y²+1÷y²=11 ,
find y-1÷y
Answers
Step-by-step explanation:
We can start this by completing the square.So,bringing 4 from the LHS to the RHS,the LHS can be written as x^2+1/x^2+y^2+1/y^2–4=0.Now,we write -4 as -2–2,and distribute each of the -2 to each of the variables(namely x and y).So,we get (x^2+1/x^2–2)+(y^2+1/y^2–2)=0.Now,on completing the squares for the two terms in the brackets in the LHS,we get (x-1/x)^2+(y-1/y)^2=0.Now,remember the sum of squares of two quantities is 0 only when each and every one of them is 0.So,we get x-1/x=0=>x^2=1.
y-1/y=0=>y^2=1 which implies x^2+y^2=2.Correct me if im wrong :)
Answer:
y-1/y=9
Squaring on both sides;
(y-1/y)^2=(9)^2
y^2-2XyX1/y+(1/y)^2=81
y^2-2+1/y^2=81
y^2+1/y^2=81+2
y^2+1/y^2=83
Squaring on both the sides, we get;
(y^2+1/y^2)^2= (83)^2
(y^2)^2+2Xy^2X1/y^2+(1/y^2)^2=6889
y^4+2+1/y^4=6889
y^4+1/y^4=6889-2
y^4+1/y^4=6887
Step-by-step explanation: