If sqrt{x}+frac{1}{sqrt{x}}=5, what will be t | vedclass

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(C) Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=5$Squaring both sides:$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} = 5^{2}$$x + 2\left(\sqrt{x}\right)\left(\

Vedclass · Accepted Answer

$527$

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(C) Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=5$Squaring both sides:$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} = 5^{2}$$x + 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) + \frac{1}{x} = 25$$x + 2 + \frac{1}{x} = 25$$x + \frac{1}{x} = 23$Now,squaring both sides again:$\left(x + \frac{1}{x}\right)^{2} = 23^{2}$$x^{2} + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^{2}} = 529$$x^{2} + 2 + \frac{1}{x^{2}} = 529$$x^{2} + \frac{1}{x^{2}} = 529 - 2 = 527$

If $\sqrt{x}+\frac{1}{\sqrt{x}}=5,$ what will be the value of $x^{2}+\frac{1}{x^{2}}$?

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