Question

Solve for x.

$\rm \longrightarrow \dfrac{x}{ \sqrt{ {x}^{2} + 1} } = {x}^{4} - x$

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Answer 1

Factorize both sides of the equation:

$\sf{\dfrac x{\sqrt{x^2+1}} = x^4 - x }$

$\sf{\dfrac x{\sqrt{x^2+1}} = x(x^3-1)}$

$\sf{\dfrac x{\sqrt{x^2+1}} - x(x^3-1) = 0}$

$\sf{x \left(\dfrac1{\sqrt{x^2+1}} - (x^3 - 1)\right) = 0}$

One immediate solution is then x = 0. This leaves us with

$\sf{\dfrac1{\sqrt{x^2+1}} - (x^3 - 1) = 0 }$

or as we had earlier,

$\sf{\dfrac1{\sqrt{x^2+1}} = x^3 - 1 }$

Take the square of both sides:

$\sf{\dfrac1{x^2+1} = \left(x^3-1\right)^2}$

$\sf{\dfrac1{x^2+1} = x^6 - 2x^3 + 1}$

Turn this into a polynomial equation:

$\sf{1 = \left(x^2+1\right) \left(x^6-2x^3+1\right)}$

Expand the right side and make it equal to zero:

$\sf{1 = x^8 + x^6 - 2x^5 - 2x^3 + x^2 + 1}$

$\sf {0 = x^8 + x^6 - 2x^5 - 2x^3 + x^2}$

Each term in the polynomial has a common factor of x². Factoring this out just gives x = 0 again as a solution.

$\sf {0 = x^2 \left(x^6 + x^4 - 2x^3 - 2x + 1\right)}$

$\sf {0 = x^6 + x^4 - 2x^3 - 2x + 1}$

You'll need a computer to solve the equation. Solving over reals, you would find two solutions, x ≈ 0.438 and x ≈ 1.18, but only x ≈ 1.18 is valid.

Answer 2

Answer:

Factorize both sides of the equation:

\sf{\dfrac x{\sqrt{x^2+1}} = x^4 - x }

x

2

+1

x

=x

4

−x

\sf{\dfrac x{\sqrt{x^2+1}} = x(x^3-1)}

x

2

+1

x

=x(x

3

−1)

\sf{\dfrac x{\sqrt{x^2+1}} - x(x^3-1) = 0}

x

2

+1

x

−x(x

3

−1)=0

\sf{x \left(\dfrac1{\sqrt{x^2+1}} - (x^3 - 1)\right) = 0}x(

x

2

+1

1

−(x

3

−1))=0

One immediate solution is then x = 0. This leaves us with

\sf{\dfrac1{\sqrt{x^2+1}} - (x^3 - 1) = 0 }

x

2

+1

1

−(x

3

−1)=0

or as we had earlier,

\sf{\dfrac1{\sqrt{x^2+1}} = x^3 - 1 }

x

2

+1

1

=x

3

−1

Take the square of both sides:

\sf{\dfrac1{x^2+1} = \left(x^3-1\right)^2}

x

2

+1

1

=(x

3

−1)

2

\sf{\dfrac1{x^2+1} = x^6 - 2x^3 + 1}

x

2

+1

1

=x

6

−2x

3

+1

Turn this into a polynomial equation:

\sf{1 = \left(x^2+1\right) \left(x^6-2x^3+1\right)}1=(x

2

+1)(x

6

−2x

3

+1)

Expand the right side and make it equal to zero:

\sf{1 = x^8 + x^6 - 2x^5 - 2x^3 + x^2 + 1}1=x

8

+x

6

−2x

5

−2x

3

+x

2

+1

\sf {0 = x^8 + x^6 - 2x^5 - 2x^3 + x^2}0=x

8

+x

6

−2x

5

−2x

3

+x

2

Each term in the polynomial has a common factor of x². Factoring this out just gives x = 0 again as a solution.

\sf {0 = x^2 \left(x^6 + x^4 - 2x^3 - 2x + 1\right)}0=x

2

(x

6

+x

4

−2x

3

−2x+1)

\sf {0 = x^6 + x^4 - 2x^3 - 2x + 1}0=x

6

+x

4

−2x

3

−2x+1

You'll need a computer to solve the equation. Solving over reals, you would find two solutions, x ≈ 0.438 and x ≈ 1.18, but only x ≈ 1.18 is valid.

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