This is a wonderful lesson even in absence of a teacher, i have liked the creator of this video and as a mathematician, he has motivated me to create my own Mathematical videos... Keep up the good work.
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What led you to pick “3” and “6” as the factors for 18, as opposed to “2” and “9”. I do realize that 2 and 9 would not have worked, but how do you decide which factors to choose BEFOREHAND?
With this method you're basically shifting the trial and improvement part of factorising to this step. Like you said, you can spot 2 and 9 won't work so you have to try another factor pair. What you are looking for is 2 factor pairs that you can pick one from each pair, multiply together then add/subtract to get the number in front of x (9 in this case). Just tried it myself with a few different examples and found it helped to list the factors in pairs then trial and error from there. Can't say this is the case for all examples but the ones I tried all used the bigger number from one factor pair multiplied by the smaller number from the other factor pair which made testing factors faster.
DUDE in India it is tough no fuss is considered hard I will surely show them this . . .> to speak frankly it tough as in Indian syllabus they give very tough probs
@Nexus of Exams IT'S Trickery, the guy knew the answer either by trying the trial & error method or after having seen the answer that is printed at the end of the book/exercise.
Last year in my Grade 8 years, I didn't get any lessons(except the cartesian plane lol) that's why this vid is a big help for me in my 9th Grade, thankyou💕
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When factoring (18)x(35), if I factor 18 as (9 x 2), then I can not find the right pair (30 and - 21). So, this method is a happy guessing. I sincerely suggest a general approach to find the real roots. The solving results in finding 2 numbers knowing their product (ac = 18 x -35 = -630), and their sum (-b = -9). First find the factor pairs of ac = -230, then find the pair that has its sum equals to (-b = -9). Proceed with a calculator: (-2, 315)(-3, 210)(-6, 105)(-18, 35)(-21, 30). This last sum is 9 = b. So, the opposite pair is (21, - 30) = -b. Now we divide these numbers by (a = 18) to find the real roots of the equation. x1 =21/a = 21/18 = 7/6, and x2 = -30/a = -30/18 = -5/3. No needs to do the lengthy factoring by grouping and solving the 2 binomials)
@@mavla2 the best approach is to multiply 18by 35 then u will get 630 after that u look for the highest number to divide 630 for me l used 30 . So 30 into 630 = 21 . So now u have ur 2 factors which 30 and 21 . So if u less 21 from 30 u will get 9 problem solved
18 can be factored into 18 x 1 or 9 x 2 or 6 x 3. When you chose 6x3, it looks like you knew the correct answer. Similarly when you wrote the factors of 35 in the order 5 and then 7, rather than 7 and then 5, you guessed the correct order. I don’t think that your method is any simpler than the traditional trial and error method that I learned in high school. Your method is trial and error too and since you knew the answers that you were looking for, it looked as though you had a foolproof method, but indeed you chose the factors and order so that they would work, but other “random” choices wouldn’t. I don’t understand the enthusiasm of your readers. I think that you tricked them!
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This method depends on selecting the correct combination of figures to equal 18. In the example, 3 and 6 give the right answer, but 2 and 9 would not,even though they also multiply to 18. So, in my opinion, using this method is not fool-proof, and users have to be careful when choosing pairs of numbers. But the method is of interest.
Well it is. You can always check that fact that 9×5 -2×7 ≠ 9. The only two conditions in that part is that it has to be multiplied to form the integer and be added to give the second term
I appreciate the approach. However, it's time consuming when thinking through the factors of 18 and 35 and further deciding the highest factor to divide them with.
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That's actually the long way, since your method still needs to determine which numbers when multipled and added together will yield the numerical value multiplying "x". You can skip your first step all together and still come up with the correct factors.
18x^2+9x-35=0 18x^2+30x-21x-35=0 6x(3x+5)-7(3x+5)=0 (3x+5)(6x-7)=0 (3x+5)=0 3x=-5 x=-5/3 (6x-7)=0 6x=7 x=7/6 Multiplied 18 * 35 to get 630 and then did a little hit and trial starting from 25 and 16 (considering have to get 9) for the middle term. Took way less than the method you are suggesting.
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Thanks a lot !!! I was wondering how to solve these kind of problems... for like months At last I got it. Your trick works like magic on every quadratic equation.
Oh dear, it's rlly vry gr8... I love this a lot, this is little bit confusing to make factors and now it's very helpful ur trick is really gr8. Loved a lot. Thnx vry much and keep it up !!!
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Guys ..pls help me...x² -5x +6 = 0 Can't they be written as (x-6)(x+1)...?. The correct ans is (x-3)(x-2) ....what is the problem with the first one...(x-6)(x+1)....? The coefficient of x² is 1 and the constant term is 6 ....1 can be written as 1× 1 and 6 as 6× 1..... to get -5 ...i took it as -6 and +1 ...-6 +1 = -5 ...right?.....but (x-6)(x+1) do not give...x² -5x +6.....any one pls help me😰
Dear Parvathy, x² -5x +6 = 0 => correct factors are (x-3)(x-2)=0 => x-3=0, x-2=0 =>our solution is: x=3, x=2 😀 However, (x-6)(x+1) are not correct factors because if you multiply them => x² -5x -6 which is wrong! Thanks for asking. All the best
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It will not give the result. The aim is to achieve +9. So be reasonable in the selection of the factors. The method is alright for higher value like this ,because one would have multipled out 18 and 35 and spent the whole day sorting the appropriate factors.
