Quadratic Inequality and Discriminant

[johnvedro72]

The question:

The solution:


What I'm stuck on:
So, I've gotten as far as arriving at 0 < k < 3, after using the discriminant to find the values of k, then plotting these values on a quadratic graph and realising that I should produce an inequality that contains all the values between these two roots, however, I'm still unsure of how to arrive at the "0 is less than or equal to k" part of the answer. I've used a red rectangle to outline the part that I don't understand, could someone care to explain it to me? I don't understand why they've plugged 0 into the original equation, or why "the equation gives 3 = 0" validates why k can be equal to 0?

Examboard: AS Pure Mathematics - Edexcel

 

[Dr.Peterson]

The discriminant is only applicable to a quadratic equation! They observed that if k=0, you don't have a quadratic, but the equation 3=0. (Just plug in k=0.) Since this is false, there is no solution in that case, too, so they included that in the answer.

 

[JeffM]

Obviously Dr. Peterson is correct, but what a weirdly phrased problem!

[MATH]0x^ - 2 * 0 * x + 3 = 0 \implies 3 = 0[/MATH],

is not a valid equation at all. Yet the question implies that it is a valid equation.

I'd have preferred if the question were worded

[MATH]\text {The equation } kx^2 - 2kx + 3 = 0, \text {where } k \text { is a non-zero constant ...}[/MATH]

 

[Dr.Peterson]

Obviously Dr. Peterson is correct, but what a weirdly phrased problem!

[MATH]0x^ - 2 * 0 * x + 3 = 0 \implies 3 = 0[/MATH],

is not a valid equation at all. Yet the question implies that it is a valid equation.

I'd have preferred if the question were worded

[MATH]\text {The equation } kx^2 - 2kx + 3 = 0, \text {where } k \text { is a non-zero constant ...}[/MATH]

I'd say that 3=0 is a perfectly valid equation. Valid doesn't have to mean it is ever true!

But I imagine the problem is intended to be tricky, in that there are two very different ways to have no real roots. They were very kind in telling us what to prove, not just asking for what values of k is has no real roots. Most of us would probably miss the zero case.

 

[JeffM]

I'd say that 3=0 is a perfectly valid equation. Valid doesn't have to mean it is ever true!

But I imagine the problem is intended to be tricky, in that there are two very different ways to have no real roots. They were very kind in telling us what to prove, not just asking for what values of k is has no real roots. Most of us would probably miss the zero case.

You and I disagree about the meaning of "valid." But there is no point in arguing about definitions. I'd say 3 = 0 is well formed.

 

[Dr.Peterson]

In any case, the problem doesn't use the word "valid"; it uses the word "equation", and 3 = 0 is certainly an equation.

 

[Steven G]

I'd say that 3=0 is a perfectly valid equation. Valid doesn't have to mean it is ever true!

I am extremely big on equality signs being valid, but I do agree with Dr Peterson saying valid or not, 3=0 is an equation.

 

[JeffM]

I am extremely big on equality signs being valid, but I do agree with Dr Peterson saying valid or not, 3=0 is an equation.

I agreed it is well formed. We are disagreeing over what is the most useful definition of the word "valid." And I admit the word "valid" does not explicitly appear in the statement of the problem.

I simply dislike math problems that involve rhetorical tricks. Probably one of my numerous personal failings.

 

[johnvedro72]

So let me get this straight,
We're looking for an inequality which shows all the set of values k can be, that agrees with the quadratic not having any real roots. Then we use the discriminant and a quadratic graph to find these values of k. Then we realise this inequality isn't the same as the one we're being asked to prove, so we have to plug 0 back into the original quadratic to see if it ends up having no real roots, instead, we're given a false equation 3=0 which isn't even a quadratic; no real roots. So k can equal to 0, and the quadratic having no real roots is still valid. Wow, I almost jumped out of my skin getting my head around all of this, HaHa!

Thank you @Dr.Peterson

 

[Dr.Peterson]

Sounds good, except that in a couple places you should replace the word "quadratic" with "equation", as it isn't necessarily a quadratic!