How do i solve this? I tried but i dont know

[Shiroe05]

Hi. This is my first post so please be nice
This is the problem:

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\(\displaystyle \mbox{(2) Let }\, a\, \mbox{ be a constant. If the constant term of }\, \left(x^3\, +\, \dfrac{a}{x^2}\right)^5\)

\(\displaystyle \mbox{is equal to }\, -270,\, \mbox{ then }\, a\, =\, \Huge{\Box}\)

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I know the answer, but i dont know how to do it. If someone can tell me how begin or what read. Thanks
Sorry about my english, i'm learning :???:

 

[Deleted member 4993]

Hi. This is my first post so please be nice :grin:
This is the problem:

---

\(\displaystyle \mbox{(2) Let }\, a\, \mbox{ be a constant. If the constant term of }\, \left(x^3\, +\, \dfrac{a}{x^2}\right)^5\)

\(\displaystyle \mbox{is equal to }\, -270,\, \mbox{ then }\, a\, =\, \Huge{\Box}\)

---

I know the answer, but i dont know how to do it. If someone can tell me how begin or what read. Thanks
Sorry about my english, i'm learning :???:

In this forum everybody is nice to guests and members - except when we detect "lazyness".

Is this problem a continuation of another problem?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[stapel]

This is the problem:

---

\(\displaystyle \mbox{(2) Let }\, a\, \mbox{ be a constant. If the constant term of }\, \left(x^3\, +\, \dfrac{a}{x^2}\right)^5\)

\(\displaystyle \mbox{is equal to }\, -270,\, \mbox{ then }\, a\, =\, \Huge{\Box}\)

---

I know the answer, but i dont know how to do it.

Hint: Binomial Theorem.

 

[Deleted member 4993]

In case interested, here's solving that for x:
http://www.wolframalpha.com/input/?i=solve+(x^3+a/x^2)^5+++270+=+0+for+x

By the way - that is a totally different problem!

OP wanted solve for "a" where

(x³ + a/x²)⁵ = x¹⁵ +....+ C + ... (a/x¹⁰) and C = -270

You, on the other hand are solving for 'x' in a different equation (problem) where (x³ + a/x²)⁵ = - 270 [or roots of (x³ + a/x²)⁵ + 270=0]

In the original case, the product of the coefficient of the 4 th term of the expansion and a³ would be -270.

 

[Shiroe05]

In this forum everybody is nice to guests and members - except when we detect "lazyness".

Is this problem a continuation of another problem?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

its not a continuation.

This is what i done, i dont know how to find "a" without the "x" value.
if you want i can send you more of my work, if that what you need to be "nice".

 

[Shiroe05]

Hint: Binomial Theorem.

Thanks, i going to search a video or something.

 

[Shiroe05]

Looking at case where x and a are integers; closest: x=-4, a=975

Then expression = -269.3893...

but i have the answer was a=-3
is the answer wrong??

 

[pka]

yi\(\displaystyle \mbox{(2) Let }\, a\, \mbox{ be a constant. If the constant term of }\, \left(x^3\, +\, \dfrac{a}{x^2}\right)^5\)

You are trying to solve:
\(\displaystyle \dbinom{5}{2}\left(x^3\right)^2\left(\dfrac{a}{x^2}\right)^3=-270\) for \(\displaystyle a\).

 

[Deleted member 4993]

its not a continuation.

This is what i done, i dont know how to find "a" without the "x" value.
if you want i can send you more of my work, if that what you need to be "nice".

The answer is correct -→ a = -3. However, the method you trying is incorrect.

Look at the response provided by pka (along with Stapel's answer) for correct method.

 

[Harry_the_cat]

Shireo05, Here's what you do:

1. Use binomial expansion to expand the expression:

\(\displaystyle (x^3+\frac{a}{x^2})^5 = (x^3)^5 + 5 (x^3)^4 * (\frac{a}{x^2}) +10(x^3)^3*(\frac{a}{x^2})^2+10(x^3)^2*(\frac{a}{x^2})^3+5(x^3)*(\frac{a}{x^2})^4 +(\frac{a}{x^2})^5 \)

2. Now, you want the constant term (ie the term with no "x" in it) to be -270. So which one of these terms will the x's all cancel out.

\(\displaystyle 10(x^3)^2*(\frac{a}{x^2})^3 = 10x^6 *\frac{a^3}{x^6} = 10a^3 \)

3. So, let \(\displaystyle 10a^3=-270\) and solve for a.

\(\displaystyle 10a^3=-270\)

\(\displaystyle a^3 = -27\)

\(\displaystyle a=-3\)

Note: When you get good at these, there's no need to write out the whole expansion. Imagine if the index was 100 instead of 5, you'd be there all day doing step 1. You need to think about what term is the one you need and focus on that one to get the coefficient and the indices correct.

