NCEA Level 2 Algebra Exam 2014

From last year’s Algebra exam: I couldn’t see how to rearrange $\large{\boldsymbol{h=rx^2 -tx}}$ to make $\large{\boldsymbol{x}}$ the subject.

You can find a copy of the exam question paper on the NZQA website here.

Question Two (c) (i): (abbreviated) Rearrange $\boldsymbol{h=rx^2 -tx}$ , where $\boldsymbol{r}$ and $\boldsymbol{t}$ are constants, to make $\boldsymbol{x}$ the subject.

The problem here is probably to do with the terminology they’ve used in asking the question. If you approach it as a typical ‘rearrange the equation’ question, and attempt to perform step-by-step rearrangement on it, you’re unlikely to get very far.

Instead, take a step back and look at the type of equation you’re dealing with. It’s a quadratic in $\boldsymbol{x}$ . We have techniques for solving quadratics and, in doing so here, we will get a solution that looks like $\boldsymbol{x}=\hdots$ , which is precisely the same thing as ‘making $\boldsymbol{x}$ the subject’.

So let’s do it. We’ll start by rearranging the equation into the form we need for the Quadratic Formula:

$\large{\boldsymbol{h = rx^2 -tx}} \\ \large{\boldsymbol{rx^2 -tx = h}} \\ \large{\boldsymbol{rx^2 -tx-h = 0}}$

Now, the Quadratic Formula tells us that:

$\large{\boldsymbol{\text{If } ax^2 +bx+c = 0\text{,}}}$

$\Large{\boldsymbol{\text{then }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}}$

In our case we have $\boldsymbol{a=r, b=-t, c=-h}$ . So let’s plug ‘em in and see what we get:

$\Large{\boldsymbol{x=\frac{-(-t)\pm\sqrt{(-t)^2-4(r)(-h)}}{2(r)}}} \\ \Large{\boldsymbol{x=\frac{t\pm\sqrt{t^2+4rh}}{2r}}}$

and we’re done.

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