In the equation below, m, c, and r are positive constants. If the equation is true for all values of $x$ and $6m - 2c = 4$, what is the value of $r$?
$(5x + m)(x + c) = 5x^2 + rx + c$
The correct answer is 6.
current priority level - does anyone “need” this one??
We wrote this question a bit ago, and we’re actually forgetting what official question inspired it. As we craft our questions, we mirror official questions, often ones that are currently in circulation, every so often ones that you can no longer find.💎
THE thing to likely take away from this question is how it could bring to life what we talk about on our Math great assessments page.
For most of us, after we read this question, we do not have a crystal clear first step, nor do we recognize what’s getting tested. Realizing that we are not certain as to how we might approach things, we make the confident and comfortable choice to simply not care about this question. We know we have the breathing room to give this question zero of our attention and brainpower. Even someone seeking to score 750+ can happily LAUGH at this one.🤣
So, we’re done here, right? Cool. See you elsewhere.🫡
Why are you still here?
….
….
Seriously.
One way to start could be to FOIL out the left side. Doing so, we arrive at $5x^2 + 5cx + mx + mc$. Keeping this set equal to what’s originally given us on the right….
….we could realize that $m$ is 1, since ‘$mc$’ and ‘$c$’ are equivalent terms. Knowing that $m$ is 1, we can determine the value of $c$ via the $6m - 2c = 4$ statement. Writing out $6(1) - 2x = 4$, we then have $6 - 2x = 4$. Subtracting 6 from both sides, we arrive at $-2x = -2$. Dividing both sides by -2, we arrive at $c = 1$.
Knowing both $c$ and $m$, we can rewrite what we had initially FOILed as $5x^2 + 5(1)x + 1x + (1)(1)$, which leads to $5x^2 + 5x + x + 1$, which leads to $5x^2 + 6x + 1$. Realizing that the ‘$6x$’ we have on the left corresponded to the $rx$ on the right, we can come to $r$ being 6.
We’re probably going to get some kind of email about this question. Should we, though?