Question

If x is a positive real number such that x square+ 1 upon x is equal to 62 then evaluate x + 1 upon x​

Answer 1

Answer:

8

Step-by-step explanation:

 Here,

⇒ x^2 + 1 / x^2 = 62

  Adding 2 on both sides:

⇒ x^2 + 1 / x^2 + 2 = 62 + 2

⇒ x^2 + 1 / x^2 + 2( 1 ) = 64

⇒ x^2 + 1 / x^2 + 2( x * 1 / x ) = 64  

    using a² + b² + 2ab = ( a + b )²

⇒ ( x + 1 / x )^2 = 64

⇒ x + 1 / x = √64

⇒ x + 1 / x = 8

     

Answer 2

$\purple{\sf{Answer:-}}$

Given:-

${\sf{ {x}^{2} + \frac{1}{{x}^{2}} = 62}}$

To find:-

${\sf{ x + \frac{1}{x}}}$

Solution:-

$⇢{\sf{ {x}^{2} + \frac{1}{{x}^{2}} = 62}} \\⇢ {\sf{ {(x + \frac{1}{x}) }^{2} - 2 \times x \times \frac{1}{x} = 62}} \\ ⇢{\sf{ {(x + \frac{1}{x}) }^{2} - 2 = 62}} \\ ⇢ {\sf{ {(x + \frac{1}{x}) }^{2} = 62 + 2}} \\⇢ {\sf{ {(x + \frac{1}{x}) }^{2} = 64}} \\⇢ {\sf{ x + \frac{1}{x} = \sqrt{64} = 8}} \\ ∴ {\sf{ x + \frac{1}{x} = 8}}$

Final answer:-

$\red{\sf{ x + \frac{1}{x}} = 8}$