#### hoosie

##### Junior Member
Applying the definition of root is one way of solving the problem. I don’t agree that graphing the parabola y = (x - 2)(x -4) makes things more complicated for the student. It helps them to see the question in its true context - that the roots of the equation (x-2)(x-4) = 0 are not just worked out in isolation but are linked in their mind to the x - intercepts of the parabola.
The student didn’t know if 0 was a root of the equation or not. My point was that a simple visualisation of the graph would have told them that the curve doesn’t pass through the origin - a condition the curve must satisfy for 0 to be a root.

#### Jomo

##### Elite Member
Applying the definition of root is one way of solving the problem. I don’t agree that graphing the parabola y = (x - 2)(x -4) makes things more complicated for the student. It helps them to see the question in its true context - that the roots of the equation (x-2)(x-4) = 0 are not just worked out in isolation but are linked in their mind to the x - intercepts of the parabola.
The student didn’t know if 0 was a root of the equation or not. My point was that a simple visualisation of the graph would have told them that the curve doesn’t pass through the origin - a condition the curve must satisfy for 0 to be a root.
This is not a problem of graphing a parabola. It is about understanding that if A*B=0, then either A and/or B is zero. Or even at a lower level whether or not (0-2)(0-4)=0.

A student must know what I wrote above. You can't argue with that, can you? There are other type of problems where they should know about x-intercepts. This is not one of them. Would you always have a student graph a quadratic to find the roots. Never by factoring, completing the square or by quadratic formula? There is a time and place for everything.

The student clearly did not know ( but answered Khan's question perfectly) to just plug in 0 for x. That immediately troubled me and I would have addressed that situation before going to another method.

#### hoosie

##### Junior Member
To find whether or not 0 is a solution to (x - 2)(x - 4) = 0 I agree a student must know that if they they replace x with 0 and the equation holds true then 0 must be a root. I would also encourage a student to use the Null Factor law to show neither of these roots is zero. I am certainly not saying they have to graph the parabola to work this out but I am saying that if a student can be encouraged to picture the graph (seeing the curve cuts the x -axis at roots 2 and 4) then they they will immediately see the curve doesn’t pass through the origin. Rather than complicating things wouldn’t being able to picture the graph act as a check and aid their understanding by helping them to place the question in context? Admittedly not all students would take up the encouragement to do this but in my experience those that do become better Math students overall.

#### Jomo

##### Elite Member
To find whether or not 0 is a solution to (x - 2)(x - 4) = 0 I agree a student must know that if they they replace x with 0 and the equation holds true then 0 must be a root. I would also encourage a student to use the Null Factor law to show neither of these roots is zero. I am certainly not saying they have to graph the parabola to work this out
You are contradicting yourself. Your post clearly instructed the student to use a graph.
Let's get right to the point where we are now. All the helpers here have made mistakes or used a less than convenient methods. This is a place for learning and this includes for the helpers. A number of qualified helpers (mathematicians, college math professors, professional mathematicians, professional engineers, financial experts, etc etc) said your method was wrong and you want to challenge all of them. You may honestly think that you are correct but I feel that a true educator is one who can learn from others. Please listen when a number of helpers here say that you're method is wrong. I myself have over 10,000 hours of tutoring experience and decades of full-time community college teaching. Try to learn from others!

#### hoosie

##### Junior Member
I did not instruct the student to graph the parabola as a sole approach to answering the question - you are clearly misinterpreting my post. I was responding to Jeff’s post and introduced graphing as a way for the student to check their answer and to consolidate their understanding of the problem by placing it in context. In this case if, after trying Jeff’s approach, the student could have simply drawn a quick sketch of the curve y = (x-2)(x-4) (or visualised it) they would have immediately seen zero is not an x-intercept and therefore not a solution to (x-2)(x-4) = 0. I believe it is important for the student to understand the link between the two. They would then know, for example ,that the equation x(x - 5) = 0 does have a zero root because the graph of y = x(x - 5) passes through the origin. Of course this approach supplements
Jeff’s method of substituting x = 0 to see if the equation holds true.

