Quadratic Equations and inequalities

anpanman SG

The curve y = (k-6)x^2 - 8x + k cuts the x-axis at 2 points and has a minimum point. find the range of values of k.

Since it cuts at 2 points, discriminant should be >0.

So I subbed using the b^2 - 4ac method...

I got -2<k<8. my friend says it should be 6<x<8. where is it that I went wrong? is it the wrong application of method or whatsoever?

SexyChin SG

the graph has a minimum point, so k must be more than 6, if not x^2 will be negative, which will make the graph to have either a min or max point. .

Ahm97sic SG

Hi Anpanman,

There are 2 conditions to satisfy for this question ie

(1) the curve cuts the x-axis at two points ie b^2 - 4ac > 0

You have done this part correctly ie the answer is -2 < k < 8

(2) the curve has a minimum point

You have not considered this condition in your answer.

y = (k-6)x^2 - 8x + k

dy/dx = 2(k-6)x - 8

d^2y/dx^2 = 2(k-6)

For minimum point, d^2y/dx^2 must be > 0

So, 2(k-6) > 0

k > 6

Now, combine the two inequalities for the overall answer.

-2 < k < 8 and k > 6.

Hence, the overall answer is 6 < k < 8.

Thank you for your kind attention.

Regards,

ahm97sic

Mikethm SG

I doubt that the setter intended it to be a quad+diff question. It is simply a basic quad question. anpanman is probably in sec 3 only and thus have no idea of differentiation.

For an quadratic equation of the form y= ax^2+bx+c...

min pt mean a>0, max pt mean a<0.

Thus all you have to do was equate a>0

ie (k-6)>0

k>6

Combine with -2<k<8 and you get 6<k<8.

Ahm97sic's answer is correct too. However he complicate the question unnecessarily.

Ahm97sic SG

Hi Mikethm,

I agree with your answer and comment.

Regards

ahm97sic

eagle SG

thanks to both ahm97sic and mikethm for the contributions :D

seems like we hav more and more maths experts coming to help in this forum :D