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})?

Video Explanation

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

Explanatory Answer: Method 2

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.