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I was asked to prove that, for all values of $m$, the equation $x^2 + 2mx + 2m^2 + m + 1$ has no real roots of $x$. In trying to prove this, I set $a = 1$, $b = 2m$, $c = 2m^2 + m + 1$ and found $$\Delta = b^2 - 4ac = 4m^2 - 8m^2 - 4m - 4$$ I am unable to use this expression to obtain a proof.

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$-4a^2-4a-4=-4(a^2+a+1)$, now you can check that $a^2+a+1$ is always positive as a function of a with the same method. For this equation, $a=1 b=1$ and $c=1$ so $b^2-4ac= 1-4 < 0$. No solutions, and is always positive, so your discriminant $-4(a^2+a+1)$ is always negative and you have no zeros on your original function.

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  • $\begingroup$ Thanks a lot! Appreciate it. $\endgroup$ – Gavish Apr 9 '17 at 17:06
  • $\begingroup$ Actually doesn't using b^2 - 4 ac for − 4a^2 − 4a - 4 give an all positive value as b^2 - 4 ac is less than 0. $\endgroup$ – Gavish Apr 9 '17 at 17:16
  • $\begingroup$ Got it this time, my bad. $\endgroup$ – Gavish Apr 9 '17 at 17:25
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Hint

you have found that the discriminant of the given equation is $-4(a^2+a+1)$.

Now you can show that $a^2+a+1>0$ for all $a \in \mathbb{R}$ ?


You can use the same reasoning as for the starting polynomial $x^2 + 2mx + 2m^2 + m + 1$. Now you have the polynomial $-4(a^2+a+1)$ and, if it has no real roots, it is always negative. It has no real root if its discriminant $\Delta_a$ is negative and we have $\Delta_a= 1-4=-3$ , so $a^2+a+1$ is always positive and $-4(a^2+a+1)$ is always negative and the starting equation has no real solutions.

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  • $\begingroup$ Thank you! I appreciate it. $\endgroup$ – Gavish Apr 9 '17 at 17:04
  • $\begingroup$ Actually may you elaborate further please, thanks. $\endgroup$ – Gavish Apr 9 '17 at 17:16
  • $\begingroup$ I've added something to my answer. $\endgroup$ – Emilio Novati Apr 9 '17 at 18:44

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