Precalculus problems, don't understand

  • Thread starter Niaboc67
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But now I understand. Thank you for clarifying it for me.I just thought that maybe it was a function like S(x), but he just forgot to put that. But now I understand. Thank you for clarifying it for me.
  • #1
Niaboc67
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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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  • #2
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.
 
  • #3
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
[tex] n = 1 + \frac{\sqrt{1+12s}}{2}.[/tex]
This is wrong; did you mean to write
[tex] n = \frac{1+\sqrt{1+12s}}{2} \, ? [/tex]
If so, you need to USE PARENTHESES, like this: n = (1 + 12√(1+12s)/2.
 
  • #4
If S is the function then just set S=0 and then solve the quadratic. That will give you your answer.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 

FAQ: Precalculus problems, don't understand

1. What exactly is precalculus?

Precalculus is a branch of mathematics that serves as a prerequisite for calculus. It covers topics such as algebra, trigonometry, and analytical geometry.

2. How is precalculus different from calculus?

Precalculus is a more basic level of mathematics that focuses on algebraic and trigonometric concepts, while calculus is a more advanced level that involves the study of limits, derivatives, and integrals.

3. Why is precalculus important?

Precalculus is important because it provides the necessary foundation for understanding and solving more complex mathematical problems in fields such as science, engineering, and economics.

4. What are some common topics covered in precalculus?

Some common topics in precalculus include functions, equations, logarithms, trigonometry, and vectors. It also includes the study of graphs and their properties.

5. What are some tips for understanding precalculus problems?

Some tips for understanding precalculus problems include practicing regularly, breaking down problems into smaller steps, seeking help from a tutor or teacher, and using online resources such as videos and practice problems.

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