Question

please answer with solution.

Answer 1

you should rationalise the denominator and then use algebraic identity

corrected answer

Answer 2

Hey There!! 
Here, 
$x=7-4\sqrt{3}$ 

So, $\frac{1}{x}=\frac{1}{7-4\sqrt{3}}$ 
Rationalising. Multiply and Divide by $7+4\sqrt{3}$ 
So, $\frac{1}{x}=\left(\frac{1}{7-4\sqrt{3}}\right) \times \left(\frac{7+4\sqrt{3}}{7+4\sqrt{3}}\right) \\ \\ \implies \frac{1}{x} = \frac{7+4\sqrt{3}}{7^2-(4\sqrt{3})^2} \\ \\ \implies \frac{1}{x} = \frac{7+4\sqrt{3}}{49-48} \\ \\ \implies \frac{1}{x}=7+4\sqrt{3}$

Now, $x=7-4\sqrt{3}$ and $\frac{1}{x}=7+4\sqrt{3}$ 
So, $x+\frac{1}{x} = 7-4\sqrt{3} + 7+4\sqrt{3} = 14$

Now, consider: 
$\left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2} + 2x\frac{1}{x} \\ \\ \implies 14^2 = x^2+\frac{1}{x^2} + 2 \\ \\ \implies x^2+\frac{1}{x^2} = 196-2 \\ \\ \implies \boxed{x^2+\frac{1}{x^2}=194}$
Hope this helps
Purva
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