3 6 + 3 6 + 3 6 + t = t
Find the real value of t satisfying the equation above.
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Can anyone tell why is it a calculus question now?
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The approach of this solution, with the iterated roots, has a lot of calculus underlying, which is why it was moved into Calculus. E.g. We have to justify why we can "Substitute it till you're content", to reach the conclusion of 3 6 + t = t .
Of course, There are purely algebraic approaches, like repeated cubing the expression. But that is very tedious. There is a better approach that looks at the graph of y = 3 6 + t − t . However, since neither of these approaches are likely to be used, we have thus moved it into Calculus instead.
You need to show that the quadratic has no real roots
How come you were able to go from the first line to the second line?
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Substitute t again and again in the first expression, you'll notice that it will go up to infinity.
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I get it now, since it's recursive. Thanks :D
That's the start. You still need to justify why the "up to infinity" step is valid.
For example, it is obvious that ∩ i = 1 n ( 0 , i 1 ) is non-empty for any n . However, if we continued it to infinity, then we get the null set.
Can you explain me the third line...... A college student here
How do you factor a cubic equation? And how do you do the second step (2nd to 3rd line)?
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Apply the Rational root theorem to identify the rational zeroes of the cubic polynomial (Note that since a cubic polynomial is a polynomial of odd degree (degree 3) and since nonreal complex roots always occur in pairs, a cubic equation with real coefficients will always have 1 real root or 3 real roots).However, if unluckily there is no nice rational root, then you can apply any root approximation method / Cardano's method to solve the cubic and identify the roots.
How did you went from the second infinite equation to the third finite equation ?
An obvious solution is when 3√(6 + t) = t as then the recursion falls out. At which point knowing 2^3 = 8 = 6+2 suffices but you could just rearrange to t^3 - t - 6 = 0 and solve.
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3 6 + 3 6 + 3 6 + t = t 3 6 + 3 6 + 3 6 + … = t 3 6 + t = t ⇒ 6 + t = t 3 t 3 − t − 6 = 0 ( t − 2 ) ( t 2 + 2 t + 3 ) = 0 Now, as the discriminant of t 2 + 2 t + 3 is negative. Therefore, t 2 + 2 t + 3 has no real roots. ∴ t = 2