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Question Number 83188 by john santu last updated on 28/Feb/20

$$\mathrm{If}\underset{\mathrm{a}=0}{\sum ^{\mathrm{n}-1}}(2\mathrm{a}+1){\mathrm{x}}^2+({\mathrm{n}}^2+4\mathrm{n}-5)\mathrm{x}+16$$$$=0\mathrm{is}\mathrm{a}\mathrm{perfect}\mathrm{square}\mathrm{such}\mathrm{that}$$$$\mathrm{n}\in {\mathbb{Z}}^{+}.\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\mathrm{x}+\mathrm{n}?$$

Answered by mind is power last updated on 01/Mar/20

$$\iff n^2{x}^2+(n^2+4n-5)x+16=0$$$$\Rightarrow {(n^2+4n-5)}^2-64n^2=0$$$$\iff (n^2-4n-5)(n^2+12n-5)=0$$$$\Rightarrow n=5$$$$x=-\frac{(n^2+4n-5)}{2n^2}=-\frac{\left(40\right)}{2.25}=-\frac{4}{5}$$$$x+n=-\frac{4}{5}+5=\frac{21}{5}$$$$$$

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