Algebra Examples

${(- 2 x - 5)}^{\frac{3}{5}} = {(x^{2} - 29)}^{\frac{3}{5}}$

Step 1

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$- 2 x - 5 = x^{2} - 29$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $x^{2}$ from both sides of the equation.

$- 2 x - 5 - x^{2} = - 29$

Step 2.2

Add $29$ to both sides of the equation.

$- 2 x - 5 - x^{2} + 29 = 0$

Step 2.3

Add $- 5$ and $29$ .

$- 2 x - x^{2} + 24 = 0$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $- 1$ out of $- 2 x - x^{2} + 24$ .

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Step 2.4.1.1

Reorder $- 2 x$ and $- x^{2}$ .

$- x^{2} - 2 x + 24 = 0$

Step 2.4.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 2 x + 24 = 0$

Step 2.4.1.3

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 24 = 0$

Step 2.4.1.4

Rewrite $24$ as $- 1 (- 24)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 24 = 0$

Step 2.4.1.5

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

$- (x^{2} + 2 x) - 1 \cdot - 24 = 0$

Step 2.4.1.6

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 24)$ .

$- (x^{2} + 2 x - 24) = 0$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} + 2 x - 24$ using the AC method.

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Step 2.4.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 2.4.2.1.2

Write the factored form using these integers.

$- ((x - 4) (x + 6)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$- (x - 4) (x + 6) = 0$

Step 2.5

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

$x - 4 = 0$

$x + 6 = 0$

Step 2.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.7

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 2.7.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 2.8

The final solution is all the values that make $- (x - 4) (x + 6) = 0$ true.

$x = 4, - 6$