 

[pka]

Shireo05, Here's what you do:
1. Use binomial expansion to expand the expression:

\(\displaystyle (x^3+\frac{a}{x^2})^5 = (x^3)^5 + 5 (x^3)^4 * (\frac{a}{x^2}) +10(x^3)^3*(\frac{a}{x^2})^2+10(x^3)^2*(\frac{a}{x^2})^3+5(x^3)*(\frac{a}{x^2})^4 +(\frac{a}{x^2})^5 \)

@Harry_the_cat, that is correct, but completely against the spirit of the question.

 

[Harry_the_cat]

@Harry_the_cat, that is correct, but completely against the spirit of the question.

Pka, Yeah I know that (and you and I would go straight to the general term) but I think it leads to a better understanding of what's going on.

Please also see the note at the end of my reply.

Your previous response didn't explain where you got the equation to solve from.

 

[Ishuda]

but i have the answer was a=-3
is the answer wrong??

Strictly speaking, yes it is wrong. However, there is an x for which the solution for a would be 3. We have
a = - x² (b + x³)
where b = 270^1/5 ~ 3.064. That x is about -1.615

 

[pka]

Pka, Yeah I know that (and you and I would go straight to the general term) but I think it leads to a better understanding of what's going on.
Your previous response didn't explain where you got the equation to solve from.

I had no intention of explaining how to solve any question. That is lazy teaching.
I am a retired university mathematics professor and department chair. I teach as I was taught, question-answer.

On the practical side, I did editing for several testing companies. This question is what is called a time consumer. If the test taker has to spend time on expanding the binomial he/she does not really understand the question. The time consumed is taken away from the other questions and hence is a penalty. Test-prep teaches people to avoid such questions, or at least delay trying them.

 

[Harry_the_cat]

I had no intention of explaining how to solve any question. That is lazy teaching.
I am a retired university mathematics professor and department chair. I teach as I was taught, question-answer.

On the practical side, I did editing for several testing companies. This question is what is called a time consumer. If the test taker has to spend time on expanding the binomial he/she does not really understand the question. The time consumed is taken away from the other questions and hence is a penalty. Test-prep teaches people to avoid such questions, or at least delay trying them.

Just feel that I have to react to your statements.

I respect your position Pka, but totally disagree with your view that this is lazy teaching. How can "explaining how to solve a question" be lazy compared to just giving the answer (or close to it)? Surely that's lazy teaching. Your statement "I teach as I was taught, question-answer" may have worked for you, as I'm sure you were a very competent mathematics student once (as, I dare say, was I), but the majority of students who ask for help on this site are not. They need explanations if they are to fully understand the process, before they are comfortable with the shortcuts which are second nature to us.

I do agree with your statement that this question is a time consumer. I would not expect this person to expand the binomial once they have a better understanding of the question (see the note in my original post) and certainly not in an exam situation. But I feel it is a good way to start the process of understanding.

 

[Harry_the_cat]

Strictly speaking, yes it is wrong. However, there is an x for which the solution for a would be 3. We have
a = - x² (b + x³)
where b = 270^1/5 ~ 3.064. That x is about -1.615

Not sure if this has anything to do with the original question that was asked.

 

[Shiroe05]

Shireo05, Here's what you do:

1. Use binomial expansion to expand the expression:

\(\displaystyle (x^3+\frac{a}{x^2})^5 = (x^3)^5 + 5 (x^3)^4 * (\frac{a}{x^2}) +10(x^3)^3*(\frac{a}{x^2})^2+10(x^3)^2*(\frac{a}{x^2})^3+5(x^3)*(\frac{a}{x^2})^4 +(\frac{a}{x^2})^5 \)

2. Now, you want the constant term (ie the term with no "x" in it) to be -270. So which one of these terms will the x's all cancel out.

\(\displaystyle 10(x^3)^2*(\frac{a}{x^2})^3 = 10x^6 *\frac{a^3}{x^6} = 10a^3 \)

3. So, let \(\displaystyle 10a^3=-270\) and solve for a.

\(\displaystyle 10a^3=-270\)

\(\displaystyle a^3 = -27\)

\(\displaystyle a=-3\)

Note: When you get good at these, there's no need to write out the whole expansion. Imagine if the index was 100 instead of 5, you'd be there all day doing step 1. You need to think about what term is the one you need and focus on that one to get the coefficient and the indices correct.

i understand the idea but im going to watch some videos about binomials to undertand better.
i have to do a math test for a scholarship but the test is harder that i thought, i dont saw anything like that in high school. so thanks , with this i can do a couple of problems of the test.
Sorry about my english.