Thread: 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 ?

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.

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


So, divide both sides by -4:


The solution set for k is
.

But I said earlier that k must be greater than -2, so intersect this with the solution set and you get
.


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