Question

if x+1/x=3 then the vaue of x^4+1/x^4​

Answer 1

Answer:17

Step-by-step explanation:form equations that is

x+1/x=3…….(i) then,

x^4+1/x^4…(ii)

now solve equation (i) to obtain the value of x;

rearranging ……(i) to obtain

x+1=3x

2x=1,dividing by 2 we obtain

x=1/2, now substituting for the value of x into equation…….(ii)

= (1/2)^4+1/(1/2)^4

= ((1/16)+1)/(1/16)

= (17/16)/(1/16)

= 17

Answer 2

Answer:

First, lets rearrange the equation into quadratic form.

x+1x=3

x2+1=3x

x2−3x+1=0

using the quadratic formula

x=32±9−4√2

case 1—assume that it is the positive square root:

x=32+5√2

you could use binomial theorem if you wanted to calculate it by hand:

x4=8116+42785√2+(42)9454+43255√8+2516

but lets just calculate (32+5√2)4 with a calculator.

x4=46.9787

now lets assume case 2—a negative square root:

x=32−5√2

x4=(32−5√2)4=0.0213

in case 1:

substitute 46.9787 into x4+1x4

46.9787+146.9787=47

in case 2:

0.0213+10.0213=47

interestingly, both solutions to the quadratic yielded the same result because the solutions to this this quadratic were reciprocals:

32+5√2=(32−5√2)−1

3+5√2=23−5√⋅3+5√3+5√

3+5√2=6+25√9−5

3+5√2=6+25√4

3+5√2=3+5√2

This makes sense because the quadratic x2−3x+1=0 has 2 irrational roots, and it can be factored into:

(x−3+5√2)(x−3−5√2)

Because its expanded form has a constant of 1, the product of the roots must be 1. Not only are the roots reciprocals, they are also conjugates. This will be true for any value of b greater than 2 in a quadratic of form x2−bx+1 How cool is that

Step-by-step explanation:

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