Solving Higher Degree Power = 2

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Ted Brown

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How do you solve (x2-5x+5)x2+4x-60=2 without graphing technology? Is there a way to solve this problem using logarithms?
 
Ted Brown said:
How do you solve (x2-5x+5)x2+4x-60=2 without graphing technology?
Click to expand...

The right-hand side is definitely meant to be 1, because those polynomials were
chosen on purpose. With a 1 on the right-hand side, there are five integer solutions
that can be solved for using the properties of bases and exponents.
 
lookagain said:
The right-hand side is definitely meant to be 1, because those polynomials were
chosen on purpose. With a 1 on the right-hand side, there are five integer solutions
that can be solved for using the properties of bases and exponents.
Click to expand...
To put this another way, this type of problem, with a 1 for the RHS, is a standard "trick" problem designed to be solvable. Such problems are unrealistic, because real-life problems typically are not designed that way. Students should be taught not only that interesting tricks can be used in special cases, but also that most equations one could write down randomly, or that arise in real life, can't be solved exactly! Teaching only what can be done when we make up the problem intentionally is poor preparation for actually using math.

So if the OP intended to say, "If I modify this problem I was given by replacing 1 with 2, can it still be solved?", the answer is No. (In that case, the number was meant to be 2, as far as the OP is concerned!) The same answer applies if (by some very odd chance) the equation came from a real-life problem. But if it was an assignment, then almost certainly @lookagain is right.

Incidentally, even graphing will not reveal all 5 solutions! One of them is invisible on a typical computer-based graph, because it is on a region where the LHS is very badly behaved.
 
Assuming 1 instead of 2, how can we get 5 solutions?
I know that you can get 2 solutions from the base and 2 from the exponent which means you can have up to 4 solutions. Where does the 5th come from?
 
Steven G said:
Assuming 1 instead of 2, how can we get 5 solutions?
I know that you can get 2 solutions from the base and 2 from the exponent which means you can have up to 4 solutions. Where does the 5th come from?
Click to expand...
The base can be 1 (2 solutions) or -1 (with an even exponent - one solution), or the exponent can be 0 (2 solutions).

It's the solution where the base is -1 that is in a badly-behaved region, as negative numbers to powers go all over the place (including complex).
 
Steven G said:
Assuming 1 instead of 2, how can we get 5 solutions?
I know that you can get 2 solutions from the base and 2 from the exponent which means you can have up to 4 solutions. Where does the 5th come from?
Click to expand...
[imath]a^b=1[/imath] when:
1) [imath]b=0, a \neq 0[/imath]
2) [imath]a=1[/imath], any value of [imath]b[/imath]
3) [imath]a=-1[/imath] and [imath]b[/imath] is even.
 
Subhotosh Khan said:
One could estimate the root through numerical methods.
Click to expand...
Thanks, I'm looking into that.
Dr.Peterson said:
The base can be 1 (2 solutions) or -1 (with an even exponent - one solution), or the exponent can be 0 (2 solutions).

It's the solution where the base is -1 that is in a badly-behaved region, as negative numbers to powers go all over the place (including complex).
Click to expand...
I graphed the equation where the right side is 1 using multiple methods, and indeed none of them showed 2 as a root. Could you elaborate on what you mean by it being in "a badly-behaved region"?
 
Ted Brown said:
I graphed the equation where the right side is 1 using multiple methods, and indeed none of them showed 2 as a root. Could you elaborate on what you mean by it being in "a badly-behaved region"?
Click to expand...
Here is what I said:
Dr.Peterson said:
It's the solution where the base is -1 that is in a badly-behaved region, as negative numbers to powers go all over the place (including complex).
Click to expand...
Try evaluating the expression for values of x near 2. For example, when x=2.1, we get (x^2-5x+5)^(x^2+4x-60) = (2.1^2-5*2.1+5)^(2.1^2+4*2.1-60) = (-1.09)^-47.19 = 0.0171334, while when x=2.01, we get (x^2-5x+5)^(x^2+4x-60) = (2.01^2-5*2.01+5)^(2.01^2+4*2.01-60) = (-1.0099)^-47.9199 reports "invalid input" on my calculator; presumably it's a complex number, which is what you get when you raise a negative number to a fractional power with an even denominator.

When I graphed it in Desmos, I defined the function as f(x) and plotted (a, f(a)) using a slider for a, and watched the dot appear and disappear as I slid it!
 
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