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 $0$?

  • $\begingroup$ 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

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 $$(11k+25)(1-k)\geq0.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.