Question

if x minus one by x is equal to 3 find the value of x 4 + 1 by x to the power 4​

Answer 1

Answer:

$x^{4}+\dfrac{1}{x^{4} }= 119$

Step-by-step explanation:

Given that-

$x - \dfrac{1}{x} = 3$

Find $x^{4} + \dfrac{1}{x^{4} } = ?$

Using formula-

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

So, $(x-\dfrac{1}{x}  )^{2}= x^{2} + \dfrac{1}{x^{2} } - 2.(x).(\dfrac{1}{x})$

$(3)^{2} = x^{2} + \dfrac{1}{x^{2} } - 2$

$9 = x^{2} + \dfrac{1}{x^{2} } - 2$

$x^{2} + \dfrac{1}{x^{2} } = 9 + 2$

$x^{2} + \dfrac{1}{x^{2} } = 11$

Squaring on both sides-

$[x^{2}+\dfrac{1}{x^{2} }  ]^{2} = (x^{2} )^{2} + (\dfrac{1}{x^{2} } )^{2} + 2.(x^{2}).(\dfrac{1}{x^{2} })$

$(11)^{2} = x^{4} + \dfrac{1}{x^{4} } + 2.(1)$

$121 = x^{4} + \dfrac{1}{x^{4}} + 2$

$x^{4}+\dfrac{1}{x^{4}} = 121 - 2$

Hence, $x^{4}+\dfrac{1}{x^{4} }= 119$

Answer 2

Step-by-step explanation:

4+x41=119

Step-by-step explanation:

Given that-

x - \dfrac{1}{x} = 3x−x1=3

Find x^{4} + \dfrac{1}{x^{4} } = ?x4+x41=?

Using formula-

(a-b)^{2} = a^{2}+b^{2}-2ab(a−b)2=a2+b2−2ab

So, (x-\dfrac{1}{x}  )^{2}= x^{2} + \dfrac{1}{x^{2} } - 2.(x).(\dfrac{1}{x})(x−x1 )2=x2+x21−2.(x).(x1)

(3)^{2} = x^{2} + \dfrac{1}{x^{2} } - 2(3)2=x2+x21−2

9 = x^{2} + \dfrac{1}{x^{2} } - 29=x2+x21−2

x^{2} + \dfrac{1}{x^{2} } = 9 + 2x2+x21=9+2

x^{2} + \dfrac{1}{x^{2} } = 11x2+x21=11

Squaring on both sides-

[x^{2}+\dfrac{1}{x^{2} }  ]^{2} = (x^{2} )^{2} + (\dfrac{1}{x^{2} } )^{2} + 2.(x^{2}).(\dfrac{1}{x^{2} })[x2+x21 ]2=(x2)2+(x21)2+2.(x2).(x21)

(11)^{2} = x^{4} + \dfrac{1}{x^{4} } + 2.(1)(11)2=x4+x41+2.(1)

121 = x^{4} + \dfrac{1}{x^{4}} + 2121=x4+x41+2

x^{4}+\dfrac{1}{x^{4}} = 121 - 2x4+x41=121−2

Hence, x^{4}+\dfrac{1}{x^{4} }= 119x4+x41=119