If $x = 7 – 4\sqrt{3}$, then find the value of $(x^2 +\frac{1}{x^2})$.
If $ x = 7 – 4\sqrt{3}$, then find the value of $ (x^2 +\frac{1}{x^2})$.
यदि $x = 7 – 4\sqrt{3}$, तो $(x^2 + \frac{1}{x^2})$ का मान ज्ञात कीजिए।
Detailed Solution & Logic
Correct Answer:
194
Given,
$x = 7 - 4\sqrt{3}$
Since,
$(7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 49 - 48 = 1$ So,
$\frac{1}{x} = 7 + 4\sqrt{3}$
Therefore,
$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$
Now,
$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2$
$= 14^2 - 2$
$= 196 - 2$
= 194
Answer: 194 (Option 2)
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