# nature of roots

• Jul 18th 2009, 05:49 AM
thereddevils
nature of roots
Determine the condition to be satisfied by k such that the expression 2x^2+6x+1+k(x^2+2) is positive for all x belongs to real numbers .

Let 2x^2+6x+1+k(x^2+2)=0

(2+k)x^2+6x+1+2k=0

The graph must be a 'floating' graph on the positive side . I am not sure if i can say it that way. Thus , it has no real roots ..

6^2-4(2+k)(1+2k)<0

-8k^2-20k+24<0

I got the wrong range of values for k . Where is my mistake ?
• Jul 18th 2009, 07:12 AM
yeongil
Quote:

Originally Posted by thereddevils
Determine the condition to be satisfied by k such that the expression 2x^2+6x+1+k(x^2+2) is positive for all x belongs to real numbers .

Let 2x^2+6x+1+k(x^2+2)=0

(2+k)x^2+6x+1+2k=0

First off, the leading coefficient must be positive in order for the expression to be positive for all x belonging to R, so
2 + k > 0 => k > -2.

Quote:

The graph must be a 'floating' graph on the positive side . I am not sure if i can say it that way. Thus , it has no real roots ..

6^2-4(2+k)(1+2k)<0

-8k^2-20k+24<0

I got the wrong range of values for k . Where is my mistake ?
The last inequality should be
$-8k^2 - 20k + {\color{red}28} < 0$

So, divide both sides by -4:
\begin{aligned}
2k^2 + 5k - 7\;&{\color{red}>} 0 \\
(2k + 7)(k - 1) &> 0
\end{aligned}

The solution set for k is
$(-\infty, -7/2) \cup (1, \infty)$.

But I said earlier that k must be greater than -2, so intersect this with the solution set and you get
$(1, \infty)\;\text{or}\;k > 1$.

01