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Question Number 199550 by hardmath last updated on 05/Nov/23

$$\mathrm{If}$$$$\mathrm{x}=\sqrt{7+4\sqrt{3}}$$$$\mathrm{Find}:$$$$\frac{{\mathrm{x}}^4-{\mathrm{x}}^3-{9\mathrm{x}}^2-2\mathrm{x}+5}{{\mathrm{x}}^2-4\mathrm{x}+3}=?$$

Answered by Rasheed.Sindhi last updated on 07/Nov/23

$$\mathrm{x}=\sqrt{7+4\sqrt{3}};\frac{{\mathrm{x}}^4-{\mathrm{x}}^3-{9\mathrm{x}}^2-5\mathrm{x}+5}{{\mathrm{x}}^2-4\mathrm{x}+3}=?$$$$\mathrm{x}=\sqrt{7+4\sqrt{3}}=2+\sqrt{3}{}^{★}$$$$\bullet \mathrm{x}=2+\sqrt{3}$$$${(\mathrm{x}-2)}^2=3$$$${\mathrm{x}}^2-4\mathrm{x}+1=0$$$$\bullet {\mathrm{x}}^2=4\mathrm{x}-1$$$$\bullet {\mathrm{x}}^3={4\mathrm{x}}^2-\mathrm{x}=4(4\mathrm{x}-1)-\mathrm{x}=15\mathrm{x}-4$$$$\bullet {\mathrm{x}}^4={15\mathrm{x}}^2-4\mathrm{x}=15(4\mathrm{x}-1)-4\mathrm{x}=56\mathrm{x}-15$$$$$$$$\frac{{\mathrm{x}}^4-{\mathrm{x}}^3-{9\mathrm{x}}^2-5\mathrm{x}+5}{{\mathrm{x}}^2-4\mathrm{x}+3}$$$$=\frac{(56\mathrm{x}-15)-(15\mathrm{x}-4)-9(4\mathrm{x}-1)-5\mathrm{x}+5}{(4\mathrm{x}-1)-4\mathrm{x}+3}$$$$=\frac{56\mathrm{x}-15-15\mathrm{x}+4-36\mathrm{x}+9-5\mathrm{x}+5}{4\mathrm{x}-1-4\mathrm{x}+3}$$$$=\frac{41\mathrm{x}-15+4-41\mathrm{x}+9-5\mathrm{x}+5}{4\mathrm{x}-1-4\mathrm{x}+3}=\frac{-11+14}{2}=\frac{3}{2}$$

Commented by Rasheed.Sindhi last updated on 06/Nov/23

$${}^{★}$$$$\mathrm{x}=\sqrt{7+4\sqrt{3}}=\mathrm{a}+\mathrm{b}\sqrt{3}$$$${\mathrm{a}}^2+{3\mathrm{b}}^2+2\mathrm{ab}\sqrt{3}=7+4\sqrt{3}$$$${\mathrm{a}}^2+{3\mathrm{b}}^2=7\wedge \mathrm{ab}=2\Rightarrow \mathrm{b}=2/\mathrm{a}$$$${\mathrm{a}}^2+3{\left(\frac{2}{\mathrm{a}}\right)}^2=7$$$${\mathrm{a}}^4-{7\mathrm{a}}^2+12$$$${\mathrm{a}}^2=\frac{7\pm \sqrt{49-48}}{2}=4,\stackrel{\times }{3}\Rightarrow \mathrm{a}=\pm 2\Rightarrow \mathrm{b}=\pm 1$$$$\mathrm{x}=\sqrt{7+4\sqrt{3}}=2+\sqrt{3},-2-\sqrt{3}\left(Rejected\right)$$

Answered by deleteduser1 last updated on 05/Nov/23

$$\frac{x^4-x^3-9x^2-2x+5=x^2(x^2-4x+3)+3x(x^2-4x+3)+5-14x}{x^2-4x+3}$$$$=x^2+3x+\frac{5-14x}{(x-1)(x-3)}$$$$=7+4\sqrt{3}+3(2+\sqrt{3})+\frac{-23-14\sqrt{3}}{2}=1.5=\frac{3}{2}$$

Commented by hardmath last updated on 05/Nov/23

$$\mathrm{Sorry}\mathrm{professor}:$$$$\frac{{\mathrm{x}}^4-{\mathrm{x}}^3-{9\mathrm{x}}^2-5\mathrm{x}+5}{{\mathrm{x}}^2-4\mathrm{x}+3}$$

Commented by hardmath last updated on 05/Nov/23

$$thankyouprofessor$$

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