Highschool Algebra: $n^2 = 18n$?

Question

I'm beginning to get into maths outside of school and at the moment I'm refreshing myself on the basics which explains why this question appears to be so simple.

I formulated this equation to find the answer to a question:

$n^2 = 18n$

I then manipulated it to:

$n^2 - 18n = 0$

So this is now a quadratic equation (right?), the problem is that there are 2 solutions.

$(n-6)(n+3) = 0$
$(n-9)(n+2) = 0$

In this case, are both solutions acceptable or am I doing something wrong?

Answer 1

There's your problem:

$$(n-6)(n+3) = n^2 - 6n - 18 \neq n^2 -18n$$

Also

$$(n-9)(n+2) = n^2 - 7n - 18\neq n^2 - 18n$$

However, you can simplify $$n^2-18n =n(n-18)$$

Answer 2

This is wrong

If you expand $(n−6)(n+3)=0$ it will become $n^2-3n-18$

Answer 3

$n^2-18n=0\\n\times n-18\times n=0$

so there are $(n-18)$ groups of $n$ and end up with

$n\times(n-18)=0\\ n(n-18)=0$

Now if two numbers multiply to get 0, either the first one or the second one is zero.

$n=0$ or $n-18=0$

$n=0$ or $n=18$

You can always test your solution by substituting back into $n$.

Answer 4

I suspect your mistake is that you've misunderstood factorising quadratic expressions. You have listed the factor pairs of $18$ and put them into brackets which is incorrect. Factorising is a bit of an art and takes a little time to get used to but the key fact here is there is no constant term. Usually this hints at taking a factor of $n$ out to give $n(n-18)=0$ I'd advise revising factoring before move on it'll help in the long run.

Hope this helps.