Find solutions to non-linear equation

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The equation (t-1)^4 + 5 = 3 simplifies to (t-1)^4 = -2, indicating there are no real solutions since a fourth power cannot be negative. This highlights that equations involving even powers cannot yield negative results. Understanding this concept is crucial for efficiently solving similar problems in tests. The discussion emphasizes the importance of recognizing the properties of exponents in determining solution sets. Overall, the equation has no solutions in the realm of real numbers.
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Question:

(t-1)^4+5=3

I know the answer to the question states that there are no solution, but I would like to understand the intuition behind this. If I had this in a test I would waste ages trying to work out why I couldn't solve it.

Any advice would be heaps appreciated.

Thanks

Mick
 
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mackhina said:
Question:

(t-1)^4+5=3

I know the answer to the question states that there are no solution, but I would like to understand the intuition behind this. If I had this in a test I would waste ages trying to work out why I couldn't solve it.
This equation is equivalent to (same solution set) (t - 1)4 = -2. There are no real numbers that can be raised to the fourth power to result in a negative number, so the solution set is empty.

The same would be true if (t - 1) were raised to any even power.
 
Start solving, you should find out what the reason is quite fast.

Please use the template next time.

Note: there is no solution in R (real numbers), it doesn't mean there is no solution at all.

Edit: beaten by Mark.
 
Thanks for your help Mark, that made it totally obvious. Much appreciated.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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