# How can I solve for this variable?

I'm only studying for the final (during thanksgiving break no less :) )

I'm in PreCalc, and I've been struggling this entire semester. -.- So any help with studying would be really helpful

I find that I don't quite no where to proceed from here on this one.

The directions on this online study quiz are:

Solve the equation for the indicated variable:

$S = n(n+1)/6$ For "n"

I got as far as: $6*S/n+1 = n$, after performing the following steps:

Multiply on both sides by six.

Divide by $n + 1$

We get: $6*S/n+1 = n$

I know that I need to get that n in the denominator to the other side so I can presumably solve a quadratic but how??? I'm not asking for the answer, just what must I do to get past this step??

• Multiply all the terms by $n$. – JB King Nov 27 '15 at 23:35

Instead of dividng by $n$, try this: \begin{equation*} \begin{aligned} S &= \frac{n(n + 1)}{6} \\ 6S &= n(n + 1) \\ 6S &= n^2 + n \\ 0 &= n^2 + n - 6S. \end{aligned} \end{equation*} So you are solving a quadratic equation in $n$. Try to finish from here.

• OK give me a few minutes I will try your suggestions! – MegaWitt Nov 27 '15 at 23:44
• I should have mentioned there are ten questions on this study quiz, I want to know if I'm on the right track before I continue on. I used the quadratic formula: $-1+sqrt(1-4(6*S))/2$ Multiplied by the conjugate: $(-1+sqrt(1-24*S)/2) * -1+sqrt(1-24*S)/-1+sqrt(1-24*S)$ and this is just getting way too difficult I must be doing something wrong :( – MegaWitt Nov 27 '15 at 23:56
• I don't see why you would "multiply by the conjugate." The quadratic formula already is the answer. But you must apply it accurately: the constant term is $-6S$, not $6S$, so you end up with $\frac12(-1 \pm \sqrt{1+24S})$. – David K Nov 28 '15 at 0:28

Hint 1: Using the distributive law, $$(x+a)(x+b) = x^2+Ax+B$$ express $A$ and $B$ in terms of $a$ and $b$ (e.g. $B=ab$, what does $A=$?). Use this model to solve $n^2+n-6S=0$.

Hint 2

Complete the square $$\begin{array}{lll} n^2+n-6S&=&0\\ 4n^2+4n-24S&=&0\\ 4n^2+4n&=&24S\\ (2n)^2+2(2n)&=&24S\\ (2n)^2+2(2n)+c&=&24S+c\\ \end{array}$$ What would $c$ have to be in order to make a perfect square (hint 2a: suppose $m=2n$)

Hint 3

derive the quadratic formula from $ax^2+bx+c=0$ by completing the square $$\begin{array}{lll} ax^2+bx+c&=&0\\ 4a(ax^2+bx+c)&=&4a(0)\\ 4a^2x^2+4abx+4ac&=&0\\ 4a^2x^2+4abx&=&-4ac\\ (2ax)^2+2b(2ax)&=&-4ac\\ (2ax)^2+2b(2ax)+d&=&d-4ac\\ \end{array}$$ for what value of $d$ would make the Left-hand-side a perfect square (suppose that $z=2ax$)? Use these results to solve the problem.

• Are there mistakes/typo's in this? Where is the "m" in $m = 2n$?? Or is that the general form for completing the square? In which case I've never seen that, just examples. I see that you multiplied by four in the second step of hint 2. Is that because you took half the coefficient of the b term squared that giving you 1/4th and then multiplied the entire equation by four to eliminate the denominator?? And jesus. The italics are really throwing me off :P – MegaWitt Nov 29 '15 at 12:15
• We multiply by four because we would like the middle coefficient to be even, and the leading coefficient to be a perfect square. The goal of eliminating fractions until the last step makes this necessary because we do not know whether $n$ is even or odd. – John Joy Nov 29 '15 at 14:17
• What happens to the equation $(2n)^2+2(2n)+c=24S+c$ when we substitute $m$ for every occurrence of $2n$? Ans. the equation becomes $m^2+2m+c=24S+c$, which has a quadratic on the LHS with no leading coefficient. What value of $c$ would make this a perfect square. – John Joy Nov 29 '15 at 14:21
• OK in case you saw that last comment I am sorry about that, this site has not been nice to me a few times and I was under the impression you were attempting to mess me up. That italic text is really confusing, and so too is the way you wrote it out. I get it though. Its more helpful from a certain perspective. I'm sure the answer is $2*root(6) -1/2 = n$ – MegaWitt Dec 9 '15 at 1:11
• Continuing from Hint 2, obviously, $c=1$ $$(2n)^2+2(2n)+1^2=24S+1$$ $$(2n+1)^2 = 24S+1$$ $$2n+1 = \pm\sqrt{24S+1}$$ $$n=\frac{-1\pm\sqrt{24S+1}}{2}$$ – John Joy Dec 9 '15 at 14:43

Hint:

It is equivalent to the quadratic equation in $n$: $$n^2+n-6S=0.$$

• I don't follow you. What is your unknown? – Bernard Nov 27 '15 at 23:44