Question

x ^ 2 + 1/(x ^ 2) = 7 find the value of x + 1/x, x > 0 .





I need answer in paper and step by step​

Answer 1

Solution:

Given That:

$\rm \longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } = 7, \: x > 0$

Adding 2 to both sides, we get:

$\rm \longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 7 + 2$

$\rm \longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 9$

Can be written as:

$\rm \longrightarrow {(x)}^{2} + \bigg(\dfrac{1}{x} \bigg)^{2} + 2 \times x \times \dfrac{1}{x} = 9$

Using identity a² + b² + 2ab = (a + b)², we get:

$\rm \longrightarrow \bigg(x + \dfrac{1}{x} \bigg)^{2} = 9$

$\rm \longrightarrow x + \dfrac{1}{x} = \sqrt{9}$

$\rm \longrightarrow x + \dfrac{1}{x} = \pm 3$

Since, we know that:

$\rm \longrightarrow x > 0$

$\rm \longrightarrow x + \dfrac{1}{x} > 0$

Therefore:

$\rm \longrightarrow x + \dfrac{1}{x} = 3$

★ Which is our required answer.

Learn More:

Algebraic Identities.

• (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