# Find all values of $k$ for $kx^2+(k+2)x-3=0$ with positive roots.

$$kx^2+(k+2)x-3=0$$

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.

• I appreciate the advice, I had already tried something but was not sure if it was the correct "path" for this problem so I didn't show it. – StrBoP Jul 9 '18 at 18:48
• You're welcome. Note that positive numbers are always real, so the "real" in "real and positive" is redundant. – Shaun Jul 9 '18 at 18:54
• Do you mean $(k+2)^2$? – Chris2018 Jul 9 '18 at 18:59
• No, I can show more work if you'd like. But after simplifying to a greater extent you get a quadratic, Since that quadratic wasnt factorable I completed the square and then I got that. Sorry if I confused you. – StrBoP Jul 9 '18 at 19:02

## 2 Answers

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 $\sqrt{60}-8 \leq k<0$.

• Might also be helpful to quote Vieta's formulas. – TheSimpliFire Jul 9 '18 at 18:47
• A positive product of the roots is not enough. – Bernard Jul 9 '18 at 18:51
• @stressed out , Thank you for the help – StrBoP Jul 9 '18 at 18:59
• How do you get $-(k+2) < 0 \iff 2 < k$? $$-(k+2) < 0 \implies -2 < k$$ From this it does not follow that $2 < k$ – gd1035 Jul 9 '18 at 19:05
• @stressedout Sorry for bothering once again, but I checked at the back of the book and the answer read: -8 +√60 </ k < 0 – StrBoP Jul 9 '18 at 19:05

You have three conditions to check:

1. This equation has real roots. It means it is a quadratic equation ($k\ne 0$) and its discriminant should be positive: $$\Delta=(k+2)^2+12k=k^2+16k+4=(k+8)^2-60>0,$$ so either $k<-8-2\sqrt{15}\:$ or $\:k>-8+2\sqrt{15}$.
2. The roots must have the same sign, i.e. their product $-\dfrac 3k>0$, which means $\:\color{red}{k<0}$.
3. This sign must be positive. If the roots have the same sign, this sign is also the sign of their sum $\:-\dfrac{k+2}k$, which is the sign of the product $-k(k+2)$. To sum it up, we need to have $k(k+2)<0$, which happens if and only if $\: \color{red}{-2<k<0}$.

Now note that$\:-8-2\sqrt{15}<-2$, and as $\;3<\sqrt{15}<4$ we have $\:-2<-8+2\sqrt{15}<0\:$.

Thus eventually the solutions are $$-8+2\sqrt{15}<k<0.$$

• The maniac downvoter struck gain! – Bernard Jul 9 '18 at 23:10