derivation of Quadratic After done by completing the Square so don't worry
But you are general the root or solution
By x=(-b(+, -) √{b^2-4ac}) /2a
Of course a not be Zero(0)
Your pq formula are irrating when p and q are fraction
You could use PQ formula as it is in the form of x^2+px+c=0
Ye he mentioned that
i got 8.321, -1.321 help
Most likely is that you missed the negative sign in the -b in the formula
@@msolec2000 oh yeah mb
I got -8.32 and 1.32, and I did it all in my head. Being a genius is not as great as it sounds.
Shush