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Question Number 56384 by Gulay last updated on 15/Mar/19

Commented by Gulay last updated on 15/Mar/19

$$\mathrm{sir}\mathrm{plz}\mathrm{help}\mathrm{me}$$

Commented by mr W last updated on 16/Mar/19

$$\alpha ,\beta =roots$$$$\alpha =\sqrt{10}+4$$$$\alpha \beta =6$$$$\Rightarrow \beta =\frac{6}{\sqrt{10}+4}=\frac{6(4-\sqrt{10})}{4^2-10}=4-\sqrt{10}$$$$C=-(\alpha +\beta )=-(\sqrt{10}+4+4-\sqrt{10})=-8$$

Answered by MJS last updated on 15/Mar/19

$$x^2+cx+6=0$$$$x=-\frac{c}{2}\pm \frac{\sqrt{c^2-24}}{2}$$$$-\frac{c}{2}=4\Rightarrow c=-8$$$$\frac{\sqrt{c^2-24}}{2}=\sqrt{10}\Rightarrow c=\pm 8$$$$\Rightarrow c=-8$$$$\Rightarrow \mathrm{the}\mathrm{other}\mathrm{solution}\mathrm{is}4-\sqrt{10}$$$$$$$$(x-x_1)(x-x_2)=(x-4-\sqrt{10})(x-4+\sqrt{10})=$$$$=x^2-8x+6$$

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