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√x = 2 - √2 Or √x = √2.√2 - √2 = √2(√2 -1) Or x = (√2)².(√2 -1)² Or x = 2(2 - 2.√2.1 + 1) Or x = 4 - 4.√2 + 2 Or x = 6 - 4.√2 ans. Do it in shorter way.
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@@Alamaths i’m watching form turkey. also currently studying applied math in college. i watch a lot of math videos i guess that’s why your video showed up to me 😅
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This is the guy that when send to count the number of horses in a herd counts the legs and divides by 4 instead of just counting the horses individually.
@@Alamaths Here's the translation: I’m learning a lot from watching your videos. I’m an ordinary working professional in my 50s living in Korea, and watching these videos brings back memories of my school days. I couldn’t leave a polite comment because I’m not very confident in English, but your videos have been a great help, and I enjoy watching them.
I too had immediately the solution being 2 & -2 ! It is nothing boss ! Complex Number Solution is again a extension of the original solution only !! It is Just Nothing for someone Who attempts n clears JEE Advanced !! And Am Not One of Them !! Hahahahahaha
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Yes. But they didn't specify whether k is real or complex. Just solve for both complex and real solutions. Thanks for the comments 👍 Are you a maths teacher 💖?
1/a +1/b = 1/11 b + a = ab/11 Assume for some form. Let us say that a is kb, where k E R. Let a = kb b + kb = kb*b/11 (k+1)b = kb^2/11 k/11b^2 - (k+1)b = 0 b[k/11*b - (k+1)] = 0 b = 0 or k/11 *b - (k+1) = 0 b != 0 bc div. by 0. k/11 *b - (k+1) = 0 --> k/11*b = k+1 b = 11(k+1)/k b = 11 + 11/k Check: b = (11 + 11/k), a = 11k + 11 1/a + 1/b = 1/(11k+11) + 1/(11 + 11/k) = 1/11 * [1/(k+1)] + 1/11 * [1/(1 + 1/k)] = 1/11 [1/(k+1) + 1/(1+ 1/k)] = 1/11 [(1+1/k + k + 1)/((k+1) * (1+1/k))] WTS: k + 1/k + 2 = denom (thereby, the non-1/11 bit would be equal to 1) Multiply out the denom: (k+1)*(1+1/k) = k + 1 + 1 + 1/k = k + 1/k + 2. So this is a valid solution. for k = 1: b = 1/(11 + 11/1) = 1/22 a = 1*1/22 = 1/22 a + b = 1/11 k = 8: 1/(11 + 11/8) + 1/(88+11) = 1/11 * [1/(1+1/8) + 1/(8 + 1)] Right bit: 1/(9/8) + 1/9 = 8/9 + 1/9 = 1 There are likely others of other forms of a = kb^n, too, probably that n E Z. idk.
@@Alamaths haha thank you but no, I’m not a professor. Because it’s two variables, and we need 2 equations to solve uniquely, I just set the restriction for myself. It was mostly for fun doing it that way, because we can do it much more easily and just set a = k, where k E R, nonzero. Then 1/b = 1/11 - 1/k := m, and taking the reciprocal, b = 1/m.
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sqr(2) + sqr(x) = 2 ---> sqr(x) = 2 - sqr(2) and now we raise to the power 2 left and right which preserves the equality of the two members, and we have: sqr(x)^2 = (2 - sqr(2)) ^ 2 ----> x = 4 - 4sqr(2) + 2 [ from the formula (a-b)^2 = (a^2 -2ab + b^2)] the (unique) solution is then x = 6 - 4sqr(2). Done!
We can express 121 as multiplication of any numbers we want, as long as N*M = 121. For example 1000 and 0.121 would work too. -i√6 and (121i√6)/6 would work too.
Semi-literate presentation. One should start by defining the domain for x. Logic and precision in formulating the problem are infinitely more important than manipulations with W-function. You do more harm than good with this flick.
@@Ofek-sb7ot The square root is a strictly increasing function, which means setting sqrt(x) = anything only has one solution (or none if the RHS is negative).
The answer is obvious. Not even a second is needed there. The solution resembles the government's efforts to deal with everything as long and complicated as possible.😂
@@Alamaths The secondary school where I studied 45 years ago was a physics-mathematics specialization. We were trained for simple solutions like in the army. By the way - if a task is given, then the = sign must be put all the time, even when continuing the solution in the next line!!!!
A quadratic equation give two solution and the cubic equation give 3 solution totally we have 6 solition in cubic equation simply you use cubic squares of unity
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