Question

If x + 1/x=6, find the value of x^4+ 1/x^4

Answer 1

EXPLANATION.

⇒ x + 1/x = 6.

As we know that,

Squaring on both sides of the equation, we get.

⇒ (x + 1/x)² = (6)².

⇒ x² + 1/x² + 2(x)(1/x) = 36.

⇒ x² + 1/x² + 2 = 36.

⇒ x² + 1/x² = 36 - 2.

⇒ x² + 1/x² = 34.

Again squaring on both sides of the equation, we get.

⇒ (x² + 1/x²)² = (34)².

⇒ x⁴ + 1/x⁴ + 2(x²)(1/x²) = 1156

⇒ x⁴ + 1/x⁴ + 2 = 1156.

⇒ x⁴ + 1/x⁴ = 1156 - 2.

⇒ x⁴ + 1/x⁴ = 1154.

Answer 2

Step-by-step explanation:

Given:-

$\tt{x + \frac{ 1}{x} = 6}$

To find:-

$\tt{The \: value \: of \: \: {x}^{4} + \frac{1}{ {x}^{4} } } = \: \huge{ ?}$

Solution:-

We have,

$\tt{x + \frac{ 1}{x} = 6}$

Squaring on both sides, we get

$\tt{ \bigg({x}^{2} + \frac{ 1}{{x}^{2} } \bigg)^{2} = {6}^{2} }$

$\tt{⬤Using \: algebraic \: Identity}$

$\tt{(a + b {)}^{2} = (a + b)(a + b) = {a}^{2} + 2ab + b}$

Now,

$⟹ \tt{{x}^{2} + 2 \times x \times \frac{1}{x} + \bigg( \frac{1}{x} \bigg)^{2} = 36 }$

$⟹ \tt{ {x}^{2} + 2 \times \cancel{x} \times \frac{1}{ \cancel{x}} + \frac{1}{ {x}^{2} } = 36 }$

$⟹ \tt{{x}^{2} + 2 + \frac{1}{ {x}^{2} } = 36}$

$⟹ \tt{{x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2}$

$⟹ {x}^{2} + \frac{1}{ {x}^{2} } = 34$

Again, squaring on both sides, we get

$\tt{ \bigg( {x}^{2} + \frac{1}{ {x}^{2} } \bigg)^{2} = (34 {)}^{2} }$

$\tt{⬤Using \: algebraic \: Identity}$

$\tt{(a + b {)}^{2} = (a + b)(a + b) = {a}^{2} + 2ab + b}$

Now,

$⟹ \tt{( {x}^{2} {)}^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } + \bigg( \frac{1}{ {x}^{2} } \bigg)^{2} = 1156 }$

$⟹ \tt{ {x}^{4} + 2 \times \cancel{{x}^{2}} \times \frac{1}{ \cancel{ {x}^{2}} } + \frac{1}{ {x}^{4} } = 1156 }$

$⟹ \tt{ {x}^{4} + 2 + \frac{1}{ {x}^{4} } = 1156}$

$⟹ \tt{ {x}^{4} + \frac{1}{ {x}^{4} } = 1156 - 2}$

$⟹ \tt{ {x}^{4} + \frac{1}{ {x}^{4} } = 1154}$

Answer:-

$\tt{Hence, the \: value \: of \: \: {x}^{4} + \frac{1}{ {x}^{4} } = 1154 }$

Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

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

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

I hope it's help you....☹️