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
Best Answer
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 $k>1$