did I solve this correctly? I get x = 1, 4 for (x^2 - 5x + 5)^3 = 1

allegansveritatem

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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.

math problem.jpg

solution.jpg
 
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The instruction: Solve for x

This could be vague, unless you also received (unwritten) instructions to find Real solutions only.

Your equation has six Complex solutions; of those, you found the two which are Real.

Are you supposed to find all solutions?
 
The left-hand side is a 6th-degree polynomial; it does not factor nicely.

If we replace 1 with 2, 3, or 4, we will still get two Real solutions, but they won't be Rational. Finding them requires more than simple algebra.

Assuming that you're only interested in Real solutions, I think your reasoning and work are excellent. :cool:
 
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.
 
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
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.
 
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.


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.
 
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.

I guess there could be more than two, but that is what I got and it is also what the key at the back of my text book gives as the correct solution for this problem.
 
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.

What is the topic of the chapter? What is the name of the book? Was there any heading to the group of exercises this was in -- challenge problems? something more specific?

Factoring is not only for quadratic equations; it can be used (where applicable) for any polynomial. But most polynomials can't be factored.

The main issue, as has been mentioned, is whether you are to find all (real and imaginary) solutions, or only real solutions; that would likely be implied by the topic of the chapter. Have you learned about complex numbers at all? Have you learned that a polynomial can have as many solutions as its degree, and in some sense will always have exactly that many? Here, you should expect 6 solutions.
 
… there was no introduction to this type of problem in the text … I guess the author--Robert Blixer-- threw this in …

I guess there could be more than two [solutions] …
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 have Blitzer's text, 'Introductory Algebra for College Students, 4th Edition'. In this edition, your exercise is listed in a subset labeled "Critical Thinking Exercises". I would expect the last few exercises in each section to be challenging. Blitzer does this to help grow the brain, forcing students to look at problems from different angles. He also likes exercises which require previous material, for memory reinforcement. Blitzer mentions these approaches to learning at the beginning of my book (after the table of contents) in the Preface and 'To The Student' sections.

Do you see this one also?

\(\displaystyle 3^{\; \mkern-3mu x^2 \; \mkern-5mu - \; \mkern-3mu 9x \; \mkern-3mu + \; \mkern-3mu 20} = 1\)

You can solve it, with reasoning similar to what you posted.

Here's a 3rd-degree polynomial equation; it can be solved by recalling some previous material.

\(\displaystyle x^3 - x^2 - 16x + 16 = 0\)


PS: I'd need to see the context of any statement about factoring "only" for quadratics, to fully understand what he meant. :cool:
 
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… what if [the right-hand side] had been equal to 2, 3, 4 etc.
Back to this question, the pair of Real solutions for

(x^2 - 5x + 5)^3 = n

look like this:

\(\displaystyle x = \dfrac{5}{2} \pm \dfrac{1}{2} \cdot \sqrt{5 + 4 \mkern-2mu \cdot \mkern-2mu \sqrt[3]{n}}\)

As I mentioned, we need more than simple algebra to get this solution by hand.

Note: this gives your solutions, when n = 1 :cool:
 
We are given so solve:

\(\displaystyle \displaystyle \left(x^2-5x+5\right)^3=1\)

We could arrange this as a difference of cubes:

\(\displaystyle \displaystyle \left(x^2-5x+5\right)^3-1^3=0\)

Factor:

\(\displaystyle \displaystyle \left(\left(x^2-5x+5\right)-(1)\right)\left(\left(x^2-5x+5\right)^2+\left(x^2-5x+5\right)(1)+(1)^2\right)=0\)

\(\displaystyle \displaystyle \left(x^2-5x+4\right)\left(x^4-10x^3+36x^2-55x+31\right)=0\)

\(\displaystyle \displaystyle (x-1)(x-4)\left(x^4-10x^3+36x^2-55x+31\right)=0\)

Okay, we have a quartic function to deal with now. One could observe that:

