Quadratic with one x-intercept

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adrianogaleno

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Question - Determine the value of k for which the following quadratic has one x-intercept - [math]kx^2-2kx+6[/math]
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I tried to isolate k, use the y-intecept (6) and rearrange the equation, but I couldn't figure out how the value of K.
 
adrianogaleno said:
Question - Determine the value of k for which the following quadratic has one x-intercept - [math]kx^2-2kx+6[/math]
---

I tried to isolate k, use the y-intecept (6) and rearrange the equation, but I couldn't figure out how the value of K.
Click to expand...
Have you learned the quadratic formula? More specifically, the discriminant.
 
adrianogaleno said:
Question - Determine the value of k for which the following quadratic has one x-intercept - [math]kx^2-2kx+6[/math]
---

I tried to isolate k, use the y-intecept (6) and rearrange the equation, but I couldn't figure out how the value of K.
Click to expand...
Can you calculate the x-intercepts of the given parabola - as functions of 'k'?

What are those?
 
Before you do any work on this problem at all you need to know have a strategy on how to determine if a quadratic has one x-intercept. I would really concentrate on what BigBeachBanana said.
 
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adrianogaleno said:
Question - Determine the value of k for which the following quadratic has one x-intercept - [math]kx^2-2kx+6[/math]
---

I tried to isolate k, use the y-intecept (6) and rearrange the equation, but I couldn't figure out how the value of K.
Click to expand...
After your suggestions, I believe I found the answer. Let me know otherwise :)

Thank you all!
 

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One thing I need to point out.
You had 4k^2 = 24k. Then you divided both sides by 4k. Doing so caused you to miss one solution to your equation.

4k^2 =24k
4k^2 - 24k = 0
4k(k-6)=0
So 4=0 or k=0 or k-6=0
4 never equals 0, k=0 when k=0 and k-6 =0 when k=6
So if 4k^2 = 24 k, then k=0 or k=6.
Do you think that you should change your answer? Why?
 
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Steven G said:
One thing I need to point out.
You had 4k^2 = 24k. Then you divided both sides by 4k. Doing so caused you to miss one solution to your equation.

4k^2 =24k
4k^2 - 24k = 0
4k(k-6)=0
So 4=0 or k=0 or k-6=0
4 never equals 0, k=0 when k=0 and k-6 =0 when k=6
So if 4k^2 = 24 k, then k=0 or k=6.
Do you think that you should change your answer? Why?
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Actually, Steven, there is only one valid answer.

[math]f(x) \text { is a quadratic function } \iff f(x) = ax^2 + bx + c \text { and } a \ne 0.[/math]
Nor does the function f(x) = 6 have any x-intercept.

But you are correct in that the OP does not appear to have consciously considered and then rationally rejected the possibility that k = 0. In other problems of this sort, you may need to consider the possibility of a parameter being zero.
 
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JeffM said:
Actually, Steven, there is only one valid answer.

[math]f(x) \text { is a quadratic function } \iff f(x) = ax^2 + bx + c \text { and } a \ne 0.[/math]
Nor does the function f(x) = 6 have any x-intercept.

But you are correct in that the OP does not appear to have consciously considered and then rationally rejected the possibility that k = 0. In other problems of this sort, you may need to consider the possibility of a parameter being zero.
Click to expand...
Yes, I know there is one answer to the problem. However there is two answer to the equation which I commented on.
 
It's been almost 4 days now and I suspect that the OP will not respond. I hope that I am wrong.
 
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