Math CentralQuandaries & Queries


Question from claire, a parent:

Find the range of values of k for which k(x^2+2x+3) - 4x - 2 is never negative.

Here is what I did

kx^2+2kx+3k-4x-2 > 0
kx^2+x(2k-4)+3k-2 >0

As k(x^2+2x+3)-4x-2 is never negative b^2-4ac<0

(2k-4)^2 - 4(k)(3k-2) <0
-8k^2-8k+16 < 0
8k^2 +8k-16>0
k^2 + k - 2 > 0

(k+1)(k-2) > 0

After sketching the graph curve,

I got k<-1 or k>2 (could these be wrong because the given answer is 0 < k =or< 1)

Hi Claire,

Well, I disagree with both your answers. You have an error in the last step.

k2 + k - 2 = (k + 2)(k - 1)

and hence the roots are -2 and 1. Also you are looking for values of k so that the expression is never negative so it could be positive or zero and hence you want values of k so that the discriminant is ≤ 0, that is

-8(k + 2)(k - 1) ≤ 0.

The solution is

k ≤ -2 or k ≥ 1.

But this can't be correct! If k is negative then y = k(x2+2x+3) - 4x - 2 = kx2+x(2k-4)+3k-2 is a parabola that opens downward and hence there must be values of x where y is negative. k must be positive! so the correct answer is k ≥ 1.

k ≤ -2 are the values of k for which k(x^2+2x+3) - 4x - 2 is never positive.

I hope this helps,
Harley Weston

About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS