Discriminant equation: Find exact values of k so y = 1-k-x is tangent to y = kx^2 + x + 2k)

[einar]

Hello can someone direct me the way to generate the answers.

I started solving by setting the equations as 1-k-x = kx^2+X+2k because the tangent and curve intersect at one point. Then I try to solve it by forming b^2- 4ac = 0. But the answer looks odd to me. Therefore, I want to justify whether my method is right or not.

 

[Dr.Peterson]

The method looks right, and the answer is ugly (two irrational solutions), so you're probably right.

What answer did you get? I checked mine out by graphing the line and parabola on Desmos, and they worked. If yours doesn't please show your work so we can see what might be wrong.

 

[khansaheb]

Hello can someone direct me the way to generate the answers.

I started solving by setting the equations as 1-k-x = kx^2+X+2k because the tangent and curve intersect at one point. Then I try to solve it by forming b^2- 4ac = 0. But the answer looks odd to me. Therefore, I want to justify whether my method is right or not.
View attachment 37973

What is the slope of the given straight line? ................................(1)

What would be the slope of the tangent-line at the point of tangency ? ...........................(2)

What should be the relationship between (1) and (2)?

 

[einar]

The method looks right, and the answer is ugly (two irrational solutions), so you're probably right.

What answer did you get? I checked mine out by graphing the line and parabola on Desmos, and they worked. If yours doesn't please show your work so we can see what might be wrong.

Mine was 1±√13 /6

 

[The Highlander]

What is the slope of the given straight line? ................................(1)

What would be the slope of the tangent-line at the point of tangency ? ...........................(2)

What should be the relationship between (1) and (2)?

Surely (1) and (2) are both the same?

Did you mean to write: "What would be the slope of the curve be at the point of tangency ? ...........................(2)"???

(Still not sure how it helps to know that. )

 

[The Highlander]

Mine was 1±√13 /6

I'm afraid that's not what I get!

Your answer gives this...

​

when you should be getting this...

​

Please show us your working so we can identify where your error has crept in.
(A picture of your handwritten work will do, as long as it's clear & legible!)

Regards,
TH.

 

[mario99]

Mine was 1±√13 /6

I got the same result

 

[MaxWong]

Mine was 1±√13 /6

If you meant [imath]\dfrac{1 \pm \sqrt{13}}6[/imath], your answer is the same as mine.
([imath]1\pm\sqrt{13}/6[/imath] could mean [imath]1 \pm \dfrac{\sqrt{13}}6[/imath].)

 

[Dr.Peterson]

What is the slope of the given straight line? ................................(1)

What would be the slope of the tangent-line at the point of tangency ? ...........................(2)

What should be the relationship between (1) and (2)?

Surely (1) and (2) are both the same?

Did you mean to write: "What would be the slope of the curve be at the point of tangency ? ...........................(2)"???

(Still not sure how it helps to know that. )

If the problem were to be solved by calculus (which it pretty clearly is not), then this would be the natural approach: Write expressions for the slope of the line and the slope of the curve (which is defined as the slope of the tangent line to the curve), set those equal, and solve. It helps a lot.

But the discriminant method is suitable for this particular problem, and is presumably the method @einar knows.

Mine was 1±√13 /6

Yes, what you intended to write, k = (1±√13)/6, is what I got. And my graphs were what TH showed as correct.

Your only error is in communicating your answer clearly.

 

[The Highlander]

I got the same result

If you got the same result then your answer was wrong too!

If you meant [imath]\dfrac{1 \pm \sqrt{13}}6[/imath], your answer is the same as mine.
([imath]1\pm\sqrt{13}/6[/imath] could mean [imath]1 \pm \dfrac{\sqrt{13}}6[/imath].)

"1±√13 /6" does mean: 1 ± [imath] \dfrac{\sqrt{13}}6[/imath]!
BOMDAS or BIDMAS (or whatever you call them in your neck of the woods) rules apply!

So there was, indeed, an "error" in the OP's response (regardless of the the possibility that it may only have been "in communicating your answer clearly").

The very first response that the OP got was to show their "work" (not just their answer) and, if the OP had shown their working (or had posted their answer as: "(1±√13) /6"), then I would have had nothing further to say other than, perhaps, to point out that the final answer needed to be made more precise (by bracketing the numerator) if it was shown in their working as "1±√13 /6".

However, when I took the trouble of doing the exercise myself to check their answer (the only way to "justify whether my method is right or not" even though we were only given a very brief insight into exactly what that method was or how (well) it had been executed) and found it to be wrong, I was then forced to go to the bother of graphing their wrong results against the correct ones to illustrate the discrepancies.

I would appreciate it, therefore, if others refrained from adding posts after mine that appear to suggest that it was I who made a mistake here!

 

[Otis]

1±√13 /6

It's interesting how people parse things. When I saw the space after 1±√13, I thought (1±√13)/6.

Then again, I'd already known the quadratic equation was in play, so it was natural for me to jive einar's result with that form.

 

[Dr.Peterson]

A couple of us have a tendency to emphasize what a student has done wrong (often technicalities like notation or lack of units) rather than interpreting generously and pointing out where they are right first (and then, if it seems helpful, pointing out what could have been written more clearly). Students are sometimes left very confused, and subsequent explanations are needed.