Question

If x2 + 1/ x2=23, evaluate( i) x+ 1/ x ( ii)x-1/x​

Answer 1

Given: x² + 1/x² = 23

Formula used:

• (a+b)²=a²+2ab+b²
• (a-b)²=a²-2ab +b²

Solutions:

( i ) x + 1/x

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

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

$\rightarrow \: (x + \frac{1}{x} ) {}^{2} = 23+ 2$

$\rightarrow \: (x + \frac{1}{x} ) {}^{2} = 25$

$\rightarrow \: (x + \frac{1}{x} ) {}^{2} = 25$

$\rightarrow \: x + \frac{1}{x} = \sqrt{25}$

$\rightarrow \: x + \frac{1}{x} = 5$

(ii) x - 1/x

$\rightarrow \: (x - \frac{1}{x} ) {}^{2} = {x}^{2} - 2 \times x \times \frac{1}{x} + ( \frac{1}{x} ) {}^{2}$

$\rightarrow \: (x - \frac{1}{ {x}} ) {}^{2} = {x}^{2} - 2 + \frac{1}{ {x}^{2} }$

$\rightarrow \: (x - \frac{1}{x} ) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2$

$\rightarrow \: (x - \frac{1}{x} ) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2$

$\rightarrow \: (x - \frac{1}{x} ) {}^{2} = 23- 2$

$\rightarrow \: (x - \frac{1}{x} ) {}^{2} = 21$

$\rightarrow \: x - \frac{1}{x} = \sqrt{21}$

Extra:

• ( a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (a+b)³ = a³ + b³ + 3ab(a+b)

• (a-b)³ = a³ - b³ - 3ab(a-b)

• (a³+b³) = (a+b) (a²- ab + b²)

• (a³-b³) = (a-b) (a²+ ab + b²)

Answer 2

Given:

x² + 1/x² = 23

Formula used:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

Solutions:

(1) (x + 1/x)

(x + 1/x)² = x² + 2 + 1/x²

(x + 1/x)² = x² + 2 + 1/x²

(x + 1/x)² = 23 + 2

(x + 1/x)² = 25

x + 1/x = √25

x + 1/x = 5

(2) (x - 1/x)

(x - 1/x)² = x² - 2 + 1/x + (1/x)²

(x - 1/x)² = x² - 2 + 1/x²

(x - 1/x)² = x² + 1/x² - 2

x - 1/x = 23 - 2

x - 1/x = 21

x - 1/x = √21

Extra:

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ + 3ab(a - b)

• (a³ + b³) = (a + b)(a² - ab + b²)

• (a³ + b³) = (a - b)(a² + ab + b²)

silentlover45.❤️