Question 79

If $$x^{2}-6x+1=0 $$ then $$x^{2}+ \frac{1}{x^{2}}=$$

Solution

The roots of a quadratic equation are given by $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Hence, the roots are = $$ \frac{6\pm\sqrt{36-4}}{2}$$ = $$3 \pm \sqrt{8}$$
From the eqn, product of roots = c/a =1/1 = 1
Hence, the roots are reciprocals.
Thus, 1/(3+√8) = 3-√8
$$x^{2}+ \frac{1}{x^{2}}$$ = $$(3+\sqrt{8})^2 + (3-\sqrt{8})^2$$ = 9 + 8 +6√8 + 9 + 8 - 6√8 = 34

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2013 RRB Allahabad

If $$x^{2}-6x+1=0 $$ then $$x^{2}+ \frac{1}{x^{2}}=$$

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