Solve for x x^4=5x^3-7x^2

$x^{4} = 5 x^{3} - 7 x^{2}$

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

$5 x^{3} - 7 x^{2} = x^{4}$

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

$5 x^{3} - 7 x^{2} - x^{4} = 0$

Factor $- x^{2}$ out of $5 x^{3} - 7 x^{2} - x^{4}$ .

Reorder the expression.

Move $- 7 x^{2}$ .

$5 x^{3} - x^{4} - 7 x^{2} = 0$

Reorder $5 x^{3}$ and $- x^{4}$ .

$- x^{4} + 5 x^{3} - 7 x^{2} = 0$

Factor $- x^{2}$ out of $- x^{4}$ .

$- x^{2} x^{2} + 5 x^{3} - 7 x^{2} = 0$

Factor $- x^{2}$ out of $5 x^{3}$ .

$- x^{2} x^{2} - x^{2} (- 5 x) - 7 x^{2} = 0$

Factor $- x^{2}$ out of $- 7 x^{2}$ .

$- x^{2} x^{2} - x^{2} (- 5 x) - x^{2} \cdot 7 = 0$

Factor $- x^{2}$ out of $- x^{2} (x^{2}) - x^{2} (- 5 x)$ .

$- x^{2} (x^{2} - 5 x) - x^{2} \cdot 7 = 0$

Factor $- x^{2}$ out of $- x^{2} (x^{2} - 5 x) - x^{2} (7)$ .

$- x^{2} (x^{2} - 5 x + 7) = 0$

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

$x^{2} = 0$

$x^{2} - 5 x + 7 = 0$

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

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

$x^{2} = 0$

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

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $x^{2} - 5 x + 7$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5 x + 7$ equal to $0$ .

$x^{2} - 5 x + 7 = 0$

Solve $x^{2} - 5 x + 7 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 5$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 7)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 7}}{2 \cdot 1}$

Multiply $- 4$ by $7$ .

$x = \frac{5 \pm \sqrt{25 - 28}}{2 \cdot 1}$

Subtract $28$ from $25$ .

$x = \frac{5 \pm \sqrt{- 3}}{2 \cdot 1}$

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

$x = \frac{5 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

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

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

$x = \frac{5 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = \frac{5 + i \sqrt{3}}{2}, \frac{5 - i \sqrt{3}}{2}$

The final solution is all the values that make $- x^{2} (x^{2} - 5 x + 7) = 0$ true.

$x = 0, \frac{5 + i \sqrt{3}}{2}, \frac{5 - i \sqrt{3}}{2}$

Solve for x x^4=5x^3-7x^2

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