# Finding all real solutions of algebraic equations

1.) Solve the equation for the indicated variable: $S = n(n + 1)/3$; for $n$.

My answer: $n = 1+\sqrt{ 1+12s}/2$ , $1+\sqrt{ 1+12s}/2$. For some reason I keep getting this as incorrect?

2.) Find all real solutions of the quadratic equation: $7y^2 − y − 1/7 = 0$

My answer: $y= 1/14 (1-3\sqrt{53})$ , $1/14(1+3\sqrt{53})$

3.) Find all real solutions of the equation. $x^{4/3} − 5x^{2/3} + 6 = 0$

My answer: $2\sqrt{2}$ , $3\sqrt{3}$.

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

• For the first one the solutions are $\dfrac{-1\pm\sqrt{1+12S}}{2}$. – André Nicolas Sep 9 '14 at 2:50
• @AndréNicolas That is what the OP obtained. – user122283 Sep 9 '14 at 2:51
• @SDevalapurkar: My display is poor, but I do not see a $-1$. – André Nicolas Sep 9 '14 at 2:53
• @AndréNicolas I'd dismiss that as a typo, but, technically, you're right. – user122283 Sep 9 '14 at 2:54
• @SDevalapurkar: Grading programs are not as forgiving of typos as you and I are. – André Nicolas Sep 9 '14 at 2:56

## 2 Answers

For the first, rewrite as follows: $$n^2+n=3S\implies n^2+n-3S=0$$ For the last, write: $$(x^{2/3})^2+5x^{2/3}+6=0$$ For all of them, use the quadratic formula:

If $ax^2+bx+c=0$, then $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

I think you messed up on the formula; redo the problems again with this formula - you should get the right answers.

After reworking it the answers I got were:

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: √(3s+1/4)-1/2 , -1/2-√(3s+1/4)

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 + √5/14, 1/14 - √5/14

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: 3√3, 2√2

How does this all look?

• Your answers are correct. However, you can simplify $$-\frac{1}{2} \pm \sqrt{3s + \frac{1}{4}}$$ by obtaining a common denominator, then extracting a factor of $1/2$ from the square root. $$-\frac{1}{2} \pm \sqrt{3s + \frac{1}{4}} = -\frac{1}{2} \pm \sqrt{\frac{12s + 1}{4}} = -\frac{1}{2} \pm \frac{\sqrt{1 + 12s}}{2} = \frac{-1 \pm \sqrt{1 + 12s}}{2}$$ – N. F. Taussig Sep 9 '14 at 3:45