\(\displaystyle \displaystyle \left(x+\frac{5}{2}\right)^4-10\left(x+\frac{5}{2}\right)^3+36\left(x+\frac{5}{2}\right)^2-55\left(x+\frac{5}{2}\right)+31=\frac{16x^4-24x^2+21}{16}\)

This is a quadratic in \(\displaystyle x^2\), and so we may state:

\(\displaystyle \displaystyle x^2=\frac{24\pm\sqrt{24^2-4(16)(21)}}{2\cdot16}=\frac{24\pm16\sqrt{3}i}{32}=\frac{3\pm2\sqrt{3}i}{4}\)

Thus:

\(\displaystyle \displaystyle x=\pm\sqrt{\frac{3\pm2\sqrt{3}i}{4}}= \pm\frac{\sqrt{3\pm2\sqrt{3}i}}{2}\)

Now, we need to shift these roots back 5/2 units to the right:

\(\displaystyle \displaystyle x=\frac{5\pm\sqrt{3\pm2\sqrt{3}i}}{2}\)

And so the six roots to the original equation are:

\(\displaystyle \displaystyle x\in\left\{1,4,\frac{5\pm\sqrt{3\pm2\sqrt{3}i}}{2}\right\}\)
 
\(\displaystyle \quad\)⁞

\(\displaystyle \quad\)⁞

the six roots to the original equation are:

\(\displaystyle \displaystyle x\in\left\{1,4,\frac{5\pm\sqrt{3\pm2\sqrt{3}i}}{2}\right\}\)
MarkFL deserves a round of applause for generating that demonstration. What a trooper! :)
 
did I solve this correctly? I get x = 1, 4 for (x^2 - 5x + 5)^3 = https://www.freema1

Hi,
The equation (x^2 - 5x + 5)^3 => x6 - 125x3 + 124 = 0
My (program) solution is :

(x - 1)
Irrational/repeating root(s) :
between
4.986630952 and 4.986630953

Hope this helps
Pete.TheOAP
 
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

Your first line appears to be based on the assumption that (a + b + c)^3 = a^3 + b^3 + c^3. This is not valid. You need to expand (a + b + c)(a + b + c)(a + b + c), which has 27 terms.

If the rest of your work is based on the first line, then it has to be incorrect.

And it is. If you check your claimed answer, you find that x^2 - 5x + 5 = 4.93333349, whose cube is 120.0663819, not 1.
 
did I solve this correctly? I get x = 1,4...

This is what I get :

(x^2 - 5x + 5)^3 = 1
A = x^2 - 5x + 5
B = -1

=> A^3 - B^3
=> (A - B)(A^2 + AB + B^2)

A - B = x^2 - 5x + 5 + 1
= (x + 2)(x - 3)
A^2 = (x^2 - 5x + 5) ^ 2
= X^4 - 10x^3 + 40x^2 - 25x + 25
AB = X^2 - 5x + 5
B^2 = 1

Calculation :
A^2 : x^4 - 5x^3 + 5x^2
- 5x^3 + 25x^2
+ 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)
 
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)

Did you check whether x = -2 and 3 are solutions?

Main lesson: Always check your work and your result before posting an answer.
 
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
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.
So if it had been (x^2 -5x +5)^3 =2, then you'll have x^2 -5x + 5 = cube root of 3. Now solve this the same as you already did. Please note that this only works for finding the real roots.
 
did I solve this correctly? I get x = 1,4...

Did you check whether x = -2 and 3 are solutions?

Main lesson: Always check your work and your result before posting an answer.

Dr Peterson,
Lesson learnt!
Sorry. Being 82 and blind in 1 eye I (too) often make mistakes.

Expanding the equation and bringing 1 across :
(x^2 - 5x + 5)(x^2 - 5x + 5)(x^2 - 5x + 5) - 1 = 0
=> x^6 - 15x^5 + 90x^4 - 275x^3 + 450x^2 - 375x + 124
Using Synthetic Divisiion
=> (x - 1)(x - 4)(x4 - 10x3 + 36x2 - 55x + 31)
=> No solution(s) for x4 - 10x3 + 36x2 - 55x + 31

Regards
 
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