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Question Number 116358 by bemath last updated on 03/Oct/20

$$\mathrm{Given}\mathrm{that}\sqrt[3]{17-\frac{27}{4}\sqrt{6}}\mathrm{and}\sqrt[3]{17+\frac{27}{4}\sqrt{6}}$$$$\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\mathrm{x}}^2-\mathrm{ax}+\mathrm{b}=0.\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{ab}.$$

Answered by MJS_new last updated on 03/Oct/20

$$\sqrt[3]{17\pm \frac{27\sqrt{6}}{4}}=2\pm \frac{\sqrt{6}}{2}$$$$(x-2+\frac{\sqrt{6}}{2})(x-2-\frac{\sqrt{6}}{2})=x^2-4x+\frac{5}{2}$$$$\Rightarrow ab=+10$$

Commented by bemath last updated on 03/Oct/20

$$\mathrm{typo}\mathrm{sir}.\mathrm{the}\mathrm{equation}{\mathrm{x}}^2-\mathrm{ax}+\mathrm{b}=0$$$$\mathrm{then}\mathrm{a}=4\mathrm{\&}\mathrm{b}=\frac{5}{2}\mathrm{sir}$$

Commented by MJS_new last updated on 03/Oct/20

$$\mathrm{yes}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$

Commented by bobhans last updated on 03/Oct/20

$$\mathrm{i}\mathrm{like}\mathrm{your}\mathrm{method}\mathrm{prof}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{great}$$

Answered by bobhans last updated on 03/Oct/20

$$\mathrm{Let}{\mathrm{x}}_1=\sqrt[3]{17-\frac{27}{4}\sqrt{6}}\mathrm{and}{\mathrm{x}}_2=\sqrt[3]{17+\frac{27}{4}\sqrt{6}}$$$$\mathrm{Then}{\mathrm{x}}_1.{\mathrm{x}}_2=\sqrt[3]{(17-\frac{27}{4}\sqrt{6})(17+\frac{27}{4}\sqrt{6})}$$$$=\sqrt[3]{\frac{125}{8}}=\frac{5}{2}$$$${\mathrm{x}}_1+{\mathrm{x}}_2=a\Rightarrow a^3=x_1^3+x_2^3+3x_1{x}_2(x_1+x_2)$$$$a^3=34+3ab\Rightarrow a^3=34+\frac{15}{2}a$$$$\Rightarrow 2a^3-15a-68=0$$$$\Rightarrow (a-4)(2a^2+8a+17)=0,\mathrm{for}a\in \mathbb{R}$$$$\mathrm{we}\mathrm{get}a=4\mathrm{and}\mathrm{thus}ab=4\times \frac{5}{2}=10$$

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