Question

if p+1/p = 3 , find the value of p-1/p​

Answer 1

Given that,

$\sf \: p + \dfrac{1}{p} = 3$

Squaring on both sides,

$\implies \sf \: (p + \dfrac{1}{p}) {}^{2} = 3 {}^{2} \\ \\ \implies \sf \: {p}^{2} + \dfrac{1}{ {p}^{2} } + 2 = 9 \\ \\ \implies \sf \: {p}^{2} + \dfrac{1}{ {p}^{2} } = 7$

Now,

$\sf \:( p - \dfrac{1}{p} ) {}^{2} = {p}^{2} + \dfrac{1}{ {p}^{2} } - 2 \\ \\ \longrightarrow \sf \: \:( p - \dfrac{1}{p} ) {}^{2} = 5 \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: \: p - \dfrac{1}{p} = \sqrt{5}}}$

Answer 2

Question :-

If p + 1/p = 3 , find the value of p - 1/p.

Given :-

p + 1/p = 3

To find :-

p - 1/p.

Solution :-

p + 1/p = 3

Squaring both side,

➺ ( p + 1/p )² = (3)²

➺ p² + 1/p² + 2 = 9

➺ p² + 1/p² = 9 - 2

➺ p² + 1/p² = 7

Now,

( p - 1/p )² = p² + 1/p² - 2

➺ ( p - 1/p )² = 7 - 2

➺ ( p - 1/p )² = 5

$\large\boxed{\implies\green{p - 1/p = √5}}$

Formula used :-

( a + 1/a )² = a² + 1/a² + 2

( a - 1/a )² = a² + 1/a² - 2

Additional Information :-

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)
• (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 + c)² = a² + b² + c² + 2(ab + bc + ca)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)