## Friday, September 26, 2014

### So Smart They're Dumb?

I was talking recently to a friend of mine who teaches calculus at an "academically solid" school, and he was mortified that not one of his calculus students got the correct answer set to the following equation:

3x^3-9x=0

Not one.

What do you think was the most common mistake made?

maxutils said...

No question. Not being able to factor.

maxutils said...

And more specifically, now that I looked at it more closely, not being able to recognize the difference of two squares. PeggyU said...

3x(x^2 - 3)

x = 0; x = sqrt 3; x = -sqrt 3 ... so, I dunno.

I'm guessing they added 9x to both sides, then divided the x out ...? Anonymous said...

not using 0 as a solution?

sdsdfd said...

Let's see - putting on my "typical student" hat, I'll take a stab:

* Subtract 9x from both sides
* Divide both sides by 3
* Take the cube root of both sides

Answer: x = cube root of x

What could be simpler ;) ?

(I've seen this essential approach taken with quadratic trinomials that don't have a middle term, so I'm guessing they could utilize the same rationale with a cubic that has no middle terms ~ )

sdsdfd said...

Rather, I've seen this approach taken with trinomials that DO have a middle term ~

gotta love typos ~ Anonymous said...

They did this:

3x^3=9x
3x^2=9
x^2=3
x=sqrt(3)

They certainly missed the zero solution, and probably half of them missed -sqrt(3). (Yes, I work with HS students)

Teacher gardener said...

will you post the answer later? Anonymous said...

Not eyeballing it to see if x=0 worked? Ie, not thinking about what they were doing.

-Mark Roulo Anonymous said...

3x^3 - 9x = 0

3x (x^2 - 3) = 0

3x (quadrtratic equation goes here) = 0, or
x = 0; sqrt(3); -sqrt(3)

Did they miss the negative solution to sqrt(3) ?

-Mark Roulo

Darren said...

The most common incorrect answer was to add 9x to both sides and, when dividing both sides by x, forget the stipulation about checking if x=0 is a solution.

Of course the "best" way to solve it would be to factor a 3x out of the left side, leaving 0 and +/-root three as solutions.

maxutils said...

Leave it to me not to come up with the worst method ... :)
Anoymous -- no need for the quadratic on (x^2-3) ... it factors easily to (x + rt3)(x- rt3)...

Auntie Ann said...

I could easily see not getting the + or - part. That's easy to forget.

maxutils said...

Auntie Ann ... that's the beauty of factoring ... you don't have to remember that rule. It presents itself to you... it also means you don't have to check for 0 being an answer, because it is also right there. Dividing both sides by x without checking to see if 0 is a possible solution ... I can't even imagine doing that. Anonymous said...

"Dividing both sides by x without checking to see if 0 is a possible solution ... I can't even imagine doing that."

I don't understand the thinking/mistake behind dividing both sides by x at all. You've got the equation in a form with all the xs on one side and the other side equal to zero. This is what you want, right?

What sort of mistaken idea would lead you to divide by x here (rather than factoring out the x)?

-Mark Roulo

Auntie Ann said...

There's also the simple check that should be one of the first things a person thinks about: a 3rd power polynomial likely has 3 roots.

maxutils said...

Because you CAN divide both sides by 3 ... because 3 can't be zero. So it can be appealing to try the same thing with x ...