$$kx^2+(k+2)x-3=0$$
This quadratic has roots which are real and positive.
Find all possible values of $k$.
I had already tried using the discriminant and reached this point
$$ Δ = (k+8)^2-60 $$
$$ ==> (k+8)^2-60>0 $$
$$ k>2\sqrt{15}\ – 8 $$
and
$$ k<-2\sqrt{15}\ -8 $$
However this didn't look right. Any suggestions?
Edit: I think I might have figured it out.
Since we know,
$$ k\not= 0 $$
$$ \frac{-3}{k}>0 $$
$$ \therefore k<0 $$
What I got earlier was not wrong, but rather incomplete.
$$ k<-2\sqrt{15} -8 $$ This can be ruled out since, $$ k<0 $$
$$ \therefore 2\sqrt{15} -8 < k < 0 $$
If someone could please still check my work that would be nice.
Best Answer
Hint:
Using Vieta's formulas, you can see that the multiplication of the roots of a quadratic $ax^2+bx+c$ is given by $\frac{c}{a},$ which is equal to $\frac{-3}{k}$ in your case.
Since both roots are positive, this means that $\frac{-3}{k}$ must be positive and $k$ must be negative. Also, the sum of the roots (given by $-\frac{b}{a}$) must be positive too: $-\frac{k+2}{k} > 0$.
Since we already know that $k$ is negative, we get $-(k+2) < 0 \iff -2 < k $. Hence, $-2<k<0$ so far.
Now in order for this quadratic to have real roots, its discriminant must be non-negative. After you write it down, you see that you should solve the inequality $$(k+2)^2-4k(-3) =(k+2)^2+12k \geq 0$$ to find the range of values for $k,$ and then intersect it with $$-2 < k <0$$ at the end.
$$(k+2)^2+12k \geq 0 \iff k\geq \sqrt{60}-8 \text{ or } k\leq -8 -\sqrt{60}$$
Intersecting this range with $-2 < k < 0$ gives $\sqrt{60}-8 \leq k<0$.