Question

x + 1 upon x = 28 find x square + 1 upon x square​

Answer 1

Answer:

$\large{\underline{\boxed{\sf x^2 + \frac{1}{x^2} = 782}}}$

Step-by-step explanation:

Given that -

$\sf x + \frac{1}{x} = 28 \\ \\ \bf on \: squaring \: both \: sides : \\ \\ \sf (x + \frac{1}{x} ) {}^{2} = (28) {}^{2} \\ \\ \implies \sf x {}^{2} + \frac{1}{ {x}^{2} } + 2 \: . \: x \: .\: \frac{1}{x} = 784 \\ \\ \implies \sf x {}^{2} + \frac{1}{ {x}^{2} } + 2 = 784 \\ \\ \implies \sf x {}^{2} + \frac{1}{ {x}^{2} } = 784 - 2 \\ \\ \implies \sf x {}^{2} + \frac{1}{ {x}^{2} } = 782$

Hence, the answer is $\boxed{\bf x^2 + \frac{1}{x^2} = 782}$

_______________________

★ Identity Used:

• (a + b)² = a² + b² + 2ab

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{x^2 + \dfrac{1}{x^2} = 782}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x + \dfrac{1}{x} = 28}$

To find :- $\sf{x^2 + \dfrac{1}{x^2}}$

Solution :-

$\sf{x + \dfrac{1}{x} = 28}$

By squaring on both the sides :-

$\sf{{(x + \frac{1}{x})}^{2} = {28}^{2} }$

We know that (a + b)² = a² + b² + 2ab

Here a = x , b = 1/x

By substituting the values in the identity we have,

$\sf{{x}^{2} + {(\frac{1}{x})}^{2} + 2 \times x \times \frac{1}{x} =784}$

$\sf{x^2 + \dfrac{1}{x^2} + 2 = 784}$

$\sf{x^2 + \dfrac{1}{x^2} = 784 - 2}$

$\sf{x^2 + \dfrac{1}{x^2} = 782}$

$\boxed{\bf{x^2 + \dfrac{1}{x^2} = 782}}$

$\mathfrak{\large{\underline{\underline{Identity\:Used:-}}}}$

(a + b)² = a² + b² + 2ab

$\mathfrak{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (a - b)² = a² + b² - 2ab

[2] (a + b)(a - b) = a² - b²

[3] (x + a)(x + b) = x² + (a + b)x + ab