Question

$\sf \longrightarrow \: If \: x - \: \dfrac{1}{x} = 4 \: then \: find \: the \: value \: of \: x + \dfrac{1}{x}$

Don't dare to spam!!!! ​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm \longrightarrow x - \dfrac{1}{x} = 4$

We know that:

$\rm \longrightarrow (x + y)^{2} = {(x - y)}^{2} + 4xy$

Similarly:

$\rm \hookrightarrow \bigg(x + \dfrac{1}{x} \bigg)^{2} = \bigg(x - \dfrac{1}{x} \bigg)^{2} + 4 \times x \times \dfrac{1}{x}$

$\rm \hookrightarrow \bigg(x + \dfrac{1}{x} \bigg) =\sqrt{ {4}^{2} + 4}$

$\rm \hookrightarrow \bigg(x + \dfrac{1}{x} \bigg) =\pm 2\sqrt{5}$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + 3ab(a + b) + b³
• (a - b)³ = a³ - 3ab(a - b) - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• (x + a)(x + b) = x² + (a + b)x + ab
• (x + a)(x - b) = x² + (a - b)x - ab
• (x - a)(x + b) = x² - (a - b)x - ab
• (x - a)(x - b) = x² - (a + b)x + ab

Answer 2

Answer:

see the image for explanation.

Step-by-step explanation:

Thank you.