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Find the range of values of $k$ for which $3x^2-4(k-x)+2$ is always positive for all real values of $x$.

I've tried simplifying, until I got to: $3x^2-4k+4x+2$.

Since it must always be positive, the discriminant, $b^2-4ac$ must be negative, ie $b^2-4ac<0$.

$b^2-4ac<0$

$(4)^2-4(3)(2-4k)<0$

$16-24+48k<0$

$48k<8$

$k<\frac 16$

I got stuck here. Is anyone able to help me continue?

EDIT: Realised that there was a typo, the discriminant should be negative, thank you Deepak.

Matt Chua
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  • Haven't you already arrived at the answer? – Deepak Mar 02 '19 at 07:41
  • @Abhinav What other constraints are you thinking of? The leading coefficient is already given to be positive. – Deepak Mar 02 '19 at 07:49
  • I think I have misread the question as to find positive roots. Sorry for the inconvenience. – Abhinav Mar 02 '19 at 07:53
  • As I've written in the edit to my answer, the discriminant must be negative. – Deepak Mar 02 '19 at 08:03
  • @MattChua the discriminant has to be positive. You solved for negative. – user376343 Mar 02 '19 at 08:33
  • @user376343. No, the discriminant has to be negative to ensure the quadratic expression has no real roots (and therefore always returns positive values since the lead coefficient is positive). The key is that the quadratic must have no (real) zeroes. A non-negative discriminant implies at least one real zero. – Deepak Mar 02 '19 at 10:37
  • My comment was related to this part in OP: "... the discriminant, $b^2−4ac$ must be positive, ie $b^2−4ac<0.$" – user376343 Mar 02 '19 at 12:15
  • @user376343 Asker has fixed it. It's correct now. – Deepak Mar 02 '19 at 12:51

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You seem to have already got the answer (which is simply $k < \frac 16$). There is only one additional modification I would make to your answer. That is to assert that the coefficient of $x^2$ is positive (in this case $3$). If it were negative, you would never be able to satisfy the required condition since for high enough values of $x$, the quadratic expression would always be negative.

EDIT: just noticed you wrote that the discriminant must be positive. I hope that's just a typo, and not a conceptual error. It seems to have been just a typo because you set up the inequality correctly. Anyway, remember that you need complex roots for a quadratic curve to lie wholly on one side of the $x$-axis or the other (which side depends on the sign of the leading coefficient). Complex roots means the discriminant is negative).

Deepak
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