TrueTears question thread

[TrueTears]

What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?


EDIT:

But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b)    (or any function)

Because in a very old VCAA exam, it had a question that says "...find the quadratic EQUATION that has the ROOTS and ..."

In the examiners report they specifically said a lot of people lost marks because they did not set it to zero.

But I did what you did, I wrote y(x) = (x - )(x - ).

[Damo17]

What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?

They are very similar to the point where one would usually say they are equivalent. However there is a subtle difference.
A solution is finding the values in an equation that satisfy the given condition, and finding roots generally implies finding the zeros (a type of solution) that satisfy the given conditions. 'Solution' is a bit broader than 'root'.

[TonyHem]

sound the same to me.

[TrueTears]

What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?

They are very similar to the point where one would usually say they are equivalent. However there is a subtle difference.
A solution is finding the values in an equation that satisfy the given condition, and finding roots generally implies finding the zeros (a type of solution) that satisfy the given conditions. 'Solution' is a bit broader than 'root'.

Oh I get it, so find the roots to a quadratic equations implies setting it to 0 and solving it.

However finding the solutions to a quadratic equation implies that it is already set to something (like an 'initial condition').

So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

However if you said find the solutions to (x-a)(x-b) = y, 'y' must be set to something, so it might be something such as find the solutions to (x-a)(x-b) = 4 (not necessarily set to 0).

But doesn't that mean if a question said find the "roots" to [this] quadratic equation you need to answer it by setting your expression to 0.

But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b)    (or any function)

??

[Damo17]

Oh I get it, so find the roots to a quadratic equations implies setting it to 0 and solving it.

However finding the solutions to a quadratic equation implies that it is already set to something (like an 'initial condition').

So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

However if you said find the solutions to (x-a)(x-b) = y, 'y' must be set to something, so it might be something such as find the solutions to (x-a)(x-b) = 4 (not necessarily set to 0).

Exactly 

But doesn't that mean if a question said find the "roots" to [this] quadratic equation you need to answer it by setting your expression to 0.

Yes you would have to let it equal zero. In what instances are you doubting this?

[TrueTears]

oh just /0's statement went opposite to  my thinking lol, but he's a bright lad so I thought I was wrong.

[TrueTears]

Also if the question said this instead: "find the quadratic function that has the roots a and b"

I thought that since roots mean zero (according to wiki) that means it implies the function MUST be set to 0

so answer would just be f(x) = (x-a)(x-b)

[Damo17]

Also if the question said this instead: "find the quadratic function that has the roots a and b"

I thought that since roots mean zero (according to wiki) that means it implies the function MUST be set to 0

so answer would just be f(x) = (x-a)(x-b)

Yes, that would be the answer.
If you had f(x)=a(x-b)(x-c)=0, then this would be a quadratic equation.
If you had f(x)=a(x-b)(x-c) , then this would be a quadratic function.

[Over9000]

oh just /0's statement went opposite to  my thinking lol, but he's a bright lad so I thought I was wrong.

yeh I think his wrong there, it should be set to 0 in my opinion

[TrueTears]

Thanks all!!!

[/0]

So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

Wait, aren't you saying that and are 'roots' of the equation ?

[TrueTears]

When a question says "...find the quadratic EQUATION that has the ROOTS b and a..."

But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b)    (or any function)

if it says find the roots of an equation then it must be 0=...

[/0]

It has an equals sign so it's an equation (tick)
It has roots a, b (tick)
It is a quadratic equation (tick)

[kamil9876]

has roots   is a clear way of saying that if i plug in x=a or x=b I get an equation that holds.

i.e: just like we say has a integers roots (x,y,z)=(5,12,13) rather than has zero's/roots etc.

Likewise, Pell's equation is called Pell's equation, not Pell's function because we care more about solving it than differentiating it.


I don't know, maybe what you said is technically true according to what's on wiki but my reply to TT on msn was done with haste because I was in a rush and quite annoyed by his numerous nudges and trivial questions. But for aesthetic reasons posted above and observation of what's commonly used, I will always chose my alternative when seeing "equations" and "roots".

Also I think the EQUATION f(x)=(x-a)(x-b) has infinitely many roots since we can rearrange it to:

0=(x-a)(x-b)-f(x)  and what do we define f(x) as?

Whereas the FUNCTION f(x)=(x-a)(x-b) has roots(I will assume this is synonymous to zeros as wiki says, I myslef would use zero's instead) a and b only. (that I am sure is true).


edit: some links:

http://www.sosmath.com/algebra/quadraticeq/root/root.html

http://www.thefreedictionary.com/Root+of+an+equation  (basically the definition I adhere to)

[TrueTears]

ok.

Then according to /0, the VCAA examiners were wrong.