Food for thought for both teachers and students

Food for thought:

It's rare that teachers will give something that are truly questioning and cultivating students' interest in math and science subjects. If math educators shift their teaching methodology towards more inquiry based teaching, and let students to take part by start thinking, improve their ability to think abstractly and multi-dimensionally, this will build strong foundation of math for students. This strong foundation will enable students to acquire higher-order skills needed for the 21st century and to becoming highly competent and extraordinary intelligent students that could always plan ahead, see the future consequences of an action, provide alternative explanation of events and reason more effectively.

The thinking process is often a good way to turn on our ideas light.

In today blog post, we will focus on the expression $b^4-2b^2-4b+70$ as it has lots to offer if we milk it for all it's worth...

Note that

$\begin{align*}(b^2-2)^2&=b^4-4b^2+4\\&=b^4-2b^2-4b+7-2b^2+4b-3\\&=(b^4-2b^2-4b+7)-2(b-1)^2-1\end{align*}$

1.

What can you say about the quantities $(b^2-2)^2$ and $b^4-2b^2-4b+7$?

Ah, it's easy and self-explanatory, isn't it? Since $-2(b-1)^2-1\lt  0$ for all $b\in R$, we can deduce that

$(b^2-2)^2\lt b^4-2b^2-4b+7$ for all $b\in R$.

Now, if we're told $b>3$, and $b$ is an integer, and given

$\begin{align*}(b^2+1)^2&=b^4+2b^2+1\\&=b^4-2b^2-4b+7+4b^2+4b-6\\&=(b^4-2b^2-4b+7)+4\left(b-\dfrac{1}{2}\right)^2-7\end{align*}$

2.

What first you can say about $4\left(b-\dfrac{1}{2}\right)^2-7$ when $b$, an integer, is greater than $3$?

If it is a bit hard for you to picture the scenario mentally, you could work for some cases and take $b=4,\,5,\,6$ to figure what you would wind up with...

$4\left(4-\dfrac{1}{2}\right)^2-7=42$

$4\left(5-\dfrac{1}{2}\right)^2-7=74$

$4\left(6-\dfrac{1}{2}\right)^2-7=114$ and so on and so forth...

So, we can conclude that $4\left(b-\dfrac{1}{2}\right)^2-7\gt0$ for $b\gt 3$, and $b$ is an integer.

3.

Which is bigger, $(b^2+1)^2$ or $b^4-2b^2-4b+7$?

Since $(b^2+1)^2=(b^4-2b^2-4b+7)+4\left(b-\dfrac{1}{2}\right)^2-7$ and $4\left(b-\dfrac{1}{2}\right)^2-7\gt0$ for $b\gt 3$, it's safe to say that

$(b^2+1)^2\gt b^4-2b^2-4b+7$

Now, we can safely position the above three expressions $b^4-2b^2-4b+7,\,(b^2+1)^2,\,(b^2-2)^2$ in the real number line, as shown below:

Here comes the most difficult question:

Do you think $b^4-2b^2-4b+50$ or $b^4-2b^2-4b+70$ lie outside the range between $((b^2-2)^2,\,(b^2+1)^2)$?

If you can't tell off the top of your head, please don't lose hope, you can still try out numerically, you know if $b=4$, we have

$\begin{align*}((b^2-2)^2,\,(b^2+1)^2)&=(14^2,\,17^2)\\&=(196,\,289)\end{align*}$

And

$b^4-2b^2-4b+7=215,\,b^4-2b^2-4b+50=258,\,b^4-2b^2-4b+70=278$

Therefore we get

$(196,\,215,\,258,\,278,\,289)$.

Thus, $b^4-2b^2-4b+50$ or $b^4-2b^2-4b+70$ lie inside the range between $((b^2-2)^2,\,(b^2+1)^2)$.

This blog post is going to be a big help to solve for the problem that is going to be another big hit soon...