Question

if x-1÷×=10,find (x+1÷×)^2​

Answer 1

$\huge{\underline{\underline{\sf{Answer \ :}}}}$

Given:

$\sf{x - \frac{1}{x} = 10 }..........(1)$

To find:

$\sf{(x + \frac{1}{x}) {}^{2} }$

Squaring equation (1) on both sides,

$\sf{x {}^{2} + \frac{1}{x {}^{2}} - 2 = 100 } \\ \\ \rightarrow \: \underline{\boxed{\sf{x {}^{2} + \frac{1}{x {}^{2} } = 102}}}$

Now,

$\sf{(x + \frac{1}{x}) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2 } \\ \\ \implies \: \sf{(x + \frac{1}{x}) {}^{2} = 102 + 2 } \\ \\ \implies \: \boxed{ \sf{(x + \frac{1}{x}) {}^{2} = 104 }}$

Answer 2

Given;

(x - $\dfrac{1}{x}$) = 10

To find ;

(x + $\dfrac{1}{x}$)²

Solution ::

(x - $\dfrac{1}{x}$) is in the form of (a-b) where x is a and $\dfrac{1}{x}$ is b.

→ But we have to find (a+b)²

→ To find the (a + b)² we have to find its (a-b)²

→ (a-b) = a² + b² - 2ab

The (a-b)² form of (x - $\dfrac{1}{x}$)² :-

= (x - $\dfrac{1}{x}$)²

= x² + $\dfrac{1}{x}$² - 2.x.$\dfrac{1}{x}$ = (10)²

= x² + $\dfrac{1}{x}$² = 100 + 2

= x² + $\dfrac{1}{x}$² = 102 ................ (1)

(a+b)² = a² + b² = 2ab

Let us substitute the value of a² and b² as per equation 1.

(x + $\dfrac{1}{x}$)² = x² + ($\dfrac{1}{x}$)² + 2

= (x + $\dfrac{1}{x}$)² = 102 + 2

=(x + $\dfrac{1}{x}$)² = 104.

∴ Your answer is 104.