#### Jomo

##### Elite Member
I believe it is important for the student to understand the link between the two.
This is exactly what we said is wrong. But I guess you must always be correct. Fine by me.

#### JeffM

##### Elite Member
I was responding to Jeff’s post and introduced graphing as a way for the student to check their answer and to consolidate their understanding of the problem by placing it in context. In this case if, after trying Jeff’s approach, the student could have simply drawn a quick sketch of the curve y = (x-2)(x-4) (or visualised it) ... to see Jeff’s method of substituting x = 0 to see if the equation holds true.
You are brilliant. You responded to my post, # 6, in post # 5. Sometimes, it's just best to admit that you got off on the wrong foot. Every helper here has gone off on a tangent a few times.

#### Jomo

##### Elite Member
You are brilliant. You responded to my post, # 6, in post # 5. Sometimes, it's just best to admit that you got off on the wrong foot. Every helper here has gone off on a tangent a few times.
That's an excellent point!

#### hoosie

##### Junior Member
I accept my mistake Jeff - in post #11 I should not have made reference to your previous post..

With hindsight my first post should have mentioned the graphing approach could be used as a check against the root definition method.

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#### Cubist

##### Junior Member
I personally think the criticism of Hoosie's post#5 has been overly harsh.

I agree that posts #4, #6 and #8 are the best and fastest way to let most people find the answer. But given that OP had stopped posting their previously quick responses then I think it was very reasonable for Hoosie to try a different approach. (In all likelihood the OP had understood at post#4 and had simply gone away)

I might be a bit biased because I have a very visual memory! I like the idea of illustrating graphically the one-to-one relationship between x-axis crossings and the individual factors of a polynomial (not just a quadratic). It is easy to see the polynomial is not zero between any of the factors. This is less direct then post #8, but it might be a powerful illustration for some people - BUT ONLY while learning (I'm certainly not suggesting a graph should be done every time!)

#### JeffM

##### Elite Member
I personally think the criticism of Hoosie's post#5 has been overly harsh.

I agree that posts #4, #6 and #8 are the best and fastest way to let most people find the answer. But given that OP had stopped posting their previously quick responses then I think it was very reasonable for Hoosie to try a different approach. (In all likelihood the OP had understood at post#4 and had simply gone away)

I might be a bit biased because I have a very visual memory! I like the idea of illustrating graphically the one-to-one relationship between x-axis crossings and the individual factors of a polynomial (not just a quadratic). It is easy to see the polynomial is not zero between any of the factors. This is less direct then post #8, but it might be a powerful illustration for some people - BUT ONLY while learning (I'm certainly not suggesting a graph should be done every time!)
Cubist and Hoosie

I should have said in post # 6 "makes no pedagogic sense." Leaving out the word "pedagogic" may have contributed to an overly harsh tone, and for that I apologize. Nevertheless, I want to clarify the pedagogic point that I was trying to make.

If you are asked whether a is a root of f(x), the general method to determine that question is to see whether f(a) = 0. It is general because it derives directly from the definition of "root." So to focus on the specific facts about this function, namely that it is a quadratic and that therefore its graph is a parabola with at most two x intercepts, completely loses sight of the general lessons that this problem could teach, namely to pay close attention to definitions and to learn general methods that will apply to whole classes of problems.

Now I agree that if you have a graphing calculator to hand, it may just as easy to determine the roots of a specific function as to calculate the value of that function at a given point. But to use the graph to do that, you still have to grasp the definition of root. That crystal clear focus on definitions was, for me at least, the brilliance of SK's answer, and, to all appearances, the OP got that message virtually instantaneously. To have a later post say, "it may make more sense" to graph than to understand a definition when the graph will help only if you already understand the definition really does make no pedagogic sense to me.

Now, in a later post, hoosie has said that he intended it as a method for checking the answer suggested by SK. Obviously, I have no objection to checking answers. Nor do I have any objection whatsoever to reinforcing an idea by visual means. What I originally objected to was the apparent failure to see that any method for solving the specific problem depended on understanding the general definition of root and that that understanding gives rise to a conceptually obvious way to solve any problem of this kind.