Ranges of k values in quadratic for it to be positive

Question

Here is the problem

$2x^2+6x+1+k(x^2+2)$, find the condition that must be satisfied by k in order that the expression may be positive for all real values of x.

The quadratic of the form $ax^2+bx+c$ will be positive for all real values of x, if $\displaystyle{\frac{4ac-b^2}{4a} > 0}$.

Therefore, $\displaystyle{\frac{4(2+k)(2k+1)-36}{4(2+k)}>0}$
$\displaystyle{\frac{(k-1)(2k+7)}{(2+k)}}>0$

Considering the 4 ranges, with $k$ values of -14/4, -2 and 1 The inequality is true when $k<-2$ and $k>1$.

However the book answer is only $k>1$. What have I done wrong?

Thank you

Answer 1

For the quadratic expression $ax^2 + bx+c$ to be greater than zero for all real $x$, two conditions must be satisfied: $a>0$ and the discriminant $b^2 - 4ac<0$

Here $a = k+2, b = 6, c = 2k+1$

The first condition gives $a > 0 \implies k+2 > 0 \implies k > -2$

The second condition gives $b^2 - 4ac < 0 \implies 36 - 4(k+2)(2k+1)<0 \implies 2k^2 + 5k - 7 > 0 \implies (2k+7)(k-1) > 0 \implies k<-\frac 72$ or $k>1$

However, note that $k<-\frac 72$ violates the previously derived condition $k>-2$

Hence the only valid condition is $k>1$

Answer 2

The condition $\frac{4ac-b^2}{4a}\gt 0$ is not correct because this includes the case when $a\lt 0$ and $b^2-4ac\gt 0$. (Imagine the graph of a parabola $y=-x^2+x+1$, for example.)

First of all, we need that the coefficient of $x^2$ is positive, i.e. $k+2\gt 0$. (note that $k=-2$ does not satisfy the condition.) Under this condition, you need to have $b^2-4ac\lt 0$, i.e. $9-(2+k)(2k+1)\lt 0$, i.e. $k\lt -\frac 27$ or $k\gt 1$. Thus, we have $k\gt 1$. This is sufficient.

Answer 3

we need a case work. 1. $k=-2$ doesn't work.
2. for $k>-2$ we have $\left(x+\frac{3}{k+2}\right)^2+\frac{2k+1}{k+2}-\frac{9}{(k+2)^2}>0$ for all $x$ The last two summands are $$\frac{(2k+7)(k-1)}{(2+k)^2}$$ this is positive if $k>1$ and the case $k<-\frac{7}{2}$ doesn't exist. 3. the case $k<-2$ is not possible. (Why?)$