Question

`0.5x^2+bx+8` How to choose `b` so that 6 children will be able to solve different inequalities and get different answers?

Teacher has explained, that we can choose whether the given equation will be greater than zero, less, or even equal to zero. And this is how we get an inequality.

Answer 1

See below:

Explanation:

Starting with

`0.5x^2+bx+8`

we're allowed to set this equal (or inequal) to 0. Let's keep this part simple and simply set it equal for now:

`0.5x^2+bx+8=0`

With that, what are potential values for `b`?

Keep in mind that when we factor, we end up with:

`(mx+n)(px+q)=0` and if we distribute this out, we get:

`(mpx^2+mqx+npx+nq)`

To have this more general formula fit into what we're looking for in our search for `b`, we'll have:

`overbrace(mpx^2)^(0.5x^2)+overbrace(mqx+npx)^(bx)+overbrace(nq)^8`

Or in other words:

`mp=0.5`
`mq+np=b`
`nq=8`

In this way, we can see that `m,p` are factors of `0.5`, `n,q` are factors of 8, and `mq+np` will give us our `b`.

We need 6 different combinations of these factors, so let's set them up and take a look at what they will generate:

`((m,p,n,q,mq,np,mq+np=b),(1,0.5,8,1,1,4,5),(1,0.5,4,2,2,2,4),(0.5,1,8,1,0.5,8,8.5),(0.5,1,2,4,2,2,4),(-1,-0.5,8,1,-1,-4,-5),(-1,-0.5,4,-2,-2,-2,-4),(-0.5,-1,8,1,-0.5,-8,-8.5))`

(I kept in the 4th line even though it led to a duplicate `b`).

And so we can have, as factors:

`(x+8)(0.5x+1)=0.5x^2+5x+8=>b=5` From line 1

and so on through the 6 different lines (skipping line 4 for being a duplicate).