Question

if x-1/x=4then evaluate x^2-1x^2 and x^4-1/x^4​

Answer 1

Solution :-

x - 1/x = 4 ---eq(1)

Squaring on both sides

⇒ (x - 1/x)² = 4²

⇒ (x)² + (1/x)² - 2(x)(1/x) = 16

[ Because (a - b)² = a² + b² - 2ab ]

⇒ x² + 1/x² - 2 = 16

⇒ x² + 1/x² = 16 + 2

⇒ x² + 1/x² = 18 --- eq(2)

Adding 2(x)(1/x) on both sides

⇒ x² + (1/x)² + 2(x)(1/x) = 18 + 2(x)(1/x)

⇒ (x + 1/x)² = 18 + 2

[ Because (a + b)² = a² + b² + 2ab ]

⇒ (x + 1/x)² = 20

⇒ x + 1/x = √20 = √4 * √5 = 2√5

⇒ x + 1/x = 2√5 ---eq(3)

Multiplying eq(1) & eq(3)

⇒ (x - 1/x)(x + 1/x) = 4 * 2√5

⇒ (x)² - (1/x)² = 8√5

[ Because (a - b)(a + b) = a² - b² ]

⇒ x² - 1/x² = 8√5 ---eq(4)

Multiplying eq(2) and eq(4)

⇒ (x² + 1/x²)(x² - 1/x²) = 18 * 8√5

⇒ ( x² )² - ( 1/x² )² = 144√5

[ Because (a + b)(a - b) = a² - b²]

⇒ x^4 - 1/x^4 = 144√5

Hence, the value of x² - 1/x² is 8√5 and the value of x^4 - 1/x^4 is 144√5.

Answer 2

$\Large\underline{\underline{\sf{Given}:}}$

• (x - 1/x ) = 4

$\Large\underline\mathfrak{Question}$

• (x² - 1/x²)
• (x⁴ - 1/x⁴)

$\large\star{\underline{\tt{\red{Answer}}}}\star$

Let ,

(x - 1/x) = 4 ---------------Equation (1)

Squaring both sides of Equation (1) we get,

→ (x - 1/x)² = 16

[ using (a-b)² = a² + b² - 2ab in LHS ]

→ (x² + 1/x² - 2x*1/x) = 16

→ (x²+1/x²) = 16+2

→ (x² + 1/x²) = 18 ------------------Equation(2)

Now, adding 2 both sides of Equation we get,

→ (x² + 1/x² + 2) = 18+2

[ using (a+b)² = a² + b² + 2ab in LHS ]

→ (x+1/x)² = 20

[ square root both sides ]

→ (x + 1/x) = √20 = 2√5 ------------Equation(3)

Mulitply Equation(1) and Equation(3) ,

and using [(a+b)(a-b) = a²-b²] we get,,

→ (x+1/x)(x-1/x) = 4×2√5

$\pink{\large\boxed{\bold{ {x}^{2} - \frac{1}{ {x}^{2} } = 8 \sqrt{5} } }}$

_______________________________

Now, multiply Value of (x² - 1/x²) and Equation(2)

and, again using [(a+b)(a-b) = a²-b²] we get,,

→ (x² - 1/x²)(x² + 1/x²) = 8√5 * 18

$\green{\large\boxed{\bold{ {x}^{4} - \frac{1}{ {x}^{4} } = 144 \sqrt{5} } }}$

____________________________

$\large\underline\textbf{Hope it Helps You.}$