swethasabu
Answered

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if x= 1/2-√3 , prove that x³-2x²-7x+5=3

Sagot :

[tex]x=\frac{1}{2}-\sqrt3\\\\x^3-2x^2-7x+5=3\\\\(\frac{1}{2}-\sqrt3)^3-2(\frac{1}{2}-\sqrt3)^2-7(\frac{1}{2}-\sqrt3)+5\\\\=(\frac{1}{2})^3-3\cdot(\frac{1}{2})^2\cdto\sqrt3+3\cdot\frac{1}{2}\cdot(\sqrt3)^2-(\sqrt3)^3-...\\\\...-2[(\frac{1}{2})^2-2\cdot\frac{1}{2}\cdot\sqrt3+(\sqrt3)^2]-\frac{7}{2}+7\sqrt3+5[/tex]

[tex]=\frac{1}{8}-\frac{3\sqrt3}{4}+\frac{9}{2}-3\sqrt3-2(\frac{1}{4}-\sqrt3+3)-\frac{7}{2}+7\sqrt3+5\\\\=\frac{1}{8}+\frac{9}{2}-\frac{1}{2}-6-\frac{7}{2}+5-\frac{3}{4}\sqrt3-3\sqrt3+2\sqrt3+7\sqrt3\\\\=\frac{1}{8}+\frac{1}{2}-1+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{1}{2}+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{4}{8}+5\frac{1}{4}\sqrt3\\\\=-\frac{3}{8}+5\frac{1}{4}\sqrt3\neq3[/tex]
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