if x² - 1/x²= 5 find the value of (i) x²+1/x² (ii)x⁴+ 1/x⁴ (iii) x³ - 1/x³
Answers
EXPLANATION.
⇒ (x² - 1/x²) = 5.
To find the value of,
(1) = x² + 1/x².
⇒ (x² - 1/x²) = 5.
Squaring on both sides, we get.
⇒ (x - 1/x)² = (5)².
⇒ (x² + 1/x² - 2(x)(1/x)) = 25.
⇒ x² + 1/x² - 2 = 25.
⇒ x² + 1/x² = 25 + 2.
⇒ x² + 1/x² = 27.
(2) = x⁴ + 1/x⁴.
As we know that,
⇒ x² + 1/x² = 25.
Squaring on both sides, we get.
⇒ (x² + 1/x²)² = (25)².
⇒ (x⁴ + 1/x⁴ + 2(x²)(1/x²)) = 625.
⇒ x⁴ + 1/x⁴ + 2 = 625.
⇒ x⁴ + 1/x⁴ = 625 - 2.
⇒ x⁴ + 1/x⁴ = 623.
(3) = x³ - 1/x³.
As we know that,
⇒ (x - 1/x)² = x² + 1/x² - 2(x)(1/x).
⇒ (x - 1/x)² = x² + 1/x² - 2.
Put the value of x² + 1/x² in equation, we get.
⇒ (x - 1/x)² = 27 - 2.
⇒ (x - 1/x)² = 25.
⇒ (x - 1/x)² = (5)².
⇒ (x - 1/x) = 5.
We get the value of = x - 1/x = 5.
Now, we can cube on both sides, we get.
⇒ (x - 1/x)³ = (5)³.
⇒ (x³ - 3(x²)(1/x) + 3(x)(1/x²) - 1/x³) = 125.
⇒ x³ - 3x + 3/x - 1/x³ = 125.
⇒ x³ - 1/x³ - 3(x - 1/x) = 125.
⇒ x³ - 1/x³ - 3(5) = 125.
⇒ x³ - 1/x³ - 15 = 125.
⇒ x³ - 1/x³ = 125 + 15.
⇒ x³ - 1/x³ = 140.
Answer:
Given :
x + 1/x = 3
To find :
value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Solution :
We have x + 1/x = 3 ……..(1)
On squaring eq 1 both sides,
(x + 1/x)² = 3²
By Using Identity : (a + b)² = a² + b² + 2ab
x² + 1/x² + 2 x × 1/x = 9
x² + 1/x² + 2 = 9
x² + 1/x² = 9 - 2
x² + 1/x² = 7 ………….(2)
On squaring eq 2 both sides,
(x² +1/x² )² = 7²
(x²)² + (1/x²)² + 2 x² × 1/x² =7²
x⁴ + 1/x⁴ + 2 = 49
x⁴ + 1/x⁴ = 49 - 2
x⁴ + 1/x⁴ = 47
On Cubing eq 1 both sides :
(x + 1/x)³ = 3³
By Using Identity : (a + b)³ = a³ + b³ + 3ab(a + b)
(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27
x³ + 1/x³ + 3(x + 1/x) = 27
x³ + 1/x³ + 3(3) = 27
x³ + 1/x³ + 9 = 27
x³ + 1/x³ = 27 - 9
x³ + 1/x³ = 18
Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is 18 & x⁴ + 1/x⁴ is 47.
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