Question 87

If 2x + (9/x) = 9, then what is the minimum value of $$x^{2}+(\frac{1}{x^{2}})$$ ?

Solution

$$\Rightarrow$$ 2x + (9/x) = 9

$$\Rightarrow$$ $$2x^2 - 9x + 9 = 0$$

$$\Rightarrow$$ $$x = 1.5, 3$$

So at x = 1.5, value of $$x^{2}+(\frac{1}{x^{2}})$$ = $$1.5^2 + \dfrac{1}{1.5^2}$$ = $$\dfrac{97}{36}$$

So at x = 3, value of $$x^{2}+(\frac{1}{x^{2}})$$ = $$3^2 + \dfrac{1}{3^2}$$ = $$\dfrac{82}{9}$$

Clearly, $$\dfrac{97}{36}$$ is lesser than $$\dfrac{82}{9}$$.

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