One could estimate the root through numerical methods.How do you solve (x2-5x+5)x2+4x-60=2 without graphing technology? Is there a way to solve this problem using logarithms?
If the RHS were 1 then it's possible to solve algebraically. I don't see it with 2.How do you solve (x2-5x+5)x2+4x-60=2 without graphing technology? Is there a way to solve this problem using logarithms?
How do you solve (x2-5x+5)x2+4x-60=2 without graphing technology?
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.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.
The base can be 1 (2 solutions) or -1 (with an even exponent - one solution), or the exponent can be 0 (2 solutions).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?
[imath]a^b=1[/imath] when: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?
Thanks, I'm looking into that.One could estimate the root through numerical methods.
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"?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).
Here is what I 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"?
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.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).