if x^2+1\x^2=7 and x>0then find x^3+1\x^3
Answers
Answered by
1
(x+1/x)³=x³+1/x³+3(x+1/x)
(x+1/x)²=x²+1/x²+2
Replacing the x²+1/x² by 7
We get;
(x+1/x)²=9
x+1/x=(+/-)3
∴ x+1/x=3 ∨ x+1/x=-3
As x>0 ⇒ x+1/x=3
∴27=x³+1/x³+9
∴ x³+1/x³=18
(x+1/x)²=x²+1/x²+2
Replacing the x²+1/x² by 7
We get;
(x+1/x)²=9
x+1/x=(+/-)3
∴ x+1/x=3 ∨ x+1/x=-3
As x>0 ⇒ x+1/x=3
∴27=x³+1/x³+9
∴ x³+1/x³=18
0lovesenergyp9mq81:
because I am new here!!
you try yourself!!
wait
By writing an equation for (x-y)^4 and then you would get the value of x+y and then to find x-y which easy...!!
the value of xy in our case is a^2.... so it's easy!!
i have sent u a request please accept it
okk help me in other question
if x=3-2root2 find rootx+1/rootx
post this question in your timeline I would comment an answer as a pic!!
By the way it's 2root2
Similar questions