Solve for x x^2 = square root of x

$x^{2} = \sqrt{x}$

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{x} = x^{2}$

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x}}^{2} = {(x^{2})}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} = {(x^{2})}^{2}$

Simplify the left side.

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} = {(x^{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = {(x^{2})}^{2}$

Rewrite the expression.

$x^{1} = {(x^{2})}^{2}$

Simplify.

$x = {(x^{2})}^{2}$

Simplify the right side.

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = x^{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = x^{4}$

Step 4

Solve for $x$ .

Subtract $x^{4}$ from both sides of the equation.

$x - x^{4} = 0$

Factor the left side of the equation.

Factor $x$ out of $x - x^{4}$ .

Raise $x$ to the power of $1$ .

$x - x^{4} = 0$

Factor $x$ out of $x^{1}$ .

$x \cdot 1 - x^{4} = 0$

Factor $x$ out of $- x^{4}$ .

$x \cdot 1 + x (- x^{3}) = 0$

Factor $x$ out of $x \cdot 1 + x (- x^{3})$ .

$x (1 - x^{3}) = 0$

Rewrite $1$ as $1^{3}$ .

$x (1^{3} - x^{3}) = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 1$ and $b = x$ .

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

Factor.

Simplify.

One to any power is one.

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

Multiply $x$ by $1$ .

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

Remove unnecessary parentheses.

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$1 - x = 0$

$1 + x + x^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $1 - x$ equal to $0$ and solve for $x$ .

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Set $1 + x + x^{2}$ equal to $0$ and solve for $x$ .

Set $1 + x + x^{2}$ equal to $0$ .

$1 + x + x^{2} = 0$

Solve $1 + x + x^{2} = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

The final solution is all the values that make $x (1 - x) (1 + x + x^{2}) = 0$ true.

$x = 0, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Solve for x x^2 = square root of x

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