Question

Find the value of x²+1/x²? if x+1/x=8?

Answer 1

x+1/x=8

both side square

(x+1/x)^2=8^2

Answer 2

ANSWER:

• Value of the above expression is 62.

GIVEN:

$\implies \: x + \dfrac{1}{x} = 8$

TO FIND:

$value \: of \: x {}^{2} + \dfrac{1}{x {}^{2} }$

SOLUTION:

$\implies \: x + \dfrac{1}{x} = 8 \\ \\ squaring \: both \: sides \: we \: get. \\ \\ \implies \: (x + \dfrac{1}{x} ) {}^{2} = (8) {}^{2} \\ \\ \implies \: x {}^{2} + \dfrac{1}{x {}^{2} } + 2 \: x \times \dfrac{1}{x} = 64 \\ \\ \implies \: x {}^{2} + \dfrac{1}{x {}^{2} } = 64 - 2 \\ \\ \implies \: x {}^{2} + \frac{1}{x {}^{2} } = 62$

Value of the above expression is 62.

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab