Question

find the value if ×=
$\sqrt{3} + 1$
$find \\ \frac{ {x}^{2} }{4} + \frac{3}{ {x}^{2} }$

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Answer 1

Answer:

$\frac{ {x}^{2} }{4} + \frac{3}{ {x}^{2} } = \frac{7}{4}$✔✔

Step-by-step explanation:

$value \: of \: x = \sqrt{3} + 1 \\ then \: we \: have \:to \: find \: the \: value \: of \: given \: question \\ \frac{ {x}^{2} }{4} + \frac{3}{ {x}^{2} } \\ putting \: value \: of \: x \\ = \frac{ ({ \sqrt{3} + 1 })^{2} }{4} + \frac{3}{ ({ \sqrt{3} + 1 })^{2} } \\ = \frac{3 + 1}{4} + \frac{3}{3 + 1} \\ = \frac{4}{4} + \frac{3}{4} \\ = \frac{4 + 3}{4} \\ = \frac{7}{4}$

Answer 2

$value \: of \: x = \sqrt{3} + 1 \\ \frac{ {x}^{2} }{4} + \frac{3}{ {x}^{2} } \\ = \frac{ ({ \sqrt{3} + 1 })^{2} }{4} + \frac{3}{ ({ \sqrt{3} + 1 })^{2} } \\ = \frac{3 + 1}{4} + \frac{3}{3 + 1} \\ = \frac{4}{4} + \frac{3}{4} \\ = \frac{4 + 3}{4} \\ = \frac{7}{4}$