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Range of $k$ for which both the roots of the equation $(k-2)x^2+(2k-8)x+3k-17=0$ are positive.

Try: if $\alpha,\beta>0$ be the roots of the equation. Then $$\alpha+\beta=\frac{8-2k}{k-2}>0\Rightarrow k\in(2,4)$$

And $$\frac{3k-17}{k-2}>0\Rightarrow k\in(-\infty,2)\cup \bigg(\frac{17}{3},\infty\bigg)$$

I have got $k=\phi$. I did not understand where i am wrong , please explain, Thanks

DXT
  • 11,241

3 Answers3

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You're perfectly right.

Assuming $k\ne2$, Descartes’ rule of signs tells you it must be $$ \begin{cases} \dfrac{2k-8}{k-2}<0\\[6px] \dfrac{3k-17}{k-2}>0 \end{cases} $$ The top inequality yields $2<k<4$, the bottom one yields $$ k<2\qquad\text{or}\qquad k>\frac{17}{3} $$ No value of $k$ satisfies both inequalities.

Thus it's not even necessary to discuss the discriminant (which it would if there were common solutions for the inequalities above).

egreg
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Use quadratic formula. $$(k-2)x^2+(2k-8)x+3k-17=0$$ gives $$x=\frac{-(2k-8)\pm\sqrt{(2k-8)^2-4(k-2)(3k-17)}}{2(k-2)}=-\frac{k-4}{k-2}\pm\frac{\sqrt{-8k^2+60k-72}}{2(k-2)}$$ Hence $$x=-1+\frac2{k-2}\pm\frac{\sqrt{(k-6)(3-2k)}}{k-2}$$ Now we need only find when the negative root is positive since the positive one would then be positive as well.

So $$\frac2{k-2}>1+\frac{\sqrt{(k-6)(3-2k)}}{k-2}\implies \cdots$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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You do not have to compute the roots and solve irrational inequations. Just a little thinking to use theorems on quadratic polynomials.

  • First, it has to be a quadratic equation, which means $k\ne 2$.
  • This condition being satisfied, it must have real roots, i.e. its reduced discriminant has to be non-negative: $$\Delta'=(k-4)^2-(3k-17)(k-2)=-2k^2+15k-18\ge 0 $$ $\Delta'$ is a quadratic polynomial in $k\mkern1mu$; its roots are $6$ and $-3/2$ and the leading coefficient is negative, so $$ \Delta'\ge 0 \iff -\frac 32\le k \le 6. $$
  • Last, these roots must have the same sign, which means their product $p$ is positive: $$p=\frac{3k-17}{k-2} > 0 \iff (3k-17)(k-2)> 0 \iff k >\frac{17}3\quad\text{or}\quad k < 2,$$ and this common sign is positive, which is equivalent to their sum $s$ is positive: $$s=-\frac{2(k-4)}{k-2}> 0\iff(k-2)(k-4) < 0 \iff 2 < k < 4. $$ The last two conditions are incompatible, so there's no $k$ such that both roots are positive
Bernard
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  • It seems that Durgesh Tiwari has alreday given the correct answer in his question, but didn't trust his solution ("I did not understand where I am wrong"). In fact, he starts by assuming the existence of two positive roots, and derives two necessary conditions (Vieta's formulas) on $k$ which are incompatible. – Paul Frost Apr 26 '18 at 21:50
  • @PaulFrost: Yes. What I wanted to show was how to organise the arguments for thse who don't know well how to do that. I explained the way I learnt it when I was in high school. – Bernard Apr 26 '18 at 21:59