allegansveritatem
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Your intuition thatThis problem is from one of those listed at the end of a chapter on solving quadratic equations by factoring. Presented below is the only way I could find to solve this problem. My way worked this time because the cubed quadratic was equal to one. But what if it had been equal to 2, 3, 4 etc.
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Your intuition that
\(\displaystyle \alpha \in \mathbb R \text { and } \alpha^3 = 1 \implies \alpha = 1\)
can be proved true formally, so good job. Please note, however, as mmm pointed out, the restriction that we are dealing with real numbers.
But your question does not really make much sense. If the cubic had been equal to anything other than 1, 0, or minus 1, you would have had to solve a polynomial of degree 6. There is no general, closed form solution to such equations (although special cases, such as this one. are soluble and approximate answerx can be found in the general case by using numerical methods).
At another site there recently was a discussion on whether problems that are soluble only because they are special cases are helpful to students. My feeling is that it depends on how frequent the special case is and whether it is presented in the context of special cases. This one seems to me to fall in the not helpful category because your question clearly implies that you did not see, and had not been alerted to, special cases where the quadratic can be used to solves equations that are not themselves quadratic.
Interesting.
I asked two different Computer Algebra Systems (CAS) to symbolically solve this equation for x:
(x^2 - 5x + 5)^3 = n
Wolfram Alpha reports five solutions, and Maple reports six.
I'm sure there's a reason for this, but I don't have time right now to puzzle over it.
No, there was no introduction to this type of problem in the text. I looked high and low in the chapter and could not find. I guess the author--Robert Blixer-- threw this in--it was the second to last exercise out of 93--with the thought that if anyone got that far in the exercises and was still alive he/she would bring it up in class for clarification. The book does state that factoring is only for 2nd degree polynomials.
… there was no introduction to this type of problem in the text … I guess the author--Robert Blixer-- threw this in …
Now you know there are, but you can ignore the four solutions having an imaginary component because you're interested only in the two that don't (i.e., the Real solutions). I'm not sure how your course progresses, but you can expect to work with both higher-degree polynomials and the imaginary unit in future courses.I guess there could be more than two [solutions] …
Back to this question, the pair of Real solutions for… what if [the right-hand side] had been equal to 2, 3, 4 etc.
MarkFL deserves a round of applause for generating that demonstration. What a trooper!\(\displaystyle \quad\)⁞
\(\displaystyle \quad\)⁞
the six roots to the original equation are:
\(\displaystyle \displaystyle x\in\left\{1,4,\frac{5\pm\sqrt{3\pm2\sqrt{3}i}}{2}\right\}\)
You mean x^6 - 125x^3 + 125, right?The equation (x^2 - 5x + 5)^3 => x6 - 125x3 + 124 = 0
The equation (x^2 - 5x + 5)^3 = 1 => x^6 - 125x^3 + 124 = 0
My (program) solution is :
(x - 1)
Irrational/repeating root(s) :
between
4.986630952 and 4.986630953
This is what I get :
(x^2 - 5x + 5)^3 = 1
A = x^2 - 5x + 5
B = -1 <-- you meant 1, I believe (you took it as such for AB, but not for A-B), since the negative sign is being taken as subtraction
=> A^3 - B^3
=> (A - B)(A^2 + AB + B^2)
A - B = x^2 - 5x + 5 + 1
= (x + 2)(x - 3) <-- no, this is x^2 - x - 6, not x^2 - 5x + 6! Did you mean to type (x - 2)(x - 3)?
A^2 = (x^2 - 5x + 5) ^ 2
= X^4 - 10x^3 + 40x^2 - 25x + 25 <-- no, two terms are wrong
AB = X^2 - 5x + 5
B^2 = 1
Calculation :
A^2 : x^4 - 5x^3 + 5x^2
- 5x^3 + 25x^2 - 25x
+ 5x^2 - 25X + 25
-------------------------------
A^2 = x^4 - 10x^3 + 40x^2 - 25x + 25
AB = x^2 - 5x + 5
B^2 = + 1
----------------------------------
x^4 - 10x^3 + 41x^2 - 30x + 31
----------------------------------
Result :
(x + 2)(X - 3)(x^4 - 10x^3 + 41x^2 - 30x + 31)
Yes 1^3 =1. You should know that (cube root of 2)^3 = 2 and (cube root of 3)^3 = 3 and .... and (cube root of n)^3 = n.This problem is from one of those listed at the end of a chapter on solving quadratic equations by factoring. Presented below is the only way I could find to solve this problem. My way worked this time because the cubed quadratic was equal to one. But what if it had been equal to 2, 3, 4 etc.
View attachment 9980
View attachment 9981
Did you check whether x = -2 and 3 are solutions?
Main lesson: Always check your work and your result before posting an answer.