Question

(c) If x + (1/x) = 5, find the value of [x - (1/x)]^2.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

21

Step-by-step explanation:

Given :

$\mathrm{x + \dfrac{1}{x} = 5}$

To find :

$\mathrm{\bigg(x - \dfrac{1}{x}\bigg)^2}$

Solution :

Given condition is,

$\longmapsto \mathrm{x + \dfrac{1}{x} = 5}$

Squaring on both sides

$\implies \mathrm{\bigg(x + \dfrac{1}{x}\bigg)^2 = 5^2}$

We know that,

$\boxed{\mathrm{(a + b)^2 = a^2 + 2ab + b^2}}$

$\implies \mathrm{\bigg(x^2 +2\not x. \dfrac{1}{\not x} + \dfrac{1}{x^2}\bigg) = 25}$

$\implies \mathrm{\bigg(x^2 +2 + \dfrac{1}{x^2}\bigg) = 25}$

$\implies \mathrm{x^2 + \dfrac{1}{x^2} = 25 - 2}$

$\implies \mathrm{x^2 + \dfrac{1}{x^2} = 23}$ ___ [1]

Now, given question is,

$\longmapsto \mathrm{\bigg(x - \dfrac{1}{x}\bigg)^2}$

We know that,

$\boxed{\mathrm{(a - b)^2 = a^2 - 2ab + b^2}}$

Here, a = x ; b = 1/x, so,

$\implies \mathrm{\bigg(x^2 -2\not x. \dfrac{1}{\not x} + \dfrac{1}{x^2}\bigg)}$

$\implies \mathrm{x^2 -2+ \dfrac{1}{x^2}}$

$\implies \mathrm{x^2+ \dfrac{1}{x^2} - 2}$

From [1]

$\implies \mathrm{23 - 2}$

$\implies 21$

$\therefore \underline{\boxed{\mathbf{\bigg(x - \dfrac{1}{x}\bigg)^2 = 21}}}$

Hope it helps!!