1. ## function problem

I am having problem doing this problem it is asking to find the condition that must be satisfied by k in order that the expression $2x^2 +6x +1 +k(x^2+2)$ may be positive for all real values of x.
Can some help me with this problem?

2. Originally Posted by scrible
I am having problem doing this problem it is asking to find the condition that must be satisfied by k in order that the expression $2x^2 +6x +1 +k(x^2+2)$ may be positive for all real values of x.
Can some help me with this problem?

That's a parabola , so it'll be always positive iff (1) it is a "smiling" or convex upwards parabola (i.e., its higher coefficient is positive), and (2) its discriminant is negative, i.e.: iff it has no real roots, so:

$2x^2 +6x +1 +k(x^2+2)\Longrightarrow (k+2)x^2+6x+(2k+1)>0\,\,\forall\,x\in \mathbb{R}\Longleftrightarrow$ $\,(1)\,\,k+2>0\,\,and\,\,(2)\,\,\Delta=b^2-4ac=6^2-4(k+2)(2k+1)<0$:

$(1)\,\,k+2>0\,\Longrightarrow\,k>-2\,;$

$(2)\,\,\Delta=36-8k^2-20k-8<0\,\Longrightarrow$ $\,-8k^2-20k+28<0\,\,(divide\,\,by\,\,-4)\,\,\Longrightarrow\,2k^2+5k-7>0$ $\Longrightarrow\,(2k+7)(k-1)>0\,\Longrightarrow\,k<-\frac{7}{2}\,\,or\,\,k>1$

Taking together both conditions for k on (1)-(2) the solution is immediate.

Tonio

3. Originally Posted by tonio
That's a parabola , so it'll be always positive iff (1) it is a "smiling" or convex upwards parabola (i.e., its higher coefficient is positive), and (2) its discriminant is negative, i.e.: iff it has no real roots, so:

$2x^2 +6x +1 +k(x^2+2)\Longrightarrow (k+2)x^2+6x+(2k+1)>0\,\,\forall\,x\in \mathbb{R}\Longleftrightarrow$ $\,(1)\,\,k+2>0\,\,and\,\,(2)\,\,\Delta=b^2-4ac=6^2-4(k+2)(2k+1)<0$:

$(1)\,\,k+2>0\,\Longrightarrow\,k>-2\,;$

$(2)\,\,\Delta=36-8k^2-20k-8<0\,\Longrightarrow$ $\,-8k^2-20k+28<0\,\,(divide\,\,by\,\,-4)\,\,\Longrightarrow\,2k^2+5k-7>0$ $\Longrightarrow\,(2k+7)(k-1)>0\,\Longrightarrow\,k<-\frac{7}{2}\,\,or\,\,k>1$

Taking together both conditions for k on (1)-(2) the solution is immediate.

Tonio
Thank you very much