Precalculus problems, don't understand

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The discussion revolves around solving various precalculus problems, particularly focusing on the correct methods to find solutions for given equations. The first equation, S = n(n + 1)/3, requires careful attention to parentheses in the solution for n, which is crucial for accuracy. The second problem, a quadratic equation, has incorrect solutions provided, prompting a request for the detailed work behind the calculations. The third problem has correct answers but lacks additional solutions, indicating a need for thoroughness in finding all possible answers. Overall, the thread emphasizes the importance of clarity in mathematical expressions and the necessity of showing work to validate solutions.
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1.) Solve the equation for the indicated variable. (Enter your answers as a comma-separated list.)

S = n(n + 1)/3; for n

My answer: n = 1+√(1+12s)/2 , 1+√(1+12s)/2

For some reason I keep getting this as incorrect?

2.) Find all real solutions of the quadratic equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
7y^2 − y − 1/7 = 0

My answer: y= 1/14 (1-3√53) , 1/14(1+3√53)

3.) Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x^(4/3) − 5x^(2/3) + 6 = 0

My answer: 2√2 , 3√3

I don't understand why I keep getting these wrong. Please explain, thank you in advance.
 
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Niaboc67 said:
1.) Solve the equation for the indicated variable. (Enter your answers as a comma-separated list.)

S = n(n + 1)/3; for n

My answer: n = 1+√(1+12s)/2 , 1+√(1+12s)/2
The two answers you wrote are both the same.
Show us the quadratic equation you're working with.
Niaboc67 said:
For some reason I keep getting this as incorrect?

2.) Find all real solutions of the quadratic equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
7y^2 − y − 1/7 = 0

My answer: y= 1/14 (1-3√53) , 1/14(1+3√53)
Both of these are wrong. Show us the work you did to solve for y.
Niaboc67 said:
3.) Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x^(4/3) − 5x^(2/3) + 6 = 0

My answer: 2√2 , 3√3
These two solutions are correct, but there are two more solutions. Show us what you did.
Niaboc67 said:
I don't understand why I keep getting these wrong. Please explain, thank you in advance.
 
Niaboc67 said:
1.) Solve the equation for the indicated variable. (Enter your answers as a comma-separated list.)

S = n(n + 1)/3; for n

My answer: n = 1+√(1+12s)/2 , 1+√(1+12s)/2

For some reason I keep getting this as incorrect?

2.) Find all real solutions of the quadratic equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
7y^2 − y − 1/7 = 0

My answer: y= 1/14 (1-3√53) , 1/14(1+3√53)

3.) Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x^(4/3) − 5x^(2/3) + 6 = 0

My answer: 2√2 , 3√3

I don't understand why I keep getting these wrong. Please explain, thank you in advance.

For the first question: what you wrote is
n = 1 + \frac{\sqrt{1+12s}}{2}.
This is wrong; did you mean to write
n = \frac{1+\sqrt{1+12s}}{2} \, ?
If so, you need to USE PARENTHESES, like this: n = (1 + 12√(1+12s)/2.
 
If S is the function then just set S=0 and then solve the quadratic. That will give you your answer.
 
dumplump said:
If S is the function then just set S=0 and then solve the quadratic. That will give you your answer.
No it won't. The goal of the exercise is to solve for n, which will give an expression that involves S.
 
Mark44 said:
No it won't. The goal of the exercise is to solve for n, which will give an expression that involves S.
I did not see that from his Original post. He just said that he needed to solve the quadratic.
 
dumplump said:
I did not see that from his Original post. He just said that he needed to solve the quadratic.
Yes, but where S is an unknown parameter. You can't just set it to zero. It must appear in the answer.
 
haruspex said:
Yes, but where S is an unknown parameter. You can't just set it to zero. It must appear in the answer.
I just thought that maybe it was a function like S(x), but he just forgot to put that.
 
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