Solve x^5 + 7x^3 + 6x < 5x^4 + 7x^2 + 2.

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

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`x^5 + 7x^3 + 6x < 5x^4 + 7x^2 + 2`

`x^5 -5x^4 + 7x^3- 7x^2 + 6x - 2 < 0`

 Let us solve the equation:

`x^5 - 5x^4 + 7x^3 - 7x^2 + 6x -2 = 0`

We will substitute with x= 1

==> 1 - 5 + 7 - 7 + 6 -2 = 0

==> 2 - 2 = 0

==> 0 = 0

Then x= 1 is a root for the equation.

==> (x-1) is a factor:

`x^5 -5x^4 +7x^3 -7x^2 + 6x -2= (x-1)(x^4 -4x^3 +3x^2-4x +2)`

Now we will factor the last terms.

First we will rearrange terms.

`x^4 - 4x^3 + 3x^2 - 4x +2 = (x^4+3x^2 +2) + (-4x^3-4x)`

`==> (x^4+3x^2 + 2)= (x^2+1)(x^2+2)`

`==> (x^2+1)(x^2+2) -4x(x^2+1)`

Now we will factor x^2+1

`==> (x^2+1)(x^2+2 - 4x) = (x^2 +1)(x^2-4x+2)`

Now we will rewrite the inequality:

`==> (x-1)(x^2+1)(x^2-4x+2) < 0`

`` We know that `x^2+1 `  is always greater than zero.

`==> (x-1)(x^2-4x+2) < 0`

`` Now we will find the roots.

==> x1= 1

==> `x2= (4+sqrt(16-8))/ 2 = (4+sqrt(8))/2 = 2+sqrt2`

`==> x3= 2- sqrt2`

`` ==> Now we have the intervals:

`(-oo, 1), (1, 2-sqrt2) , (2-sqrt2 , 2+sqrt2) , and (2+sqrt2, oo)`

`( -oo, 1) ==> +`

`(1, 2-sqrt2) ==> -`

`(2,2-sqrt2, 2+sqrt2) ==> +`

`(2+sqrt2, oo) ==> -`

Then, solution is:

x `in ( 1, 2-sqrt2) U (2+sqrt2, oo)`

 

``

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