When first introduced to exponents, many students think that
n 2 = 2 n .
Find the positive integer n for which the equation is true.
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If a student is an idiot than yes indeed.
simple method ,,i liked
or you could divide both sides by n and you still left with n=2
n^2 = 2n => n^2 - 2n = 0 => n(n-2) = 0 = > n = 0 V n-2 = 0; n = 2, 2 is possible answer
*PLZ. NOTE : The second solution is deprecated and inappropriate for solving these kind of problems(because as equations keep getting bigger, it gets more complex). Since, n^2 = 2n is quite special as it tells a unique property of 2, I thought I'd post it for the benifit of the younger members of this BRILLIANT website!!! ;) *
n^2 = 2n <=> n . n = 2 . n <=> n = 2
Is n = 2 the only solution to your equation?
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Really, this is equation is same : n 2 − 2 n = 0 , where I passed the 2 n to the first member, so using Girard we have. → sum = a − b = 1 2 ; → product = a c = 0 , so the roots are 2 and 0
n square = 2 square = 4
and 2n=2*2 = 4
It's better if you use LaTex Code next time, as it makes the solution more elegant and pleasant to look at. :)
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Yup! I agree(I love using LateX and read over my answers again and again after using it !!!!).
n^2=2n
then,
n=2
2^2=2*2
4=4.
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Given the equation, n 2 = 2 × n ⇒ n = n 2 × n ⇒ n = 2 Thus, n = 2