SOLUTION: Please solve the equation
y^2-6y+8=0
Hence find the roots of the equation
(x+1/x)^2 -6(x+1/x)+8=0 x is the denominator
This is a tough one
Algebra.Com
Question 365991: Please solve the equation
y^2-6y+8=0
Hence find the roots of the equation
(x+1/x)^2 -6(x+1/x)+8=0 x is the denominator
This is a tough one. I tried to do it but it's the first time I've seen this.
Answer by amoresroy(361) (Show Source): You can put this solution on YOUR website!
Please solve the equation
y^2-6y+8=0
By factoring, you get
(y-2) (y-4) = 0
So you have 2 values for y
y = 2
y = 4
Hence find the roots of the equation
(x+1/x)^2 -6(x+1/x)+8=0 x is the denominator
let y = x+1/x
So the equation is equal to
y^2-6y+8=0
where
y=2
y=4
Using y=2, then
2 = x+1/x
multiply both sides by x
2x = x^2+1
subtract 2x to both sides
0 = x^2-2x+1
By factoring, you get
(x-1) (x-1) = 0
So x = 1
Using y=4, then
4 = x+1/x
multiply both sides by x
4x = x^2+1
subtract 4x to both sides
0 = x^2-4x+1
Solve using quadratic formula
x = [4 +/- (16-4)^.5]/2
= [4 +/- 2(3)^.5] /2
RELATED QUESTIONS
I have to find all roots of X^6 + 9X^3 + 8 = 0.
I first let Y=X^3, then my equation... (answered by josmiceli)
Find the real solutions of each equation. Please help me solve this equation... (answered by Fombitz)
I need your help please.
1. Write the equation in the form ax+by+c=0
A. Y=3x+1
B.... (answered by Alan3354,rothauserc)
Solve the equation 4x^2 +12x=0
f(x) =4x^2+12x=0 where c is a constant
And part b... (answered by solver91311)
Find the roots of the given equation x(x-1)^2=0
Show work below
(answered by vleith)
solve the equation
1/8(x-6)^3/2=1
please help me solve this... (answered by edjones)
Find the real roots of the equation 9/x^4+8/x^2=1
(answered by math_helper,MathTherapy,ikleyn)
please solve the equation or find the roots of... (answered by nerdybill)
Find the exact number of real roots of the equation x^6-x^3+2x^2-3x-1=0... (answered by lwsshak3)