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the quadratic equation $3(k+2)x^2+(k+5)x+k=0$ has real roots

show $(k-1)(11k+25) \geq 0 $


If $\Delta$ greater than $0$ it has real roots so,

$$\Delta = (k+5)^2 - 4 \cdot (3(k+2))\cdot k$$

$$k^2+10k+25-4(3k+6)\cdot k = (?)-12k^2-24k$$

which doesn't help and the answer is not the same for Wolfram|Alpha.

So how can I prove this is greater than $0$?

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  • $\begingroup$ Is this $\Delta = (k+5)^2 - 4 * (3(3k+2))*k$ meant to be $\Delta = (k+5)^2 - 4 * (3(k+2))*k$? $\endgroup$ – shuttle87 Dec 11 '13 at 8:12
  • $\begingroup$ that's an example of one sort of mistake i make all the time. i call them "mental short-circuits". can cause distress, and certainly wastes a lot of effort. don't know how i can curb the tendency. maybe a change of diet? $\endgroup$ – David Holden Dec 11 '13 at 8:19
  • $\begingroup$ If you expand $b^2 - 4ac = (k + 5)^2 - 4 \cdot 3(k+2) \cdot k$, you get $-11k^2 + 14k - 25$, which is the opposite of your $(k - 1)(11k + 25)$. $\endgroup$ – Henry Swanson Dec 11 '13 at 8:20
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Well, I think there are some mistakes in your working and perhaps your question isn't right. Here is mine: \begin{align*} \Delta&=(k+5)^2 - 4k(3(k+2))\\ &=k^2+10k+25-12k^2-24k\\ &=25-11k^2-14k\\ &=(11k+25)(1-k) \end{align*} Since there are real roots, we know that $\Delta\geq0$, hence we have $$(11k+25)(1-k)\geq0.$$

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