Question

okay 1 more question
Solve (x+5/x^2)=(1/x^2+x)-(x-6/x+1)
please show the steps with the answer so I can learn from it

Answer 1

and any non-permissible values if its not too much trouble

Answer 2

I find your parentheses suspicious
maybe:
(x+5) / (x^2)=1/(x^2+x) - (x-6) / (x+1)

(x+5) (x+1) / [ (x^2) (x+1) ] = 1 / [x(x+1)] - x(x-6) / [ x(x+1)]

(x+5) (x+1) / [ (x^2) (x+1) ] = x / [x^2(x+1)] - x^2(x-6) / [ x^2(x+1)]
or
(x+5) (x+1) = x - x^2(x-6)
x^2 + 6 x + 5 = x - x^3 + 6 x^2
x^3 -5 x^2 +5 x + 5 = 0

by the way that denominator
[ (x^2) (x+1) ] better not be zero so x not 0 and x not -1

Answer 3

The Anonymous solution is not correct.

x + 5 / x² = 1 / x² + x - ( x - 6 / x + 1 )

x + 5 / x² = 1 / x² + x - x + 6 / x - 1

x + 5 / x² = 1 / x² + 6 / x - 1

Multiply both sides by x²

x³ + 5 = 1 + 6 x - x²

Subtract ( 1 + 6 x - x² ) to both sides

x³ + 5 - ( 1 + 6 x - x² ) = 0

x³ + 5 - 1 - 6 x + x² = 0

x³ + 4 - 6 x + x² = 0

x³ + x² - 6 x + 4 = 0

To find zeros for polynomials of degree 3 or higher you use Rational root test.

If the polynomial has a rational zero then it must be a fraction p / q
​
where

p = factor of the trailing constant

q = factor of the leading coefficient.

The factor of the leading coefficient ( 1 ) is 1

The factors of the constant term ( 4 ) are 1 , 2 , 2 , 4

Possible roots:

± 1 / 1 , ± 2 / 1 , ± 2 / 1 , ± 4 / 1

± 1 / 1 , ± 2 / 1 , ± 4 / 1

Substitute the possible roots one by one into x³ + x² - 6 x + 4 to find the actual roots.

x = 1

x³ + x² - 6 x + 4 = 1³ + 1² - 6 ∙ 1 + 4 = 1 + 1 - 6 + 4 = 0

x = - 1

x³ + x² - 6 x + 4 = ( -1 )³ + ( - 1 )² - 6 ∙ ( - 1 ) + 4 = - 1 + 1 + 6 + 4 = 10

x = 2

x³ + x² - 6 x + 4 = 2³ + 2² - 6 ∙ 2 + 4 = 8 + 4 - 12 + 4 = 4

x = - 2

x³ + x² - 6 x + 4 = ( - 2)³ + ( - 2 )² - 6 ∙ ( - 2 ) + 4 = - 8 + 4 + 12 + 4 = 12

x = 4

x³ + x² - 6 x + 4 = 4³ + 4² - 6 ∙ 4 + 4 = 64 + 16 - 24 + 4 = 60

x = - 4

x³ + x² - 6 x + 4 = ( - 4 )³ + ( - 4 )² - 6 ∙ ( - 4 ) + 4 = - 64 + 16 + 24 + 4 = - 20

You obtain for x = 1 , x³ + x² - 6 x + 4 = 0

To find remaining zeros use Factor theorem.

If p / q is root of the polynomial then this polynomial
can be divided with q x - p

In this case:

p / q = 1 / 1

p = 1 , q = 1

q x - p = 1 ∙ x - 1

q x - p = x - 1

So x³ + x² - 6 x + 4 is divisible by x - 1

Using synthetic division you get:

( x³ + x² - 6 x + 4 ) / ( x - 1 ) = x² + 2 x - 4

You can write

x³ + x² - 6 x + 4 = ( x - 1 ) ( x² + 2 x - 4 )

Now find:

x³ + x² - 6 x + 4 = 0

( x - 1 ) ( x² + 2 x - 4 ) = 0

The equation will be equal to zero when the values in parentheses are equal to zero.

You have two conditions:

First condition:

x - 1 = 0

The solution is:

x = 1

Second condition:

x² + 2 x - 4 = 0

The solutions are:

x = - 1 - √5

and

x = - 1 + √5

So your equation has three solutions:

x = 1

x = - 1 - √5

x = - 1 + √5