What sort of question is this? I was solving quadratic but factoring does not work. Reddit r/GCSE

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Күн бұрын

How to solve the quadratic equation x^2+7x-11=0 by using the quadratic formula. This question is from Reddit r/GCSE / bgfnko1z2z
"This problem deceived me so hard, am I stupid?": • This problem deceived ...
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Пікірлер: 63

@bprpmathbasics 5 ай бұрын

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@auztenz 5 ай бұрын

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@Jitesh-ek5xf 5 ай бұрын

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@onethegogd5783 5 ай бұрын

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@bprpmathbasics 5 ай бұрын

Thank you!

@the-traveler-8 5 ай бұрын

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@Fenamer 5 ай бұрын

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@nilsalmgren4492 5 ай бұрын

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.

@hyperbroli6672 5 ай бұрын

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?

@Shark9637 3 ай бұрын

for me, i just use factorization unless question states to give answer in decimal places

@rivenoak 11 күн бұрын

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 :)

@yurenchu Ай бұрын

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)!

@RemecArt 5 ай бұрын

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@evilsmileyface 5 ай бұрын

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@RemecArt 5 ай бұрын

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@RemecArt 5 ай бұрын

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@evilsmileyface 5 ай бұрын

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@RemecArt 5 ай бұрын

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@afuyeas9914 4 ай бұрын

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.

@yurenchu Ай бұрын

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.

@leaDR356 5 ай бұрын

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)

@picup30296 5 ай бұрын

I am not sure if I understand correctly, but let me try. Let d be a 2-digit number in the form of 10d_0+d_1, where d_0 and d_1 is in the interval 0-9 inclusively and not necessarily distinct with each other and a. then if d_0

@lagomoof 5 ай бұрын

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.

@yurenchu Ай бұрын

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.

@dd-di3mz 5 ай бұрын

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.

@Kaleidosium 5 ай бұрын

A modified version is very useful to quickly calculate Eigenvalues of 2x2 Matrices

@ThisIsntReallyMyRealName 5 ай бұрын

•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.

@Copemaxer 4 ай бұрын

Pq is super easy and is really easy to simplify 2.

@yurenchu Ай бұрын

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").

@dd-di3mz Ай бұрын

@@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.

@stevencarr4002 5 ай бұрын

I always prefer to have the 2c on the top and the -b +-sqrt(b^2-4ac) on the bottom of the formula.

@uggupuggu 5 ай бұрын

great so you'll get 1/x

@stevencarr4002 5 ай бұрын

@@uggupuggu No, I will get x .

@yurenchu Ай бұрын

But what if c = 0 ? 😉

@stevencarr4002 Ай бұрын

@@yurenchu You get 0 and -c/b as the answers.

@yurenchu Ай бұрын

@@stevencarr4002 No, you don't. Consider the equation (3x² - 21x + 0) = 0 , for example.

@mohammadhuzaifkhan7828 5 ай бұрын

You could use PQ formula as it is in the form of x^2+px+c=0

@catboyrice 5 ай бұрын

Ye he mentioned that

@technicallightingfriend4247 5 ай бұрын

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