What is one of the possible values of x²+\;{1}{x²} ?

Correct Answer is : 829\;82/9982

Correct Answer is : 829\;82/9982

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UPSC CDS 2 2022 Math Paper

Consider the following for the next two (02) items that follow :

Consider the equation

6 x^{2} - 25 x + \;\frac{6}{x^{2}} + \;\frac{25}{x} + 12 = 0

Question : 52 of 100

Marks: +1, -0

x^{2}+\;\frac{1}{x^{2}}

Solution: 👈: Video Solution

n-\;\frac{1}{n}=p

then

n^{2}+\;\frac{1}{n^{2}}=p^{2}+2

6x^{2}-25x+\;\frac{6}{x^{2}}+\;\frac{25}{x}+12=0

\Rightarrow 6\left(x^{2}+\;\frac{1}{x^{2}}\right)-25\left(x-\;\frac{1}{x}\right)+12=0

Let,

x-\;\frac{1}{x}

be 'a'

From the above concept,

\left(x^{2}+\;\frac{1}{x^{2}}\right)=a^{2}+2

\Rightarrow 6\left(a^{2}+2\right)-25a+12=0

\Rightarrow 6a^{2}+12-25a+12=0

\Rightarrow 6a^{2}-25a+24=0

\Rightarrow 6a^{2}-16a-9a+24=0

\Rightarrow 2a\left(3a-8\right)-3\left(3a-8\right)=0

\Rightarrow \left(2a-3\right)\left(3a-8\right)=0

\Rightarrow a=\;\frac{3}{2}\;\text{ or }\;\frac{8}{3}

Case:1 a

=\;\frac{3}{2}

\Rightarrow \left(x^{2}+\;\frac{1}{x^{2}}\right)=\left(a^{2}+2\right)=\left(\frac{3}{2}\right)^{2}+2

\Rightarrow \left(x^{2}+\;\frac{1}{x^{2}}\right)=\left(\frac{9}{4}+2\right)=\frac{17}{4}\;\text{ (not in the option) }\;

\;\text{ Case:2 }\; a=\;\frac{8}{3}

\Rightarrow \left(x^{2}+\;\frac{1}{x^{2}}\right)=\left(\frac{8}{3}\right)^{2}+2

\Rightarrow \left(x^{2}+\;\frac{1}{x^{2}}\right)=\left(\frac{64}{9}+2\right)=\frac{82}{9}

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