Question

If
${x}^{2} + \frac{1}{ {25x}^{2} } = 43/5 find the value of: [tex]x + \frac{1}{5x}$
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Answer 1

$Given \: \: \: question \: \: is \\ \\ x {}^{2} + \frac{1}{(25x) {}^{2} } = \frac{43}{5} \\ \\ find \: \: x + \frac{1}{5x} \\ \\ Answer \: \\ \\ (x + \frac{1}{5x} ) {}^{2} - 2(x)( \frac{1}{5x} ) = \frac{43}{5} \\ \\ (x + \frac{1}{5x} ) {}^{2} - \frac{2}{5} = \frac{43}{5} \\ \\ (x + \frac{1}{5x} ) {}^{2} = \frac{43}{5} + \frac{2}{5} \\ \\ (x + \frac{1}{5x} ) {}^{2} = \frac{45}{5} \\ \\ (x + \frac{1}{5x} ) {}^{2} = 9 \\ \\ (x + \frac{1}{5x} ) = \sqrt{9} \\ \\ (x + \frac{1}{5x} ) = 3 \: \: \: or \: \: \: (x + \frac{1}{5x} ) = - 3$

Answer 2

Answer:

± 3

Step-by-step explanation:

$\sf ATQ: \ x^2 + \dfrac{1}{25x^2} = \dfrac{43}{5}\\ \\ \\\sf To \ Find: \ \ x + \dfrac{1}{5x}$

$\Longrightarrow \sf x + \dfrac{1}{5x}\\ \\ \\\sf Squaring \ on \ both \ sides,\\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2\\ \\ \\ \\\boxed{\sf Using \ the \ identity \ (a+b)^2 = a^2 + 2ab + b^2}\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = x^2 + 2\left(x\right) \left(\dfrac{1}{5x}\right) + \left(\dfrac{1}{5x}\right)^2\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = x^2 + \dfrac{2}{5} + \dfrac{1}{25x^2}$

$\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = x^2 + \dfrac{1}{25x^2} + \dfrac{2}{5}\\ \\ \\ \\\sf Substitute \ x^2 + \dfrac{1}{25x^2} = \dfrac{43}{5} \ above.\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = \dfrac{43}{5} + \dfrac{2}{5}\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = \dfrac{43 + 2}{5}\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = \dfrac{45}{5}\\ \\ \\ \\\Longrightarrow \sf \left(x + \dfrac{1}{5x}\right)^2 = 9$

$\Longrightarrow \sf x + \dfrac{1}{5x} = \sqrt{9}\\ \\ \\ \\\Longrightarrow \sf x + \dfrac{1}{5x} = \pm 3$

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$\sf Answer: \sf x + \dfrac{1}{5x} = \pm 3$