What is the answer to x/(sqrt(x^2+1))=x^4-x ?

The answer to x/(sqrt(x^2+1))=x^4-x is x=0,x\approx 1.18077…

Solve for x: x/(sqrt(x^2+1))=x^4-x

The solution is x=0,x\approx 1.18077…

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\frac{x}{x ^{2} + 1} = x^{4} - x

x = 0, x \approx 1.18077 \dots

Solution steps

\frac{x}{x ^{2} + 1} = x^{4} - x

Multiply both sides by

x^{2} + 1

\frac{x}{x ^{2} + 1} x^{2} + 1 = x^{4} x^{2} + 1 - x x^{2} + 1

Simplify

x = x^{4} x^{2} + 1 - x x^{2} + 1

Subtract

x^{4} x^{2} + 1 - x x^{2} + 1

from both sides

x - (x^{4} x^{2} + 1 - x x^{2} + 1) = x^{4} x^{2} + 1 - x x^{2} + 1 - (x^{4} x^{2} + 1 - x x^{2} + 1)

Simplify

x - x^{2} + 1 x^{4} + x^{2} + 1 x = 0

Factor

x - x^{2} + 1 x^{4} + x^{2} + 1 x : x (1 - x^{2} + 1 x^{3} + x^{2} + 1)

x (1 - x^{2} + 1 x^{3} + x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 1 - x^{2} + 1 x^{3} + x^{2} + 1 = 0

Solve

1 - x^{2} + 1 x^{3} + x^{2} + 1 = 0 : x \approx 1.18077 \dots

x = 0, x \approx 1.18077 \dots

Verify Solutions:

x = 0

True

, x \approx 1.18077 \dots

True

The solutions are

x = 0, x \approx 1.18077 \dots

y, x - y = 1

5 - x = 5 - x^{2}

x - 3 + 9 = x

5 x - 2 = x + 6

3 \cdot (x - 1)^{0.5} + 11 = 2 x

What is the answer to x/(sqrt(x^2+1))=x^4-x ?
The answer to x/(sqrt(x^2+1))=x^4-x is x=0,x\approx 1.18077…
Solve for x: x/(sqrt(x^2+1))=x^4-x
The solution is x=0,x\approx 1.18077…