Question

If x=
$\frac{2 - \sqrt{3} }{2 + \sqrt{3} }$


, find the value of
$2 {x}^{2} - 7x + 5$
​

Answer 1

Answer:

$\red {Value \:of \:2x^{2}-7x+5}$

$\green {= 147 - 84\sqrt{3}}$

$\green {= 7(21-12\sqrt{3})}$

Step-by-step explanation:

$Given \: x = \frac{2-\sqrt{3}}{2+\sqrt{3}}$

$\implies x = \frac{(2-\sqrt{3})}{(2+\sqrt{3})}\times \frac{(2-\sqrt{3})}{(2-\sqrt{3})}$

$= \frac{(2-\sqrt{3})^{2}}{2^{2} - (\sqrt{3})^{2}}$

$= \frac{ 2^{2} + (\sqrt{3})^{2} - 2\times 2\times \sqrt{3} }{4-3}\\= \frac{4+3-4\sqrt{3}}{1}\\= 7-4\sqrt{3}\:--(1)$

$Now , \: Value \:of \:2x^{2}-7x+5\\=2(7-4\sqrt{3})^{2} -7(7-4\sqrt{3})+5 \:[from \:(1)]$

$= 2(49+48-56\sqrt{3} )- 49 + 28\sqrt{3} + 5\\=2(97-56\sqrt{3}) - 47+28\sqrt{3}\\= 194 - 112\sqrt{3}-47+28\sqrt{3}\\= 147 - 84\sqrt{3}\\= 7(21-12\sqrt{3})$

Therefore.,

$\red {Value \:of \:2x^{2}-7x+5}$

$\green {= 147 - 84\sqrt{3}}$

$\green {= 7(21-12\sqrt{3})}$

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