# Extra Questions For CBSE Class 9 Math Chapter 1 | Q8

Question 8: If x = $$frac{3 - {\sqrt{13}}}{2}$, what is the value of $x^2 + \frac{1}{x^2}$? ## Target Centum in CBSE 10th Maths #### Free Online CBSE Coaching online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer: Method 1 Computing x2 and its reciprocal and adding is one way of solving this question$we will do it in method 2).
However, many of these questions can be solved efficiently if we rewrite $x^2 + $frac{1}{x^2}$ in terms of ${$x - $frac{1}{x}$}^2) or ${$x + $frac{1}{x}$}^2). Which one to choose ${$x - $frac{1}{x}$}^2) or ${$x + $frac{1}{x}$}^2)? That depends on how the numbers pan out. The first few steps are the same for both. Try which one of the two leads to a more friendly number to calculate and proceed with that. In this method we will express $x^2 + \frac{1}{x^2}$ in terms of ${$x - $frac{1}{x}$}^2) $x^2 + \frac{1}{x^2}$ = ${$x - $frac{1}{x}$}^2 + 2) ${$x - $frac{1}{x}$}) = $\frac{3 - \sqrt{13}}{2}$ - $\frac{2}{3 - \sqrt{13}}$ = $\frac{${3 - $sqrt{13}}$^2 - 4}{2${3 - $sqrt{13}})}) = $\frac{$9 + 13 - 6$sqrt{13}$ - 4}{2${3 - $sqrt{13}})}) = $\frac{$18 - 6$sqrt{13}$}{2${3 - $sqrt{13}})}) Take 6 outside as a common term in the numerator, we get ${$x - $frac{1}{x}$}) = $\frac{6$3 - $sqrt{13}$}{2${3 - $sqrt{13}})}) Cancelling ${$3 - $sqrt{13}$}) in the numerator and denominator we get ${$x - $frac{1}{x}$}) = 3 Now this is really a friendly number to work with. $x^2 + \frac{1}{x^2}$ = ${$x - $frac{1}{x}$}^2) + 2 $x^2 + \frac{1}{x^2}$ =$3)2 + 2 = 11

In this method let us find x2 and its reciprocal and add it and check whether we get the same answer

x2 = ${$$frac{3 - $sqrt{13}}{2}$^2}) x2 = ${$$frac{9 + 13 - 6{\sqrt{13}}}{4}$}) = ${$$frac{22 - 6{\sqrt{13}}}{4}$}) = ${$$frac{11 - 3\sqrt{13}}{2}$}) $\frac{1}{x^2}$ = ${$$frac{2}{11 - 3\sqrt{13}}$}) $x^2 + \frac{1}{x^2}$ = ${$$frac{11 - 3\sqrt{13}}{2}$}) + ${$$frac{2}{11 - 3\sqrt{13}}$}) = $\frac{$11 - 3$sqrt{13}$^2 + 4}{2${11 - 3$sqrt{13}})}) = $\frac{$121 + 117 - 66{$sqrt{13}}$ + 4}{2${11 - 3{$sqrt{13}}})}) = $\frac{$242 - 66$sqrt{13}$}{2${11 - 3$sqrt{13}})}) Take 22 as a term common in the numerator and rewrite the expression $x^2 + \frac{1}{x^2}$ = $\frac{22$11 - 3$sqrt{13}$}{2${11 - 3$sqrt{13}})}) Cancel ${$11 - 3\sqrt{13}$}) in both the numerator and denominator
$x^2 + \frac{1}{x^2}$ = $\frac{22}{2}$ = 11.

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