Solve the equation [x/(x+1)]^2 + [(x+1)/x]^2 = 17/4

4 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

[x/(x+1)]^2 + [(x+1)/x]^2 = 17/4

We notice that the terms are inveres for each other.

The, ley x/(x+1) = y

==> (x+1)/x = 1/y

==> y^2 + 1/y^2 = 17/4

Multiply by 4y^2:

==> 4y^4 + 4 = 17y^2

==> 4y^4 - 17y^2 + 4 = 0

==> (4y^2 -1)(y^2 -4) = 0

== y1= 1/2 

==> y2= -1/2 

==> y3= 2

==> y4= -2

but y= x/(x+1)

==> 1/2 = x/(x+1) ==> x+1 = 2x ==> x1= 1

==> -1/2 = x/(x+1) ==> -x-1 = 2x ==> x2= -1/3

==> 2= x/(x+1) ==> 2x+2 = x ==> x3= -2

=> -2 = x/(x+1) ==> -2x-2 = x ==> x4= -2/3

==> x= { 1, -1/3, -2, -2/3}

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

{x/(x+1)}^2 +{(x+1)/x}^2 = 17/4. To solve for x.

Solution:

Put   (x+1)/x = y.

y^2+1/y^2 =  4+1/4 = 2^2 +1/2^2.

By comparing both sides:

y = 2 Or 1/2.

Therefore (x+1)/x = 2.

x+1 = 2x

1 = x .

Therefore x = 1.

 

 

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

To solve this equation let's take Y= [x / (1+x)]^2

So [x/(x+1)]^2 + [(x+1)/x]^2 = 17/4

=> Y + 1/ Y = 17 / 4

=> Y^2 + 1 = 17Y / 4

=> 4Y^2 + 4 - 17Y = 0

=> 4Y^2 - 17Y + 4=0

=> 4Y^2 - 16Y - Y +4 =0

=> 4Y ( Y-4) -1 ( Y-4) =0

=> (4Y -1)(Y-4)=0

Y = 1/4 and Y= 4

First lets take  Y = 1/4

=>[x / (1+x)]^2 = 1/4

=> [x / (1+x)] = 1/2 or -1/2

If [x / (1+x)] = 1/2

=> 2x = 1+x

=> x = 1

If If [x / (1+x)] = -1/2

=> 2x = -1-x

=>3x = -1

=> x= -1/3

Now Y = 2

=>[x / (1+x)]^2= 4

=> [x / (1+x)]= 2 and  [x / (1+x)] = -2

For  [x / (1+x)]= 2

=> x = 2+ 2x

=> x= -2

For [x / (1+x)] = -2

=> -x = 2+2x

=> -3x =2

=> x = -2/3

Therefore we get four values for x, x = 1 , x= -1/3, x = -2 and x= -2/3

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll notice that the product of the 2 ratios x/(x+1)*(x+1)/x = 1.

If we'll square raise the product, we'll obtain:

[x/(x+1)]^2*[(x+1)/x]^2 = 1

We'll substitute the ratio [x/(x+1)]^2 = t

t = 1/[(x+1)/x]^2

We'll re-write the equation:

t + 1/t = 17/4

4t^2 + 4 = 17t

We'll subtract 17 t both sides:

4t^2 - 17t + 4 = 0

We'll apply the quadratic formula:

t1 = [17 + sqrt(289 - 64)]/8

t1 = (17+15)/8

t1 = 32/8

t1 = 4

t2 = (17-15)/8

t2 = 2/8

t2 = 1/4

Since t =  [x/(x+1)]^2, both values for t have to be positive and they are.

 [x/(x+1)]^2 = 4

x/(x+1) = 2

x = 2x + 2

-x = 2

x = -2

x/(x+1) = -2

x = -2x - 2

3x = -2

x = -2/3

x/(x+1) = 1/2

2x = x+1

x = 1

x/(x+1) = -1/2

2x = -x-1

3x = -1

x = -1/3

The roots of the equation are: {-2 , -2/3 , -1/3 , 1}.

We’ve answered 318,946 questions. We can answer yours, too.

Ask a question