Solve x^32=10(x^5)

Question

Solve the equation

x_{1} = - \frac{5}{5}, x_{2} = 0, x_{3} = \frac{5}{5}

Alternative Form

x_{1} \approx - 0.447214, x_{2} = 0, x_{3} \approx 0.447214

Evaluate

x^{3} \times 2 = 10 x^{5}

Use the commutative property to reorder the terms

2 x^{3} = 10 x^{5}

Add or subtract both sides

2 x^{3} - 10 x^{5} = 0

Factor the expression

2 x^{3} (1 - 5 x^{2}) = 0

Divide both sides

x^{3} (1 - 5 x^{2}) = 0

Separate the equation into 2 possible cases

x^{3} = 0 1 - 5 x^{2} = 0

The only way a power can be 0 is when the base equals 0

x = 0 1 - 5 x^{2} = 0

Solve the equation

More Steps

Evaluate

1 - 5 x^{2} = 0

Move the constant to the right-hand side and change its sign

- 5 x^{2} = 0 - 1

Removing 0 doesn't change the value,so remove it from the expression

- 5 x^{2} = - 1

Change the signs on both sides of the equation

5 x^{2} = 1

Divide both sides

\frac{5 x ^{2}}{5} = \frac{1}{5}

Divide the numbers

x^{2} = \frac{1}{5}

Take the root of both sides of the equation and remember to use both positive and negative roots

x = \pm \frac{1}{5}

Simplify the expression

More Steps

Evaluate

\frac{1}{5}

To take a root of a fraction,take the root of the numerator and denominator separately

\frac{1}{5}

Simplify the radical expression

\frac{1}{5}

Multiply by the Conjugate

\frac{5}{5 \times 5}

When a square root of an expression is multiplied by itself,the result is that expression

\frac{5}{5}

x = \pm \frac{5}{5}

Separate the equation into 2 possible cases

x = \frac{5}{5} x = - \frac{5}{5}

x = 0 x = \frac{5}{5} x = - \frac{5}{5}

Solution

x_{1} = - \frac{5}{5}, x_{2} = 0, x_{3} = \frac{5}{5}

Alternative Form

x_{1} \approx - 0.447214, x_{2} = 0, x_{3} \approx 0.447214

Show Solution