# Precalculus problems, don't understand

1. Sep 8, 2014

### Niaboc67

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

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

2. Sep 9, 2014

### Staff: Mentor

The two answers you wrote are both the same.
Show us the quadratic equation you're working with.
Both of these are wrong. Show us the work you did to solve for y.
These two solutions are correct, but there are two more solutions. Show us what you did.

3. Sep 9, 2014

### Ray Vickson

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.

4. Oct 11, 2014

### dumplump

If S is the function then just set S=0 and then solve the quadratic. That will give you your answer.

5. Oct 11, 2014

### Staff: Mentor

No it won't. The goal of the exercise is to solve for n, which will give an expression that involves S.

6. Oct 11, 2014

### dumplump

I did not see that from his Original post. He just said that he needed to solve the quadratic.

7. Oct 11, 2014

### haruspex

Yes, but where S is an unknown parameter. You can't just set it to zero. It must appear in the answer.

8. Oct 11, 2014

### dumplump

I just thought that maybe it was a function like S(x), but he just forgot to put that.