Question 58

If $$\frac{3(x^2 + 1) - 7x}{3x} = 6,$$ x ≠ 0, then the value of $$\sqrt{x} + \frac{1}{\sqrt{x}}$$ is:

Solution

$$\frac{3(x^2 + 1) - 7x}{3x} = 6$$
$$\frac{x[3(x + \frac{1}{x}) - 7]}{3x} = 6,$$

$$(x +\frac{1}{x}) - \frac{7}{3} = 6$$

$$x +\frac{1}{x} = \frac{25}{3}$$

$$x +\frac{1}{x} + 2= \frac{25}{3} + 2$$

($$\because(a+b)^2 = a^2 + b^2 + 2ab$$)

$$(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = \frac{31}{3}$$

$$\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{\frac{31}{3}}$$

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