Algebra Examples

$x^{2} e^{x} + x e^{x} - e^{x} = 0$

Step 1

Factor $e^{x}$ out of $x^{2} e^{x} + x e^{x} - e^{x}$ .

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Step 1.1

Factor $e^{x}$ out of $x^{2} e^{x}$ .

$e^{x} x^{2} + x e^{x} - e^{x} = 0$

Step 1.2

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

$e^{x} x^{2} + e^{x} x - e^{x} = 0$

Step 1.3

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

$e^{x} x^{2} + e^{x} x + e^{x} \cdot - 1 = 0$

Step 1.4

Factor $e^{x}$ out of $e^{x} x^{2} + e^{x} x$ .

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

Step 1.5

Factor $e^{x}$ out of $e^{x} (x^{2} + x) + e^{x} \cdot - 1$ .

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

Step 2

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

$e^{x} = 0$

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

Step 3

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

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Step 3.1

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

$e^{x} = 0$

Step 3.2

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

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Step 3.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x}) = \ln (0)$

Step 3.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 3.2.3

There is no solution for $e^{x} = 0$

No solution

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

Use the quadratic formula to find the solutions.

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

Step 4.2.2

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

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

Simplify the numerator.

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Step 4.2.3.1.1

One to any power is one.

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

Step 4.2.3.1.2

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

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Step 4.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 4.2.3.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 4.2.3.1.3

Add $1$ and $4$ .

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

Step 4.2.3.2

Multiply $2$ by $1$ .

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

Step 4.2.4

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{2}$

Step 5

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

$x = - \frac{1 - \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{2}$

Decimal Form:

$x = 0.61803398 \dots, - 1.61803398 \dots$