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Question Number 46973 by rahul 19 last updated on 03/Nov/18

Answered by MJS last updated on 03/Nov/18

$$x^3-27x+k=0$$$$D=\frac{k^2}{4}-729<0\Rightarrow -54⩽k⩽54$$$$\mathrm{but}\mathrm{we}\mathrm{need}\mathrm{integer}\mathrm{solutions}$$$$\mathrm{case}1:3\mathrm{solutions}$$$$(x-o)(x-p)(x-q)=0$$$$o+p+q=0\Rightarrow q=-(o+p)$$$$op+oq+pq=-27\Rightarrow o^2+op+p^2=27\Rightarrow p=-\frac{o}{2}\pm \frac{\sqrt{108-3o^2}}{2}$$$$-opq=k\Rightarrow k=op(o+p)$$$$\sqrt{108-3o^2}\in \mathbb{N}\Rightarrow o\in \{-6,-3,3,6\}$$$$p=-\frac{o}{2}\pm \frac{\sqrt{108-3o^2}}{2}\wedge q=-(o+p)\wedge k=op(o+p)\Rightarrow$$$$\Rightarrow \left(\begin{array}{c}o\\ p\\ q\\ k\end{array}\right)\in \{\left(\begin{array}{c}-6\\ 3\\ -3\\ 54\end{array}\right),\left(\begin{array}{c}-3\\ -3\\ -6\\ 54\end{array}\right),\left(\begin{array}{c}-3\\ 6\\ 3\\ -54\end{array}\right),\left(\begin{array}{c}3\\ -6\\ -3\\ 54\end{array}\right),\left(\begin{array}{c}3\\ 3\\ 6\\ 54\end{array}\right),\left(\begin{array}{c}6\\ -3\\ 3\\ -54\end{array}\right)\}$$$$\mathrm{case}2:2\mathrm{solutions}(q=p)$$$$(x-o){(x-p)}^2=0$$$$o+2p=0\Rightarrow p=-\frac{o}{2}$$$$2op+p^2=-27\Rightarrow o=\pm 6\Rightarrow p=\mp 3$$$$-{op}^2=k\Rightarrow k=\mp 54$$$$\Rightarrow \left(\begin{array}{c}o\\ p\\ q=p\\ k\end{array}\right)\in \{\left(\begin{array}{c}-6\\ 3\\ 3\\ 54\end{array}\right),\left(\begin{array}{c}6\\ -3\\ -3\\ -54\end{array}\right)\}$$$$\mathrm{so}\mathrm{answer}\mathrm{B}\mathrm{seems}\mathrm{right}:2\mathrm{integer}\mathrm{values}\mathrm{for}k$$

Commented by rahul 19 last updated on 04/Nov/18

thanks sir����

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