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How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots. I am unable to see how I'd start this question. A small hint, or the entire solution, both will be highly appreciated!

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If you have a look at the derivative - it is $6(x^2+x+1)>0$, which implies that your polynomial is increasing, and hence can never have $3$ roots

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    $\begingroup$ Well spotted! (+1) Apparently, a trap question. $\endgroup$ – Peter Jul 5 '18 at 11:26
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The discriminant of $2x^3+3x^2+6x+k$ is $-108 (k^2 - 5 k + 13)=-27 (2 k - 5)^2 - 729 $, which is always negative. Therefore, there is one real root and two complex conjugate roots (see Wikipedia).

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Since this is a cubic polynomial.

By using Vieta's formula we have $a+b+c=-\dfrac{3}{2}$ and $ab+bc+ca=3$. Now squaring $a+b+c$ gives $a^2+b^2+c^2+2(ab+bc+ca)=\dfrac{9}{4}$, so $a^2+b^2+c^2=-\dfrac{15}{4}$ which is always impossible if the roots are real.

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Let $f(x)=2x^3+3x^2+6x+k$ and suppose that $f$ has three distinct real zeros. Then, by Rolle, $f'$ has two distinct real zeros. But $f'(x)=6x^2+6x+6$ and the equation

$$x^2+x+1=0$$

has no real solution. A contradiction !

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