Substitute It Til You're Content

Calculus Level 2

6 + 6 + 6 + t 3 3 3 = t \large \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+ t}}} = t

Find the real value of t t satisfying the equation above.


The answer is 2.

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2 solutions

Akshat Sharda
Mar 10, 2016

6 + 6 + 6 + t 3 3 3 = t 6 + 6 + 6 + 3 3 3 = t 6 + t 3 = 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 \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+ t}}} = t \\ \ \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+ \ldots}}}=t \\ \sqrt[3]{6+t}=t \Rightarrow 6+t=t^3 \\ t^3-t-6=0 \\ (t-2)(t^2+2t+3)=0 \\ \text{Now, as the discriminant of } t^2+2t+3 \text{ is negative.} \\ \text{Therefore, } t^2+2t+3 \text{ has no real roots.} \\ \therefore t=\boxed{2}

Can anyone tell why is it a calculus question now?

Swapnil Das - 5 years, 3 months ago

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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 6 + t 3 = t \sqrt[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 = 6 + t 3 t y = \sqrt[3]{ 6 + t } - t . However, since neither of these approaches are likely to be used, we have thus moved it into Calculus instead.

Calvin Lin Staff - 5 years, 3 months ago

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I see. Thanks for replying.

Swapnil Das - 5 years, 3 months ago

You need to show that the quadratic has no real roots

Abdur Rehman Zahid - 5 years, 3 months ago

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Edited! Thanks for suggestion!

Akshat Sharda - 5 years, 3 months ago

How come you were able to go from the first line to the second line?

Thomas James Bautista - 5 years, 3 months ago

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Substitute t t again and again in the first expression, you'll notice that it will go up to infinity.

Akshat Sharda - 5 years, 3 months ago

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I get it now, since it's recursive. Thanks :D

Thomas James Bautista - 5 years, 3 months ago

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 , 1 i ) \cap_{i=1}^n (0 , \frac{1}{i} ) is non-empty for any n n . However, if we continued it to infinity, then we get the null set.

Calvin Lin Staff - 5 years, 3 months ago

Can you explain me the third line...... A college student here

Sadikur Rahman - 5 years, 3 months ago

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Cubing both sides.

Akshat Sharda - 5 years, 3 months ago

How do you factor a cubic equation? And how do you do the second step (2nd to 3rd line)?

Alex Li - 5 years, 3 months ago

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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.

Abdur Rehman Zahid - 5 years, 2 months ago

How did you went from the second infinite equation to the third finite equation ?

A Former Brilliant Member - 5 years, 2 months ago
Anthony Stewart
Mar 11, 2016

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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