18x² + 9x - 35 = 0 This is comparing with ax² + bx + c = 0 then we get a=18, b=9, c=-35 Factors should be satisfy two conditions 1) Product of factors = a * c 2) Sum of factors = b a * c = 18 * (-35) = 2 * 9 * 5 * (-7) = 9 * 5 * 2 * (-7) = 45 * (-14) Factors are 45, -14 Product of factors = 45 * (-14) = a * c Sum of factors = 45 + (-14) = 31 ≠ b *∴ It not satisfied second condition so don't take this factors.* a * c = 18 * (-35) = 2 * 9 * 7 * (-5) = 9 * 7 * 2 * (-5) = 63 * (-10) Factors are 63, -10 Product of factors = 63 * (-10) = a * c Sum of factors = 63 + (-10) = 53 ≠ b *∴ It not satisfied second condition so don't take this factors.* a * c = 18 * (-35) = 3 * 6 * 7 * (-5) = 6 * 7 * 3 * (-5) = 42 * (-15) Factors are 42, -15 Product of factors = 42 * (-15) = a * c Sum of factors = 42 + (-15) = 27 ≠ b *∴ It not satisfied second condition so don't take this factors.* a * c = 18 * (-35) = 3 * 6 * 5 * (-7) = 6 * 5 * 3 * (-7) = 30 * (-21) Factors are 30, -21 Product of factors = 30 * (-21) = a * c Sum of factors = 30 + (-21) = 9 = b *∴It satisfied both conditions so take this factors.*
Yo I was looking for a way to solve quadratic equations with ease and I'm so glad I found this video. I'm taking a test tomorrow and this video was very helpful and now I have a fighting chance. THANK YOU SO MUCH BRO. YOUR THE MAN!!!!
This is my 5th video and after 2 hours searching for help. Thank you so much you explained the very step so many professors/tutorials and apps skip over. Made it make sense. bless 🙏🏽
Most of the people comments only attendance for this year or legends and studying before one day of exam same comments in every video fed up to see these all
THANK YOU SO MUCH FOR CREATING THIS KIND OF VIDEO ITS SO HELPFUL, ESPECIALLY FOR ME THAT HARD TO UNDERSTAND IN TERMS OF MATH SOLVING, the way u solve the equation so understandable, may you Bless more...
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Hello Claire, you can reach me through cavineomondi93@gmail.com so that I may help you with your assignments on one on one basis. I hope to hear from you soon.
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Always start with small and simple factors and look for suitable factors. I'm sure you are a bright and brilliant student. Take care dear and all the best😃
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When you are studying you should never notice any of these fun fact you have written in your comment. I said these things not to notice when you are studying but when you go to job or when you talk to someone you must notice these things. Remember if you use these tips to wrong place you are going to be in loss........... So just focus on what you are doing currently..
During the time of selecting factors of the two numbers we should choose the smallest factor or the highest factor cause my calculations go wrong during the time of selecting factors in some questions so please can you help me and this video is very helpful no doubt about that but I would have really liked it if you would have chosen an even more complicated question 😅
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18x^2+9x-35=0 630-1=629 no 315-2=313 no 210-3=207 no 126-5=121 no 105-6=99 no 90-7=83 no 70-9=61 no 63-10=53 no 42-15=17 no 35-18=17 no 30-21=9 ✓ (18x+30)(18x-21)=0 (3x+5)(6x-7)=0 3x+5=0 3x=-5 x=-5/3 6x-7=0 6x=7 x=7/6
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will this work even if we take factors of 18 as 9 and 2 ? or 2 and 9 ? or if we write the factors of 35 as 7 and 5 instead of 5 and 7 ? Does the changing factor pairs or changing the order affect the answer ?
Bianca dela Cruz more than your teacher huh? Do you know what a discriminant is or what it indicates? Do you know what the vertex of a quadratic tells us? Do you understand that the solutions to a quadratic give us the x-intercepts? Do you understand that those intercepts are equidistant from the line of symmetry? I doubt you understand this better than your “teacher do.” One thing you might attempt is to use proper English, otherwise you come off as an inconsistent idiot.
I enjoy most of his videos but this is one of his worst because he is doing voodoo math. The (18x+ __)(18x +__) setup does not follow how we learned to factor the simple cases x^2+bx+c. Let me try to explain why his voodoo math works. Let’s take his problem 18x^2 + 9x -35=0 and turn it into an easier problem of the form x^2 +bx+c=0. Let’s rewrite the problem as (1/18)(18)(18x^2+9x-35)=0 by multiplying be (1/18)(18) which is 1. Now distribute the 18 to (18x^2+9x-35) but don’t compute the products. So we get (1/18)(18^2x^2+9*18x-35*18)=0 and now we can rewrite this as (1/18)( (18x)^2+9(18x)-35*18)=0. We now have succeeded to have the problem in the form x^2+bx+c where we can easily factor. So factoring, we get (1/18)(18x+30)(18x-21)=0 which now we set each factor equal to zero. 18x+30=0 or 18x-21=0 thus x=-5/3 and x=7/6. The Lesson is that any ax^2+bx+c can be rewritten as (1/a)((ax)^2+b(ax)+c*a) so we only factor (ax)^2+b(ax)+c*a ,which if it can be factored, is by (ax + __)(ax + __) where __*__=c*a and __+__=b.
One thing I love about these RU-vid tutorials is that you replay and replay the video until you understand Unlike a teacher..you ask till he/she gets annoyed and would think that you're just too stupid to get anything
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Omgosh i was so stressed that my teacher didn't show me those steps so thank u for making it so easy! I feel less pressured now thank goodness i found this video
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